Chapter 4: Q5P (page 199)

The following reaction has a value of ${ΔG}^{0} = -2.1 kJ/mol (-0.50 kcal/mol)$
${CH}_{3} {Br + H}_{2} S ⇄ {CH}_{3} SH + HBr$
(a) Calculate $K_{e q}$ at room temperature ${(25}^{0} C)$ for this reaction as written.
(b)Starting with a $1M$ solution of ${CH}_{3} Br$ and $H_{2} S$ , calculate the final concentrations of all four species at equilibrium.

Short Answer

Expert verified

a. $K_{e q} = 2.33$

b. ${[CH}_{3} Br] = 0.4 M$

${[H}_{2} S] = 0.4 M$

${[CH}_{3} SH] = 0.6 M$

$[HBr] = 0.6 M$

Step by step solution

Equilibrium constant (Keq)

Equilibrium constant $(K_{e q})$ of the reaction governs the equilibrium concentration of the reactants and products. For a general reaction of the type, $aA + bB ⇄ cC + dD$ , the equilibrium constant $(K_{e q})$ expression can be written as:

$\begin{matrix} K_{e q} = \frac{[products]}{[reactants]} \\ = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}} \end{matrix}$

Position of equilibrium

The position of equilibrium can be predicted form the value of equilibrium constant $(K_{e q})$ . The forward reaction is favored if $K_{e q} > 1$ and backward reaction is favored if $K_{e q} < 1$ .

Explanation

(a)The expression for standard Gibb’s free energy is ${ΔG}^{0} = - RTln K_{e q}$ . This expression can also be written as data-custom-editor="chemistry" ${ΔG}^{0} = - 2.303 RT(log K_{e q})$ .

As per given data,

data-custom-editor="chemistry" ${ΔG}^{0} = -2.1 kJ/mol (-0.50 kcal/mol)$

data-custom-editor="chemistry" ${T=25}^{0} C =(25+273)K = 298K$

data-custom-editor="chemistry" $R= 8.314 J/K.mol (universal gas constant)$

Find data-custom-editor="chemistry" $K_{e q}$ by putting the given values in the standard Gibb’s free energy expression.

Now,

data-custom-editor="chemistry" $\begin{matrix} {ΔG}^{0} = - 2.303 RTlog K_{e q} \\ \Rightarrow log K_{e q} = \frac{{ΔG}^{0}}{- 2.303 RT} \\ \Rightarrow log K_{e q} = \frac{- 2.1 kJ/ mol}{- 2.303 \times 8.314 J/ K . mol \times 298 K} \\ \Rightarrow log K_{e q} = \frac{2100 J}{5705.848 J} \\ \Rightarrow log K_{e q} = 0.368 \\ \Rightarrow K_{e q} = 2.33 \end{matrix}$

b) Set up an ICE table as:

data-custom-editor="chemistry" $\begin{array}{l} {________ CH}_{3} {Br + H}_{2} S ⇄ {CH}_{3} SH + HBr \\ Initial____ 1M___1M_____0_____0 \\ Change____ -x____-x______+x____+x \\ Equilibrium__(1-x)__(1-x)____x_____x \end{array}$

Now,

$\begin{array}{l} K_{eq} = \frac{[products]}{[reactants]} \\ \Rightarrow 2.33 = \frac{{{[CH}_{3} SH]}^{1} {[HBr]}^{1}}{{{[CH}_{3} Br]}^{1} {{[H}_{2} S]}^{1}} \\ \Rightarrow 2.33 = \frac{x*x}{(1-x)*(1-x)} \\ \Rightarrow 2.33 = \frac{x^{2}}{{(1-x)}^{2}} \\ \Rightarrow 2.33 = \frac{x^{2}}{{(1}^{2} {+x}^{2} - 2 x)} \\ \Rightarrow 2.33 * (1^{2} {+x}^{2} - 2 x) = x^{2} \\ \Rightarrow 2.33 + 2.33 x^{2} - 4.66 x - x^{2} = 0 \\ \Rightarrow 1.33 x^{2} - 4.66 x + 2.33 = 0 \end{array}$

Here, $a =1.33 , b = -4.66 , c = 2.33$

Use quadratic formula $\frac{-b ± \sqrt{b^{2} -4ac}}{2a}$ to solve for the value of $x$ .

$\begin{matrix} x = \frac{-b ± \sqrt{b^{2} -4ac}}{2a} \\ \Rightarrow x = \frac{-(-4.66) ± \sqrt{{(- 4.66)}^{2} -4*1.33*2.33}}{2*1.33} \\ \Rightarrow x = \frac{4.66 ± \sqrt{21.7156-12.3956}}{2.66} \\ \Rightarrow x = \frac{4.66 ± \sqrt{9.32}}{2.66} \\ \Rightarrow x = \frac{4.66 ±3.05}{2.66} \end{matrix}$