h2s(0.1mol) is added to 1l of water. what is the ph of the solution at equilibrium

Affiliate thirteen. Cardinal Equilibrium Concepts

xiii.4 Equilibrium Calculations

Learning Objectives

By the end of this section, you lot will exist able to:

• Write equations representing changes in concentration and pressure for chemical species in equilibrium systems
• Use algebra to perform various types of equilibrium calculations

We know that at equilibrium, the value of the reaction quotient of whatsoever reaction is equal to its equilibrium abiding. Thus, nosotros can utilize the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we have learned to place in which management a reaction volition shift to achieve equilibrium, we want to extend that understanding to quantitative calculations. We exercise so past evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic approach to equilibrium calculations volition be explored in this section.

Changes in concentrations or pressures of reactants and products occur equally a reaction system approaches equilibrium. In this section we will come across that we tin can chronicle these changes to each other using the coefficients in the counterbalanced chemical equation describing the arrangement. We utilize the decomposition of ammonia equally an example.

On heating, ammonia reversibly decomposes into nitrogen and hydrogen according to this equation:

[latex]2\text{NH}_3(thousand)\;{\rightleftharpoons}\;\text{N}_2(g)\;+\;3\text{H}_2(g)[/latex]

If a sample of ammonia decomposes in a airtight system and the concentration of N₂ increases by 0.11 M, the change in the N₂ concentration, Δ[N₂], the last concentration minus the initial concentration, is 0.11 M. The change is positive because the concentration of North_ii increases.

The modify in the H₂ concentration, Δ[H₂], is likewise positive—the concentration of H₂ increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H_two is three times the change in the concentration of N₂ because for each mole of Due north₂ produced, three moles of H₂ are produced.

[latex]{\Delta}[\text{H}_2] = 3\;\times\;{\Delta}[\text{N}_2][/latex]

[latex]= 3\;\times\;(0.11\;M) = 0.33\;M[/latex]

The change in concentration of NH₃, Δ[NH₃], is twice that of Δ[N₂]; the equation indicates that ii moles of NH₃ must decompose for each mole of N_ii formed. Withal, the change in the NH₃ concentration is negative because the concentration of ammonia decreases as it decomposes.

[latex]{\Delta}[\text{NH}_3] = -2\;\times\;{\Delta}[\text{Northward}_2] = -two\;\times\;(0.11\;M) = -0.22\;Chiliad[/latex]

We can relate these relationships directly to the coefficients in the equation

[latex]\begin{assortment}{ccccc} 2\text{NH}_3(thou) & {\rightleftharpoons} & \text{N}_2(g) & + & 3\text{H}_2(one thousand) \\[0.5em] {\Delta}[\text{NH}_3] = -2\;\times\;{\Delta}[\text{N}_2] & & {\Delta}[\text{N}_2] = 0.11\;Yard & & {\Delta}[\text{H}_2] = 3\;\times\;{\Delta}[\text{N}_2] \end{assortment}[/latex]

Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.

If we did non know the magnitude of the change in the concentration of N₂, nosotros could stand for it by the symbol x.

[latex]{\Delta}[\text{N}_2] = x[/latex]

The changes in the other concentrations would so be represented as:

[latex]{\Delta}[\text{H}_2] = 3\;\times\;{\Delta}[\text{N}_2] = 3x[/latex]

[latex]{\Delta}[\text{NH}_3] = -2\;\times\;{\Delta}[\text{N}_2] = -2x[/latex]

The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.

[latex]\brainstorm{array}{ccccc} 2\text{NH}_3(m) & {\rightleftharpoons} & \text{North}_2(g) & + & three\text{H}_2(yard) \\[0.5em] -2x & & x & & 3x \end{assortment}[/latex]

The simplest style for us to detect the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation. The sign of the coefficient is positive when the concentration increases; information technology is negative when the concentration decreases.

Example 1

Determining Relative Changes in Concentration
Complete the changes in concentrations for each of the following reactions.

(a) [latex]\begin{assortment}{lcccc} \text{C}_2\text{H}_2(k) & + & ii\text{Br}_2(thou) & {\rightleftharpoons} & \text{C}_2\text{H}_2\text{Br}_4(g) \\[0.5em] x & & \dominion[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]

(b) [latex]\brainstorm{array}{lcccc} \text{I}_2(aq) & + & \text{I}^{-}(aq) & {\rightleftharpoons} & \text{I}_3^{\;\;-}(aq) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{two.5em}{0.1ex} & & x \finish{array}[/latex]

(c) [latex]\begin{array}{lcccccc} \text{C}_3\text{H}_8(g) & + & 5\text{O}_2(thou) & {\rightleftharpoons} & iii\text{CO}_2(g) & + & four\text{H}_2\text{O}(g) \\[0.5em] x & & \rule[0ex]{two.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} \end{array}[/latex]

Solution
(a) [latex]\brainstorm{array}{lcccc} \text{C}_2\text{H}_2(g) & + & two\text{Br}_2(g) & {\rightleftharpoons} & \text{C}_2\text{H}_2\text{Br}_4(g) \\[0.5em] x & & 2x & & -x \stop{array}[/latex]

(b) [latex]\begin{array}{lcccc} \text{I}_2(aq) & + & \text{I}^{-}(aq) & {\rightleftharpoons} & \text{I}_3^{\;\;-}(aq) \\[0.5em] -x & & -x & & 10 \end{array}[/latex]

(c) [latex]\begin{array}{lcccccc} \text{C}_3\text{H}_8(yard) & + & v\text{O}_2(k) & {\rightleftharpoons} & 3\text{CO}_2(g) & + & 4\text{H}_2\text{O}(g) \\[0.5em] ten & & 5x & & -3x & & -4x \end{assortment}[/latex]

Check Your Learning
Complete the changes in concentrations for each of the post-obit reactions:

(a) [latex]\begin{array}{lcccc} ii\text{Then}_2(g) & + & \text{O}_2(g) & {\rightleftharpoons} & 2\text{SO}_3(yard) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & x & & \dominion[0ex]{2.5em}{0.1ex} \terminate{array}[/latex]

(b) [latex]\begin{assortment}{lcc} \text{C}_4\text{H}_8(grand) & {\rightleftharpoons} & 2\text{C}_2\text{H}_4(yard) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & -2x \stop{array}[/latex]

(c) [latex]\begin{assortment}{lcccccc} 4\text{NH}_3(one thousand) & + & 7\text{H}_2\text{O}(k) & {\rightleftharpoons} & 4\text{NO}_2(thousand) & + & 6\text{H}_2\text{O}(g) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \stop{array}[/latex]

Answer:

(a) 2ten, x, −2x; (b) x, −2x; (c) 410, 7x, −fourx, −half-dozenten or −ivx, −7x, 4x, 610

Calculations Involving Equilibrium Concentrations

Because the value of the reaction quotient of any reaction at equilibrium is equal to its equilibrium abiding, we can employ the mathematical expression for Q_c (i.e., the law of mass activity) to determine a number of quantities associated with a reaction at equilibrium. It may help if we proceed in mind that Q_c = 1000_c (at equilibrium) in all of these situations and that there are but iii bones types of calculations:

1. Calculation of an equilibrium constant. If concentrations of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated.
2. Calculation of missing equilibrium concentrations. If the value of the equilibrium constant and all of the equilibrium concentrations, except one, are known, the remaining concentration can be calculated.
3. Calculation of equilibrium concentrations from initial concentrations. If the value of the equilibrium constant and a set of concentrations of reactants and products that are non at equilibrium are known, the concentrations at equilibrium can be calculated.

A similar list could exist generated using Q_P , K_P , and partial pressure. We will expect at solving each of these cases in sequence.

Adding of an Equilibrium Constant

Since the constabulary of mass action is the only equation we accept to describe the relationship between K_c and the concentrations of reactants and products, any problem that requires u.s.a. to solve for K_c must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for One thousand_c , as it will be the but unknown.

Instance ii showed usa how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, usually chosen an Water ice chart—for Initial, Change, and Equilibrium–volition be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and we consider no shift toward equilibrium to have happened. The side by side row of information is the change that occurs equally the organization shifts toward equilibrium—do non forget to consider the reaction stoichiometry every bit described in a previous department of this chapter. The last row contains the concentrations once equilibrium has been reached.

Example two

Calculation of an Equilibrium Constant
Iodine molecules react reversibly with iodide ions to produce triiodide ions.

[latex]\text{I}_2(aq)\;+\;\text{I}^{-}(aq)\;{\rightleftharpoons}\;\text{I}_3^{\;\;-}(aq)[/latex]

If a solution with the concentrations of I_ii and I⁻ both equal to one.000 × x^−three Grand before reaction gives an equilibrium concentration of I_ii of half-dozen.61 × ten⁻⁴ Thou, what is the equilibrium constant for the reaction?

Solution
We will begin this problem past calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium abiding. Kickoff, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using −ten every bit the change in concentration of I_two.

Since the equilibrium concentration of I₂ is given, nosotros can solve for 10. At equilibrium the concentration of I₂ is 6.61 × 10^−iv M so that

[latex]1.000\;\times\;10^{-three}\;-\;x = 6.61\;\times\;x^{-4}[/latex]

[latex]x = 1.000\;\times\;x^{-three}\;-\;6.61\;\times\;10^{-4}[/latex]

[latex]= 3.39\;\times\;ten^{-4}\;Thou[/latex]

Now we can fill in the tabular array with the concentrations at equilibrium.

We now calculate the value of the equilibrium constant.

[latex]K_c = Q_c = \frac{[\text{I}_3^{\;\;-}]}{[\text{I}_2][\text{I}^{-}]}[/latex]

[latex]= \frac{three.39\;\times\;ten^{-iv}\;M}{(6.61\;\times\;10^{-4}\;M)(6.61\;\times\;10^{-iv}\;Thou)} = 776[/latex]

Cheque Your Learning
Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the aroma of some nail smooth removers.

[latex]\text{C}_2\text{H}_5\text{OH}\;+\;\text{CH}_3\text{CO}_2\text{H}\;{\rightleftharpoons}\;\text{CH}_3\text{CO}_2\text{C}_2\text{H}_5\;+\;\text{H}_2\text{O}[/latex]

When 1 mol each of C_twoH₅OH and CH_iiiCO₂H are immune to react in i L of the solvent dioxane, equilibrium is established when 1313 mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction.)

Adding of a Missing Equilibrium Concentration

If we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except i, nosotros tin can calculate the missing concentration.

Instance 3

Calculation of a Missing Equilibrium Concentration
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the equilibrium constant for the reaction, [latex]\text{North}_2(k)\;+\;\text{O}_2(chiliad)\;{\rightleftharpoons}\;two\text{NO}(thou)[/latex], is 4.1 × ten⁻⁴. Discover the concentration of NO(g) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N₂] = 0.036 mol/50 and [O_ii] 0.0089 mol/Fifty.

Solution
We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration past rearranging the equation for the equilibrium constant.

[latex]K_c = Q_c = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]}[/latex]

[latex][\text{NO}]^2 = K_c[\text{N}_2][\text{O}_2][/latex]

[latex][\text{NO}] = \sqrt{K_c[\text{Northward}_2][\text{O}_2]}[/latex]

[latex]= \sqrt{(4.ane\;\times\;10^{-4})(0.036)(0.0089)}[/latex]

[latex]= \sqrt{1.31\;\times\;10^{-7}}[/latex]

[latex]= three.6\;\times\;x^{-4}[/latex]

Thus [NO] is 3.6 × x^−four mol/L at equilibrium nether these conditions.

We tin can bank check our respond by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.

[latex]Q_c = \frac{[\text{NO}]^two}{[\text{N}_2][\text{O}_2]}[/latex]

[latex]= \frac{(3.6\;\times\;x^{-4})^2}{(0.036)(0.0089)}[/latex]

[latex]Q_c = 4.0\;\times\;x^{-4} = K_c[/latex]

The answer checks; our calculated value gives the equilibrium abiding within the error associated with the significant figures in the problem.

Check Your Learning
The equilibrium abiding for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is half dozen.00 × x⁻². Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are four.26 Thousand and 2.09 M, respectively.

Adding of Changes in Concentration

If we know the equilibrium abiding for a reaction and a prepare of concentrations of reactants and products that are not at equilibrium, we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.

1. Make up one's mind the management the reaction proceeds to come up to equilibrium.
   1. Write a counterbalanced chemical equation for the reaction.
   2. If the direction in which the reaction must proceed to reach equilibrium is non obvious, calculate Q_c from the initial concentrations and compare to K_c to determine the direction of change.
2. Determine the relative changes needed to achieve equilibrium, then write the equilibrium concentrations in terms of these changes.
   1. Ascertain the changes in the initial concentrations that are needed for the reaction to reach equilibrium. By and large, we stand for the smallest modify with the symbol x and express the other changes in terms of the smallest change.
   2. Ascertain missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration determined in (a).
3. Solve for the alter and the equilibrium concentrations.
   1. Substitute the equilibrium concentrations into the expression for the equilibrium abiding, solve for x, and check any assumptions used to observe ten.
   2. Calculate the equilibrium concentrations.
4. Cheque the arithmetic.
   1. Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium abiding.

      Sometimes a detail stride may differ from problem to problem—information technology may be more than complex in some bug and less complex in others. Nonetheless, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps.

      In solving equilibrium problems that involve changes in concentration, sometimes information technology is convenient to prepare an Ice table, as described in the previous section.

Example 4

Adding of Concentration Changes as a Reaction Goes to Equilibrium
Under certain conditions, the equilibrium constant for the decomposition of PCl₅(k) into PCl₃(g) and Cl₂(g) is 0.0211. What are the equilibrium concentrations of PCl_five, PCl₃, and Cl₂ if the initial concentration of PCl₅ was 1.00 M?

Solution
Use the stepwise process described earlier.

1. Determine the direction the reaction proceeds.

   The counterbalanced equation for the decomposition of PCl_v is

   [latex]\text{PCl}_5(grand)\;{\rightleftharpoons}\;\text{PCl}_3(g)\;+\;\text{Cl}_2(g)[/latex]

   Because we have no products initially, Q_c = 0 and the reaction will keep to the right.

2. Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.

   Let the states correspond the increase in concentration of PCl₃ past the symbol x. The other changes may be written in terms of ten by considering the coefficients in the chemical equation.

   [latex]\begin{array}{lcccc} \text{PCl}_5(one thousand) & {\rightleftharpoons} & \text{PCl}_3(one thousand) & + & \text{Cl}_2(1000) \\[0.5em] -x & & x & & x \finish{array}[/latex]

   The changes in concentration and the expressions for the equilibrium concentrations are:
   

3. Solve for the change and the equilibrium concentrations.

   Substituting the equilibrium concentrations into the equilibrium constant equation gives

   [latex]K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = 0.0211[/latex]

   [latex]= \frac{(10)(x)}{(1.00\;-\;x)}[/latex]

   This equation contains merely one variable, x, the modify in concentration. Nosotros can write the equation as a quadratic equation and solve for x using the quadratic formula.

   [latex]0.0211 = \frac{(10)(x)}{(1.00\;-\;ten)}[/latex]

   [latex]0.0211(1.00\;-\;x) = x^2[/latex]

   [latex]x^2\;+\;0.0211x\;-\;0.0211 = 0[/latex]

   Appendix B shows us an equation of the course ax ² + bx + c = 0 can be rearranged to solve for x:

   [latex]x = \frac{-b\;{\pm}\;\sqrt{b^two\;-\;4ac}}{2a}[/latex]

   In this case, a = ane, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:

   [latex]x = \frac{-0.0211\;{\pm}\;\sqrt{(0.0211)^2\;-\;4(1)(-0.0211)}}{2(i)}[/latex]

   [latex]= \frac{-0.0211\;{\pm}\;\sqrt{(4.45\;\times\;x^{-4})\;+\;(8.44\;\times\;10^{-2})}}{2}[/latex]

   [latex]= \frac{-0.0211\;{\pm}\;0.291}{2}[/latex]

   Hence

   [latex]x = \frac{-0.0211\;+\;0.291}{two} = 0.135[/latex]

   or

   [latex]x = \frac{-0.0211\;-\;0.291}{ii} = -0.156[/latex]

   Quadratic equations oft have 2 different solutions, ane that is physically possible and one that is physically incommunicable (an inapplicable root). In this example, the second solution (−0.156) is physically impossible considering we know the alter must exist a positive number (otherwise nosotros would terminate up with negative values for concentrations of the products). Thus, x = 0.135 M.

   The equilibrium concentrations are

   [latex][\text{PCl}_5] = 1.00\;-\;0.135 = 0.87\;M[/latex]

   [latex][\text{PCl}_3] = x = 0.135\;M[/latex]

   [latex][\text{Cl}_2] = x = 0.135\;M[/latex]

4. Check the arithmetics.

   Substitution into the expression for Thou_c (to check the adding) gives

   [latex]K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{(0.135)(0.135)}{0.87} = 0.021[/latex]

   The equilibrium constant calculated from the equilibrium concentrations is equal to the value of K_c given in the problem (when rounded to the proper number of pregnant figures). Thus, the calculated equilibrium concentrations bank check.

Bank check Your Learning
Acerb acrid, CH₃CO₂H, reacts with ethanol, C₂H_fiveOH, to form h2o and ethyl acetate, CH₃CO₂C₂H₅.

[latex]\text{CH}_3\text{CO}_2\text{H}\;+\;\text{C}_2\text{H}_5\text{OH}\;{\leftrightharpoons}\;\text{CH}_3\text{CO}_2\text{C}_2\text{H}_5\;+\;\text{H}_2\text{O}[/latex]

The equilibrium constant for this reaction with dioxane equally a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 M in CH₃CO₂H, 0.15 M in C_iiH_fiveOH, 0.twoscore M in CH_iiiCO₂C_twoH_v, and 0.40 M in H₂O are mixed in enough dioxane to make 1.0 L of solution?

Answer:

[CH_iiiCO₂H] = 0.36 M, [C₂H₅OH] = 0.36 M, [CH_iiiCO_twoC₂H_five] = 0.17 Yard, [H_twoO] = 0.17 M

Cheque Your Learning
A 1.00-50 flask is filled with i.00 moles of H₂ and 2.00 moles of I₂. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.five under the given conditions. What are the equilibrium concentrations of H₂, I₂, and Hullo in moles/L?

[latex]\text{H}_2(g)\;+\;\text{I}_2(g)\;{\leftrightharpoons}\;2\text{HI}(thou)[/latex]

Answer:

[H₂] = 0.06 G, [I₂] = one.06 M, [How-do-you-do] = 1.88 Chiliad

Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without actually solving a quadratic (or more complicated) equation. Commencement, even so, it is useful to verify that equilibrium can exist obtained starting from ii extremes: all (or by and large) reactants and all (or mostly) products (like to what was shown in Figure 2 in Chapter 13.ii Equilibrium Constants).

Consider the ionization of 0.150 Grand HA, a weak acid.

[latex]\text{HA}(aq)\;{\rightleftharpoons}\;\text{H}^{+}(aq)\;+\;\text{A}^{-}(aq)\;\;\;\;\;\;\;K_c = half-dozen.80\;\times\;ten^{-4}[/latex]

The nearly obvious style to determine the equilibrium concentrations would be to start with merely reactants. This could be called the "all reactant" starting point. Using ten for the amount of acid ionized at equilibrium, this is the Water ice tabular array and solution.

Setting upwards and solving the quadratic equation gives

[latex]K_c = \frac{[\text{H}^{+}][\text{A}^{-}]}{[\text{HA}]} = \frac{(x)(x)}{(0.150\;-\;x)} = 6.lxxx\;\times\;10^{-iv}[/latex]

[latex]x^ii\;+\;(6.80\;\times\;10^{-4}x)\;-\;(one.02\;\times\;ten^{-iv}) = 0[/latex]

[latex]ten = \frac{-6.80\;\times\;10^{-iv}\;{\pm}\;\sqrt{(six.80\;\times\;10^{-4})^ii\;-\;(4)(1)(-1.02\;\times\;10^{-iv})}}{(two)(1)}[/latex]

[latex]x = 0.00977\;M\;\text{or}\;-0.0104\;M[/latex]

Using the positive (physical) root, the equilibrium concentrations are

[latex][\text{HA}] = 0.150\;-\;x = 0.140\;M[/latex]

[latex][\text{H}^{+}] = [\text{A}^{-}] = x = 0.00977\;Chiliad[/latex]

A less obvious way to solve the trouble would be to presume all the HA ionizes first, and then the organization comes to equilibrium. This could exist chosen the "all product" starting betoken. Assuming all of the HA ionizes gives

[latex][\text{HA}] = 0.150\;-\;0.150 = 0\;M[/latex]

[latex][\text{H}^{+}] = 0\;+\;0.150 = 0.150\;K[/latex]

[latex][\text{A}^{-}] = 0\;+\;0.150 = 0.150\;M[/latex]

Using these every bit initial concentrations and "y" to represent the concentration of HA at equilibrium, this is the Water ice table for this starting signal.

Setting up and solving the quadratic equation gives

[latex]K_c = \frac{[\text{H}^{+}][\text{A}^{-}]}{[\text{HA}]} = \frac{(0.150\;-\;y)(0.150\;-\;y)}{(y)} = half dozen.80\;\times\;10^{-4}[/latex]

[latex]half-dozen.fourscore\;\times\;10^{-4}y = 0.0225\;-\;0.300y\;+\;y^2[/latex]

Retain a few extra significant figures to minimize rounding problems.

[latex]y^2\;-\;0.30068y\;+\;0.022500 = 0[/latex]

[latex]y = \frac{0.30068\;{\pm}\;\sqrt{(0.30068)^2\;-\;(4)(ane)(0.022500)}}{(two)(1)}[/latex]

[latex]y = \frac{0.30068\;{\pm}\;0.020210}{2}[/latex]

Rounding each solution to 3 pregnant figures gives

[latex]y = 0.160\;M\;\;\;\;\;\;\;\text{or}\;\;\;\;\;\;\;y = 0.140\;Thou[/latex]

Using the physically significant root (0.140 M) gives the equilibrium concentrations as

[latex][\text{HA}] = y = 0.140\;Thou[/latex]

[latex][\text{H}^{+}] = 0.150\;-\;y = 0.010\;Chiliad[/latex]

[latex][\text{A}^{-}] = 0.150\;-\;y = 0.010\;G[/latex]

Thus, the two approaches give the aforementioned results (to three decimal places), and show that both starting points lead to the aforementioned equilibrium conditions. The "all reactant" starting signal resulted in a relatively minor alter (x) because the system was shut to equilibrium, while the "all product" starting point had a relatively large alter (y) that was nearly the size of the initial concentrations. It can be said that a system that starts "shut" to equilibrium will require merely a "small" modify in conditions (x) to reach equilibrium.

Remember that a small K_c ways that very piffling of the reactants form products and a big K_c means that almost of the reactants form products. If the system tin be arranged so it starts "close" to equilibrium, then if the modify (10) is modest compared to any initial concentrations, it can be neglected. Pocket-sized is usually divers as resulting in an mistake of less than 5%. The following 2 examples demonstrate this.

Instance 5

Approximate Solution Starting Close to Equilibrium
What are the concentrations at equilibrium of a 0.15 1000 solution of HCN?

[latex]\text{HCN}(aq)\;{\rightleftharpoons}\;\text{H}^{+}(aq)\;+\;\text{CN}^{-}(aq)\;\;\;\;\;\;\;K_c = 4.nine\;\times\;10^{-x}[/latex]

Solution
Using "x" to stand for the concentration of each product at equilibrium gives this ICE tabular array.

The verbal solution may be obtained using the quadratic formula with

[latex]K_c = \frac{(10)(x)}{0.xv\;-\;ten}[/latex]

solving

[latex]10^two\;+\;4.nine\;\times\;10^{-10}\;-\;7.35\;\times\;10^{-11} = 0[/latex]

[latex]x = 8.56\;\times\;10^{-6}\;M\;(3\;\text{sig.\;figs.}) = 8.6\;\times\;10^{-vi}\;M\;(2\;\text{sig.\;figs.})[/latex]

Thus [H⁺] = [CN^–] = x = 8.half dozen × 10^{–half-dozen} G and [HCN] = 0.15 – ten = 0.15 Thou.

In this case, chemical intuition can provide a simpler solution. From the equilibrium abiding and the initial conditions, x must be pocket-sized compared to 0.15 M. More formally, if [latex]x\;{\ll}\;0.15[/latex], then 0.xv – ten ≈ 0.15. If this assumption is true, and then it simplifies obtaining 10

[latex]K_c = \frac{(x)(x)}{0.15\;-\;x}\;{\approx}\;\frac{10^2}{0.15}[/latex]

[latex]4.9\;\times\;10^{-10} = \frac{ten^ii}{0.xv}[/latex]

[latex]x^2 = (0.xv)(4.9\;\times\;10^{-10}) = 7.4\;\times\;10^{-11}[/latex]

[latex]x = \sqrt{7.4\;\times\;ten^{-11}} = 8.vi\;\times\;ten^{-6}\;M[/latex]

In this example, solving the exact (quadratic) equation and using approximations gave the same result to 2 significant figures. While most of the time the approximation is a fleck dissimilar from the verbal solution, as long every bit the error is less than 5%, the guess solution is considered valid. In this problem, the five% applies to IF (0.fifteen – 10) ≈ 0.15 M, so if

[latex]\frac{x}{0.15}\;\times\;100\% = \frac{eight.6\;\times\;10^{-6}}{0.xv}\;\times\;100\% = 0.006\%[/latex]

is less than 5%, as it is in this case, the assumption is valid. The approximate solution is thus a valid solution.

Check Your Learning
What are the equilibrium concentrations in a 0.25 M NH₃ solution?

[latex]\text{NH}_3(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{NH}_4^{\;\;+}(aq)\;+\;\text{OH}^{-}(aq)\;\;\;\;\;\;\;K_c = 1.8\;\times\;10^{-five}[/latex]

Assume that x is much less than 0.25 Grand and calculate the error in your assumption.

Answer:

[latex][\text{OH}^{-}] = [\text{NH}_4^{\;\;+}] = 0.0021\;Grand[/latex]; [NH_iii] = 0.25 Thou, mistake = 0.84%

The 2nd example requires that the original information be processed a chip, merely it still tin can exist solved using a small ten approximation.

Instance 6

Approximate Solution Later on Shifting Starting Concentration
Copper(II) ions form a complex ion in the presence of ammonia

[latex]\text{Cu}^{2+}(aq)\;+\;4\text{NH}_3(aq)\;{\rightleftharpoons}\;\text{Cu(NH}_3)_4^{\;\;2+}(aq)\;\;\;\;\;\;\;K_c = v.0\;\times\;10^{13} = \frac{[\text{Cu(NH}_3)_4^{\;\;ii+}]}{[\text{Cu}^{2+}(aq)][\text{NH}_3]^4}[/latex]

If 0.010 mol Cu^two+ is added to 1.00 Fifty of a solution that is 1.00 One thousand NH_iii what are the concentrations when the system comes to equilibrium?

Solution
The initial concentration of copper(II) is 0.010 M. The equilibrium constant is very large so information technology would be better to start with equally much product as possible because "all products" is much closer to equilibrium than "all reactants." Notation that Cu²⁺ is the limiting reactant; if all 0.010 M of it reacts to form production the concentrations would be

[latex][\text{Cu}^{two+}] = 0.010\;-\;0.010 = 0\;M[/latex]

[latex][\text{Cu(NH}_3)_4^{\;\;2+}] = 0.010\;Thou[/latex]

[latex][\text{NH}_3] = one.00\;-\;4\;\times\;0.010 = 0.96\;M[/latex]

Using these "shifted" values as initial concentrations with 10 as the free copper(II) ion concentration at equilibrium gives this ICE table.

Since nosotros are starting close to equilibrium, x should be small so that

[latex]0.96\;+\;4x\;{\approx}\;0.96\;Thou[/latex]

[latex]0.010\;-\;10\;{\approx}\;0.010\;M[/latex]

[latex]K_c = \frac{(0.010\;-\;x)}{x(0.96\;-\;4x)^4}\;{\approx}\;\frac{(0.010)}{x(0.96)^four} = v.0\;\times\;10^{13}[/latex]

[latex]x = \frac{(0.010)}{K_c(0.96)^iv} = 2.4\;\times\;10^{-16}\;1000[/latex]

Select the smallest concentration for the v% rule.

[latex]\frac{2.four\;\times\;ten^{-16}}{0.010}\;\times\;100\% = ii\;\times\;ten^{-12}\%[/latex]

This is much less than 5%, so the assumptions are valid. The concentrations at equilibrium are

[latex][\text{Cu}^{2+}] = x = two.4\;\times\;10^{-xvi}\;Grand[/latex]

[latex][\text{NH}_3] = 0.96\;-\;4x = 0.96\;Grand[/latex]

[latex][\text{Cu(NH}_3)_4^{\;\;ii+}] = 0.010\;-\;x = 0.010\;Thou[/latex]

Past starting with the maximum amount of product, this organisation was nearly equilibrium and the modify (x) was very small-scale. With only a small alter required to become to equilibrium, the equation for x was profoundly simplified and gave a valid result well within the five% error maximum.

Check Your Learning
What are the equilibrium concentrations when 0.25 mol Ni²⁺ is added to ane.00 Fifty of two.00 M NH₃ solution?

[latex]\text{Ni}^{2+}(aq)\;+\;6\text{NH}_3(aq)\;{\rightleftharpoons}\;\text{Ni(NH}_3)_6^{\;\;2+}(aq)\;\;\;\;\;\;\;K_c = 5.5\;\times\;ten^8[/latex]

With such a large equilibrium constant, first class as much product as possible, then assume that only a small-scale amount (ten) of the product shifts left. Calculate the error in your supposition.

Reply:

[latex][\text{Ni(NH}_3)_6^{\;\;2+}] = 0.25\;M[/latex], [NH_three] = 0.50 M, [Ni²⁺] = 2.nine × 10^–eight M, error = 1.2 × 10^–v%

Primal Concepts and Summary

The ratios of the rate of change in concentrations of a reaction are equal to the ratios of the coefficients in the balanced chemic equation. The sign of the coefficient of X is positive when the concentration increases and negative when it decreases. We learned to arroyo three basic types of equilibrium problems. When given the concentrations of the reactants and products at equilibrium, we tin solve for the equilibrium abiding; when given the equilibrium constant and some of the concentrations involved, nosotros tin can solve for the missing concentrations; and when given the equilibrium constant and the initial concentrations, we can solve for the concentrations at equilibrium.

Chemistry End of Affiliate Exercises

1. A reaction is represented by this equation: [latex]\text{A}(aq)\;+\;ii\text{B}(aq)\;{\rightleftharpoons}\;2\text{C}(aq)\;\;\;\;\;\;\;K_c = 1\;\times\;ten^iii[/latex]

   (a) Write the mathematical expression for the equilibrium abiding.

   (b) Using concentrations ≤i M, brand up ii sets of concentrations that draw a mixture of A, B, and C at equilibrium.

2. A reaction is represented by this equation: [latex]2\text{Westward}(aq)\;{\rightleftharpoons}\;\text{X}(aq)\;+\;2\text{Y}(aq)\;\;\;\;\;\;\;K_c = five\;\times\;x^{-iv}[/latex]

   (a) Write the mathematical expression for the equilibrium constant.

   (b) Using concentrations of ≤ane K, make up two sets of concentrations that describe a mixture of Westward, X, and Y at equilibrium.

3. What is the value of the equilibrium constant at 500 °C for the formation of NH₃ according to the following equation?

   [latex]\text{N}_2(chiliad)\;+\;3\text{H}_2(g)\;{\rightleftharpoons}\;2\text{NH}_3(g)[/latex]

   An equilibrium mixture of NH₃(g), H_ii(yard), and N₂(g) at 500 °C was found to contain 1.35 Yard H_two, one.15 M Northward₂, and 4.12 × ten⁻¹ M NH_three.

4. Hydrogen is prepared commercially by the reaction of marsh gas and water vapor at elevated temperatures.

   [latex]\text{CH}_4(one thousand)\;+\;\text{H}_2\text{O}(m)\;{\rightleftharpoons}\;3\text{H}_2(grand)\;+\;\text{CO}(g)[/latex]

   What is the equilibrium abiding for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH_iv, 0.126 M; H₂O, 0.242 M; CO, 0.126 M; H₂ 1.15 M, at a temperature of 760 °C?

5. A 0.72-mol sample of PCl₅ is put into a 1.00-50 vessel and heated. At equilibrium, the vessel contains 0.forty mol of PCl_three(thousand) and 0.forty mol of Cl_ii(g). Calculate the value of the equilibrium constant for the decomposition of PCl_v to PCl_iii and Cl₂ at this temperature.
6. At 1 atm and 25 °C, NO_ii with an initial concentration of 1.00 Yard is 3.three × x⁻ⁱⁱⁱ% decomposed into NO and O₂. Calculate the value of the equilibrium constant for the reaction.

   [latex]two\text{NO}_2(g)\;{\rightleftharpoons}\;ii\text{NO}(m)\;+\;\text{O}_2(m)[/latex]

7. Calculate the value of the equilibrium constant K_P for the reaction [latex]two\text{NO}(g)\;+\;\text{Cl}_2(k)\;{\rightleftharpoons}\;2\text{NOCl}(yard)[/latex] from these equilibrium pressures: NO, 0.050 atm; Cl₂, 0.thirty atm; NOCl, 1.ii atm.
8. When heated, iodine vapor dissociates according to this equation:

   [latex]\text{I}_2(g)\;{\rightleftharpoons}\;2\text{I}(g)[/latex]

   At 1274 K, a sample exhibits a partial force per unit area of I₂ of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, K_P , for the decomposition at 1274 Chiliad.

9. A sample of ammonium chloride was heated in a closed container.

   [latex]\text{NH}_4\text{Cl}(southward)\;{\rightleftharpoons}\;\text{NH}_3(g)\;+\;\text{HCl}(g)[/latex]

   At equilibrium, the pressure of NH₃(thou) was found to be 1.75 atm. What is the value of the equilibrium constant K_P for the decomposition at this temperature?

10. At a temperature of lx °C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant K_P for the transformation at threescore °C?

    [latex]\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_2\text{O}(thousand)[/latex]

11. Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.

    (a)

    [latex]\begin{array}{lcccc} 2\text{SO}_3(thou) & {\rightleftharpoons} & 2\text{SO}_2(g) & + & \text{O}_2(one thousand) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} & & +ten \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} & & 0.125\;G \stop{assortment}[/latex]

    (b)

    [latex]\begin{array}{lcccccc} 4\text{NH}_3(g) & + & three\text{O}_2(g) & {\rightleftharpoons} & two\text{Northward}_2(g) & + & 6\text{H}_2\text{O}(g) \\[0.5em] \dominion[0ex]{two.5em}{0.1ex} & & 3x & & \dominion[0ex]{2.5em}{0.1ex} & & \dominion[0ex]{ii.5em}{0.1ex} \\[0.5em] \rule[0ex]{ii.5em}{0.1ex} & & 0.24\;G & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{two.5em}{0.1ex} \end{array}[/latex]

    (c) Change in pressure:

    [latex]\begin{array}{lcccc} 2\text{CH}_4(g) & {\rightleftharpoons} & \text{C}_2\text{H}_2(1000) & + & three\text{H}_2(g) \\[0.5em] \dominion[0ex]{2.5em}{0.1ex} & & x & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & 25\;\text{torr} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]

    (d) Change in pressure:

    [latex]\begin{array}{lcccccc} \text{CH}_4(thousand) & + & \text{H}_2\text{O}(g) & {\rightleftharpoons} & \text{CO}(thou) & + & 3\text{H}_2(m) \\[0.5em] \dominion[0ex]{2.5em}{0.1ex} & & 10 & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{two.5em}{0.1ex} \\[0.5em] \dominion[0ex]{ii.5em}{0.1ex} & & 5\;\text{atm} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} \finish{assortment}[/latex]

    (eastward)

    [latex]\begin{array}{lcccc} \text{NH}_4\text{Cl}(southward) & {\rightleftharpoons} & \text{NH}_3(one thousand) & + & \text{HCl}(g) \\[0.5em] & & ten & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] & & 1.03\;\times\;10^{-4}\;M & & \rule[0ex]{two.5em}{0.1ex} \end{assortment}[/latex]

    (f) modify in pressure:

    [latex]\begin{array}{lcccc} \text{Ni}(s) & + & iv\text{CO}(g) & {\leftrightharpoons} & \text{Ni(CO)}_4(yard) \\[0.5em] & & 4x & & \dominion[0ex]{two.5em}{0.1ex} \\[0.5em] & & 0.40\;\text{atm} & & \rule[0ex]{ii.5em}{0.1ex} \end{array}[/latex]

12. Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.

    (a)

    [latex]\begin{array}{lcccc} 2\text{H}_2(g) & + & \text{O}_2(g) & {\rightleftharpoons} & 2\text{H}_2\text{O}(g) \\[0.5em] \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & &+2x \\[0.5em] \rule[0ex]{ii.5em}{0.1ex} & & \dominion[0ex]{2.5em}{0.1ex} & & ane.fifty\;One thousand \cease{array}[/latex]

    (b)

    [latex]\brainstorm{array}{lcccccc} \text{CS}_2(chiliad) & + & 4\text{H}_2(g) & {\rightleftharpoons} & \text{CH}_4(chiliad) & + & 2\text{H}_2\text{S}(yard) \\[0.5em] x & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{two.5em}{0.1ex} & & \rule[0ex]{two.5em}{0.1ex} \\[0.5em] 0.020\;G & & \rule[0ex]{two.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{two.5em}{0.1ex} \end{array}[/latex]

    (c) Change in pressure:

    [latex]\begin{array}{lcccc} \text{H}_2(yard) & + & \text{Cl}_2(g) & {\rightleftharpoons} & 2\text{HCl}(one thousand) \\[0.5em] x & & \rule[0ex]{ii.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} \\[0.5em] 1.fifty\;\text{atm} & & \dominion[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]

    (d) Change in pressure:

    [latex]\begin{array}{lcccccc} 2\text{NH}_3(g) & + & 2\text{O}_2(g) & {\rightleftharpoons} & \text{N}_2\text{O}(chiliad) & + & 3\text{H}_2\text{O}(1000) \\[0.5em] \rule[0ex]{two.5em}{0.1ex} & & \dominion[0ex]{2.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & x \\[0.5em] \rule[0ex]{two.5em}{0.1ex} & & \rule[0ex]{2.5em}{0.1ex} & & \rule[0ex]{ii.5em}{0.1ex} & & 60.six\;\text{torr} \end{array}[/latex]

    (e)

    [latex]\brainstorm{array}{lcccc} \text{NH}_4\text{HS}(s) & {\leftrightharpoons} & \text{NH}_3(g) & + & \text{H}_2\text{S}(thousand) \\[0.5em] & & x & & \rule[0ex]{2.5em}{0.1ex} \\[0.5em] & & 9.eight\;\times\;ten^{-6}\;1000 & & \rule[0ex]{2.5em}{0.1ex} \end{array}[/latex]

    (f) Change in pressure:

    [latex]\begin{array}{lcccc} \text{Iron}(southward) & + & 5\text{CO}(g) & {\leftrightharpoons} & \text{Fe(CO)}_4(g) \\[0.5em] & & \dominion[0ex]{ii.5em}{0.1ex} & & ten \\[0.5em] & & \rule[0ex]{two.5em}{0.1ex} & & 0.012\;\text{atm} \end{array}[/latex]

13. Why are there no changes specified for Ni in Exercise 11, office (f)? What property of Ni does change?
14. Why are in that location no changes specified for NH₄HS in Exercise 12, office (e)? What holding of NH₄HS does change?
15. Analysis of the gases in a sealed reaction vessel containing NH₃, N₂, and H₂ at equilibrium at 400 °C established the concentration of N₂ to be 1.2 M and the concentration of H₂ to be 0.24 M.

    [latex]\text{North}_2(g)\;+\;3\text{H}_2(g)\;{\rightleftharpoons}\;ii\text{NH}_3(g)\;\;\;\;\;\;\;K_c = 0.50\;\text{at}\;400\;^{\circ}\text{C}[/latex]

    Summate the equilibrium molar concentration of NH₃.

16. Calculate the number of moles of Hi that are at equilibrium with one.25 mol of H₂ and ane.25 mol of I_ii in a 5.00−L flask at 448 °C.

    [latex]\text{H}_2\;+\;\text{I}_2\;{\rightleftharpoons}\;2\text{Hello}\;\;\;\;\;\;\;K_c = 50.2\;\text{at}\;448\;^{\circ}\text{C}[/latex]

17. What is the pressure of BrCl in an equilibrium mixture of Cl_ii, Br₂, and BrCl if the pressure of Cl₂ in the mixture is 0.115 atm and the force per unit area of Br_two in the mixture is 0.450 atm?

    [latex]\text{Cl}_2(thou)\;+\;\text{Br}_2(g)\;{\rightleftharpoons}\;two\text{BrCl}(g)\;\;\;\;\;\;\;K_P = four.seven\;\times\;x^{-ii}[/latex]

18. What is the pressure of CO₂ in a mixture at equilibrium that contains 0.50 atm H₂, 2.0 atm of H₂O, and 1.0 atm of CO at 990 °C?

    [latex]\text{H}_2(g)\;+\;\text{CO}_2(g)\;{\rightleftharpoons}\;\text{H}_2\text{O}(g)\;+\;\text{CO}(1000)\;\;\;\;\;\;\;K_P = 1.6\;\text{at}\;990\;^{\circ}\text{C}[/latex]

19. Cobalt metal tin can be prepared by reducing cobalt(2) oxide with carbon monoxide.

    [latex]\text{CoO}(s)\;+\;\text{CO}(m)\;{\rightleftharpoons}\;\text{Co}(due south)\;+\;\text{CO}_2(grand)\;\;\;\;\;\;\;K_c = 4.90\;\times\;10^2\;\text{at}\;550\;^{\circ}\text{C}[/latex]

    What concentration of CO remains in an equilibrium mixture with [CO_ii] = 0.100 Thou?

20. Carbon reacts with h2o vapor at elevated temperatures.

    [latex]\text{C}(s)\;+\;\text{H}_2\text{O}(g)\;{\rightleftharpoons}\;\text{CO}(1000)\;+\;\text{H}_2(g)\;\;\;\;\;\;\;K_c = 0.2\;\text{at}\;1000\;^{\circ}\text{C}[/latex]

    What is the concentration of CO in an equilibrium mixture with [H_twoO] = 0.500 K at 1000 °C?

21. Sodium sulfate 10−hydrate, Na_iiSO₄·10H_twoO, dehydrates according to the equation

    [latex]\text{Na}_2\text{Then}_4{\cdot}10\text{H}_2\text{O}(s)\;{\rightleftharpoons}\;\text{Na}_2\text{So}_4(southward)\;+\;ten\text{H}_2\text{O}(g)\;\;\;\;\;\;\;K_P = four.08\;\times\;10^{-25}\;\text{at}\;25\;^{\circ}\text{C}[/latex]

    What is the force per unit area of h2o vapor at equilibrium with a mixture of Na_iiThen₄·10H₂O and NaSO_four?

22. Calcium chloride six−hydrate, CaCl₂·6H_twoO, dehydrates according to the equation

    [latex]\text{CaCl}_2{\cdot}half-dozen\text{H}_2\text{O}(south)\;{\rightleftharpoons}\;\text{CaCl}_2(s)\;+\;6\text{H}_2\text{O}(g)\;\;\;\;\;\;\;K_P = v.09\;\times\;10^{-44}\;\text{at}\;25\;^{\circ}\text{C}[/latex]

    What is the force per unit area of water vapor at equilibrium with a mixture of CaCl₂·6H₂O and CaCl₂?

23. A student solved the following problem and found the equilibrium concentrations to exist [SO₂] = 0.590 M, [O₂] = 0.0450 M, and [SO₃] = 0.260 K. How could this pupil cheque the piece of work without reworking the problem? The trouble was: For the post-obit reaction at 600 °C:

    [latex]2\text{And so}_2(thousand)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{SO}_3(g)\;\;\;\;\;\;\;K_c = 4.32[/latex]

    What are the equilibrium concentrations of all species in a mixture that was prepared with [SO₃] = 0.500 M, [And so₂] = 0 M, and [O_ii] = 0.350 M?

24. A student solved the following problem and constitute [Due north₂O_four] = 0.16 M at equilibrium. How could this student recognize that the reply was wrong without reworking the problem? The problem was: What is the equilibrium concentration of Northward₂O₄ in a mixture formed from a sample of NO₂ with a concentration of 0.10 M?

    [latex]2\text{NO}_2(g)\;{\rightleftharpoons}\;\text{N}_2\text{O}_4(g)\;\;\;\;\;\;\;K_c = 160[/latex]

25. Presume that the alter in concentration of N₂O₄ is small plenty to exist neglected in the following problem.
    (a) Calculate the equilibrium concentration of both species in ane.00 Fifty of a solution prepared from 0.129 mol of N₂O_iv with chloroform as the solvent.

    [latex]\text{N}_2\text{O}_4(g)\;{\leftrightharpoons}\;2\text{NO}_2(g)\;\;\;\;\;\;\;K_c = 1.07\;\times\;x^{-5}[/latex] in chloroform

    (b) Evidence that the modify is minor plenty to be neglected.

26. Presume that the change in concentration of COCl₂ is small enough to be neglected in the following problem.

    (a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl₂ with an initial concentration of 0.3166 M.

    [latex]\text{COCl}_2(g)\;{\rightleftharpoons}\;\text{CO}(thou)\;+\;\text{Cl}_2(g)\;\;\;\;\;\;\;K_c = 2.2\;\times\;10^{-10}[/latex]

    (b) Show that the alter is small enough to be neglected.

27. Presume that the change in force per unit area of H_twoS is small enough to be neglected in the following trouble.

    (a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H₂S with an initial force per unit area of 0.824 atm.

    [latex]2\text{H}_2\text{S}(g)\;{\rightleftharpoons}\;2\text{H}_2(thou)\;+\;\text{S}_2(g)\;\;\;\;\;\;\;K_P = 2.ii\;\times\;10^{-half-dozen}[/latex]

    (b) Show that the change is small enough to be neglected.

28. What are all concentrations after a mixture that contains [H₂O] = 1.00 M and [Cl₂O] = 1.00 M comes to equilibrium at 25 °C?

    [latex]\text{H}_2\text{O}(g)\;+\;\text{Cl}_2\text{O}(yard)\;{\rightleftharpoons}\;two\text{HOCl}(g)\;\;\;\;\;\;\;K_c = 0.0900[/latex]

29. What are the concentrations of PCl_v, PCl₃, and Cl_two in an equilibrium mixture produced by the decomposition of a sample of pure PCl₅ with [PCl₅] = ii.00 M?

    [latex]\text{PCl}_5(yard)\;{\rightleftharpoons}\;\text{PCl}_3(thou)\;+\;\text{Cl}_2(k)\;\;\;\;\;\;\;K_c = 0.0211[/latex]

30. Summate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl₂ produced when a sample of NOCl with a pressure of 10.0 atm comes to equilibrium according to this reaction:
    [latex]2\text{NOCl}(k)\;{\rightleftharpoons}\;2\text{NO}(g)\;+\;\text{Cl}_2(g)\;\;\;\;\;\;\;K_P = iv.0\;\times\;ten^{-four}[/latex]
31. Calculate the equilibrium concentrations of NO, O₂, and NO_ii in a mixture at 250 °C that results from the reaction of 0.20 M NO and 0.x Yard O_two. (Hint: Chiliad is large; presume the reaction goes to completion and so comes back to equilibrium.)

    [latex]2\text{NO}(chiliad)\;+\;\text{O}_2(k)\;{\rightleftharpoons}\;2\text{NO}_2(one thousand)\;\;\;\;\;\;\;K_c = ii.3\;\times\;x^v\;\text{at}\;250\;^{\circ}\text{C}[/latex]

32. Calculate the equilibrium concentrations that issue when 0.25 M O_two and 1.0 Chiliad HCl react and come to equilibrium.

    [latex]4\text{HCl}(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;ii\text{Cl}_2(thou)\;+\;ii\text{H}_2\text{O}(m)\;\;\;\;\;\;\;K_c = three.1\;\times\;10^{xiii}[/latex]

33. 1 of the important reactions in the formation of smog is represented by the equation

    [latex]\text{O}_3(m)\;+\;\text{NO}(g)\;{\rightleftharpoons}\;\text{NO}_2(k)\;+\;\text{O}_2(grand)\;\;\;\;\;\;\;K_P = half dozen.0\;\times\;x^{34}[/latex]

    What is the pressure of O_three remaining after a mixture of O₃ with a pressure of 1.2 × 10^−viii atm and NO with a pressure level of one.two × 10⁻⁸ atm comes to equilibrium? (Hint: K_P is large; presume the reaction goes to completion so comes back to equilibrium.)

34. Calculate the pressures of NO, Cl₂, and NOCl in an equilibrium mixture produced past the reaction of a starting mixture with 4.0 atm NO and two.0 atm Cl_ii. (Hint: K_P is small; assume the opposite reaction goes to completion so comes back to equilibrium.)

    [latex]two\text{NO}(thousand)\;+\;\text{Cl}_2(g)\;{\rightleftharpoons}\;ii\text{NOCl}(1000)\;\;\;\;\;\;\;K_P = 2.5\;\times\;10^3[/latex]

35. Calculate the number of grams of Howdy that are at equilibrium with 1.25 mol of H₂ and 63.5 g of iodine at 448 °C.

    [latex]\text{H}_2\;+\;\text{I}_2\;{\rightleftharpoons}\;2\text{HI}\;\;\;\;\;\;\;K_c = 50.ii\;\text{at}\;448\;^{\circ}\text{C}[/latex]

36. Butane exists as 2 isomers, n−butane and isobutane.
    

    K_P = 2.v at 25 °C

    What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure level of 1.22 atm?

37. What is the minimum mass of CaCO₃ required to constitute equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (M_c ) is 0.050 for the decomposition reaction of CaCO₃ at that temperature?

    [latex]\text{CaCO}_3(south)\;{\rightleftharpoons}\;\text{CaO}(s)\;+\;\text{CO}_2(g)[/latex]

38. The equilibrium constant (G_c ) for this reaction is 1.60 at 990 °C:

    [latex]\text{H}_2(g)\;+\;\text{CO}_2(g)\;{\rightleftharpoons}\;\text{H}_2\text{O}(k)\;+\;\text{CO}(chiliad)[/latex]

    Calculate the number of moles of each component in the concluding equilibrium mixture obtained from adding 1.00 mol of H_ii, 2.00 mol of CO_two, 0.750 mol of H₂O, and 1.00 mol of CO to a v.00-50 container at 990 °C.

39. At 25 °C and at 1 atm, the fractional pressures in an equilibrium mixture of N_twoO_iv and NO₂ are [latex]\text{P}_{\text{N}_2\text{O}_4} = 0.70\;\text{atm}[/latex] and [latex]\text{P}_{\text{NO}_2} = 0.30\;\text{atm}[/latex].

    (a) Predict how the pressures of NO₂ and Due north₂O_iv volition change if the full pressure increases to 9.0 atm. Will they increment, subtract, or remain the same?

    (b) Calculate the partial pressures of NO₂ and Due north₂O₄ when they are at equilibrium at 9.0 atm and 25 °C.

40. In a iii.0-Fifty vessel, the following equilibrium fractional pressures are measured: N_two, 190 torr; H₂, 317 torr; NH_three, ane.00 × 10³ torr.

    [latex]\text{N}_2(one thousand)\;+\;iii\text{H}_2(chiliad)\;{\rightleftharpoons}\;2\text{NH}_3(g)[/latex]

    (a) How volition the partial pressures of H₂, North₂, and NH₃ change if H₂ is removed from the system? Will they increase, decrease, or remain the same?

    (b) Hydrogen is removed from the vessel until the partial pressure level of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new weather condition.

41. The equilibrium constant (K_c ) for this reaction is 5.0 at a given temperature.

    [latex]\text{CO}(g)\;+\;\text{H}_2\text{O}(1000)\;{\rightleftharpoons}\;\text{CO}_2(yard)\;+\;\text{H}_2(thousand)[/latex]

    (a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.ninety mol of H₂ in a liter. How many moles of CO₂ were there in the equilibrium mixture?

    (b) Maintaining the aforementioned temperature, additional H_ii was added to the system, and some water vapor was removed past drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H₂ in a liter. How many moles of CO₂ were in the new equilibrium mixture? Compare this with the quantity in part (a), and talk over whether the 2nd value is reasonable. Explicate how information technology is possible for the water vapor concentration to exist the same in the two equilibrium solutions fifty-fifty though some vapor was removed earlier the second equilibrium was established.

42. Antimony pentachloride decomposes according to this equation:

    [latex]\text{SbCl}_5(g)\;{\rightleftharpoons}\;\text{SbCl}_3(g)\;+\;\text{Cl}_2(1000)[/latex]

    An equilibrium mixture in a 5.00-L flask at 448 °C contains three.85 g of SbCl_v, ix.fourteen g of SbCl₃, and ii.84 chiliad of Cl₂. How many grams of each volition be constitute if the mixture is transferred into a two.00-50 flask at the aforementioned temperature?

43. Consider the reaction betwixt H₂ and O_two at 1000 K
    [latex]two\text{H}_2(thousand)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;ii\text{H}_2\text{O}(g)\;\;\;\;\;\;\;K_P = \frac{(P_{\text{H}_2\text{O}})^two}{(P_{\text{O}_2})(P_{\text{H}_2})^3} = i.33\;\times\;10^{twenty}[/latex]

    If 0.500 atm of H₂ and 0.500 atm of O_two are immune to come to equilibrium at this temperature, what are the fractional pressures of the components?

44. An equilibrium is established according to the following equation

    [latex]\text{Hg}_2^{\;\;2+}(aq)\;+\;\text{NO}_3^{\;\;-}(aq)\;+\;3\text{H}^{+}(aq)\;{\rightleftharpoons}\;two\text{Hg}^{2+}(aq)\;+\;\text{HNO}_2(aq)\;+\;\text{H}_2\text{O}(l)\;\;\;\;\;\;\;K_c = 4.vi[/latex]

    What will happen in a solution that is 0.twenty Thou each in [latex]\text{Hg}_2^{\;\;2+}[/latex], [latex]\text{NO}_3^{\;\;-}[/latex], H⁺, Hg²⁺, and HNO₂?

    (a) [latex]\text{Hg}_2^{\;\;two+}[/latex] will be oxidized and [latex]\text{NO}_3^{\;\;-}[/latex] reduced.

    (b) [latex]\text{Hg}_2^{\;\;2+}[/latex] will exist reduced and [latex]\text{NO}_3^{\;\;-}[/latex] oxidized.

    (c) Hg²⁺ will be oxidized and HNO₂ reduced.

    (d) Hg²⁺ will exist reduced and HNO₂ oxidized.

    (east) There will be no change because all reactants and products have an activity of 1.

45. Consider the equilibrium

    [latex]4\text{NO}_2(1000)\;+\;half dozen\text{H}_2\text{O}(g)\;{\rightleftharpoons}\;four\text{NH}_3(1000)\;+\;vii\text{O}_2(thou)[/latex]

    (a) What is the expression for the equilibrium constant (One thousand_c ) of the reaction?

    (b) How must the concentration of NH₃ change to reach equilibrium if the reaction quotient is less than the equilibrium constant?

    (c) If the reaction were at equilibrium, how would a decrease in pressure (from an increase in the book of the reaction vessel) bear upon the pressure of NO₂?

    (d) If the change in the pressure of NO₂ is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of O₂ alter?

46. The binding of oxygen past hemoglobin (Hb), giving oxyhemoglobin (HbO₂), is partially regulated past the concentration of H_iiiO⁺ and dissolved CO_two in the claret. Although the equilibrium is complicated, information technology can be summarized as

    [latex]\text{HbO}_2(aq)\;+\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{CO}_2(g)\;{\rightleftharpoons}\;\text{CO}_2\;-\;\text{Hb}\;-\;\text{H}^{+}\;+\;\text{O}_2(g)\;+\;\text{H}_2\text{O}(l)[/latex]

    (a) Write the equilibrium constant expression for this reaction.

    (b) Explain why the production of lactic acid and CO₂ in a musculus during exertion stimulates release of O₂ from the oxyhemoglobin in the blood passing through the musculus.

47. The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-society charge per unit equation for the disappearance of sucrose.

    [latex]\text{C}_{12}\text{H}_{22}\text{O}_{11}(aq)\;+\;\text{H}_2\text{O}(l)\;{\longrightarrow}\;\text{C}_6\text{H}_{12}\text{O}_6(aq)\;+\;\text{C}_6\text{H}_{12}\text{O}_6(aq)[/latex]

    Rate = k[C₁₂H₂₂O₁₁]

    In neutral solution, k = two.i × x^−eleven/s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up past a type of catalyst called an enzyme.) (Note: That is non a fault in the equation—the products of the reaction, glucose and fructose, take the aforementioned molecular formulas, C_{half dozen}H₁₂O_six, but differ in the organisation of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × ten⁵ at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is ane.

48. The density of trifluoroacetic acid vapor was adamant at 118.ane °C and 468.v torr, and found to be ii.784 g/L. Calculate K_c for the association of the acid.
    
49. Liquid N₂O₃ is dark blue at depression temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO₂. At 25 °C, a value of K_P = i.91 has been established for this decomposition. If 0.236 moles of N_iiO₃ are placed in a 1.52-Fifty vessel at 25 °C, calculate the equilibrium partial pressures of N₂O₃(g), NO₂(g), and NO(g).
50. A 1.00-Fifty vessel at 400 °C contains the post-obit equilibrium concentrations: N_two, ane.00 M; H₂, 0.fifty M; and NH₃, 0.25 Grand. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 Grand?
51. A 0.010 M solution of the weak acid HA has an osmotic pressure (run into chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010 M solution of the weak acrid HB has an osmotic pressure of 0.345 atm under the same conditions.

    (a) Which acrid has the larger equilibrium abiding for ionization

    HA [latex][\text{HA}(aq)\;{\rightleftharpoons}\;\text{A}^{-}(aq)\;+\;\text{H}^{+}(aq)][/latex] or HB [latex][\text{HB}(aq)\;{\rightleftharpoons}\;\text{H}^{+}(aq)\;+\;\text{B}^{-}(aq)][/latex]?

    (b) What are the equilibrium constants for the ionization of these acids?

    (Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the cohabit base (A⁻ or B⁻), and the hydrogen ion (H⁺). Remember that osmotic pressure level (similar all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)

Solutions

Answers to Chemistry Terminate of Chapter Exercises

one. [latex]K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2}[/latex]. [A] = 0.1 M, [B] = 0.1 M, [C] = one Thousand; and [A] = 0.01, [B] = 0.250, [C] = 0.791.

three. M_c = half dozen.00 × 10⁻²

five. Chiliad_c = 0.l

vii. The equilibrium equation is
K_P = one.9 × 10³

9. K_P = 3.06

11. (a) −twox, 210, −0.250 M, 0.250 M; (b) 4x, −210, −sixx, 0.32 K, −0.16 M, −0.48 M; (c) −twox, 310, −50 torr, 75 torr; (d) 10, − x, −3ten, five atm, −5 atm, −fifteen atm; (e) ten, 1.03 × x⁻⁴ M; (f) x, 0.1 atm.

13. Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.

15. [NH₃] = 9.1 × 10⁻² M

17. P _BrCl = 4.9 × 10⁻² atm

19. [CO] = 2.0 × ten⁻⁴ Thou

21. [latex]P_{\text{H}_2\text{O}} = 3.64\;\times\;x^{-iii}\;\text{atm}[/latex]

23. Summate Q based on the calculated concentrations and see if it is equal to Thou_c . Considering Q does equal 4.32, the system must be at equilibrium.

25. (a) [NO_two] = 1.17 × x⁻³ Yard
[North₂O₄] = 0.128 Thousand
(b) Percent error [latex]= \frac{5.87\;\times\;x^{-four}}{0.129}\;\times\;100\% = 0.455\%[/latex]. The change in concentration of N₂O₄ is far less than the v% maximum allowed.

27. (a) [H₂Southward] = 0.810 atm
[H_ii] = 0.014 atm
[S₂] = 0.0072 atm
(b) The 2x is dropped from the equilibrium calculation considering 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2x is [latex]\frac{0.014}{0.824}\;\times\;100\% = i.vii\%[/latex], which is less than allowed by the "5% test." The fault is, indeed, negligible.

29. [PCl₃] = 1.80 Thou; [PC₃] = 0.195 Thousand; [PCl₃] = 0.195 M.

31. [NO₂] = 0.19 One thousand
[NO] = 0.0070 G
[O₂] = 0.0035 Thou

33. [latex]P_{\text{O}_3} = 4.9\;\times\;10^{-26}\;\text{atm}[/latex]

35. 507 g

37. 330 g

39. (a) Both gases must increment in force per unit area.
(b)[latex]P_{\text{N}_2\text{O}_4} = eight.0\;\text{atm\;and}\;P_{\text{NO}_2} = 1.0\;\text{atm}[/latex]

41. (a) 0.33 mol.
(b) [CO]² = 0.50 Thousand Added H₂ forms some water to recoup for the removal of water vapor and as a upshot of a shift to the left after H_ii is added.

43. [latex]P_{\text{H}_2} = 8.64\;\times\;10^{-11}\;\text{atm}[/latex]
[latex]P_{\text{O}_2} = 0.250\;\text{atm}[/latex]
[latex]P_{\text{H}_2\text{O}} = 0.500\;\text{atm}[/latex]

45. (a) [latex]K_c = \frac{[\text{NH}_3]^4[\text{O}_2]^7}{[\text{NO}_2]^4[\text{H}_2\text{O}]^six}[/latex]. (b) [NH₃] must increase for Q_c to reach K_c . (c) That decrease in pressure would subtract [NO₂]. (d) [latex]P_{\text{O}_2} = 49\;\text{torr}[/latex]

47. [fructose] = 0.15 M

49. [latex]P_{\text{N}_2\text{O}_3} = 1.xc\;\text{atm\;and}\;P_{\text{NO}} = P_{\text{NO}_2} = i.90\;\text{atm}[/latex]

51. (a) HB ionizes to a greater degree and has the larger One thousand_c .
(b) K_c (HA) = 5 × 10^−iv
G_c (HB) = 3 × ten^−three

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Source: https://opentextbc.ca/chemistry/chapter/13-4-equilibrium-calculations/

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