Question

The following reaction has a value of${\mathbf{\text{ΔG}}}^{\mathbf{\text{0}}}\mathbf{\text{=\hspace{0.17em}-2.1\hspace{0.17em}kJ/mol\hspace{0.17em}\hspace{0.17em}(-0.50\hspace{0.17em}\hspace{0.17em}kcal/mol)}}$

${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{Br\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}H}}}_{\mathbf{\text{2}}}\mathbf{\text{S\hspace{0.17em}\hspace{0.17em}}}\mathbf{\rightleftarrows }{\mathbf{\text{\hspace{0.17em}\hspace{0.17em}CH}}}_{\mathbf{\text{3}}}\mathbf{\text{SH\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}HBr}}$

(a) Calculate ${\mathit{K}}_{\mathbf{e}\mathbf{q}}$at room temperature${\mathbf{\text{(25}}}^{\mathbf{\text{0}}}\mathbf{\text{C)}}$ for this reaction as written.

(b)Starting with a$\mathbf{\text{1M}}$ solution of ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}\mathbf{\text{Br}}$and ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{S}}$, calculate the final concentrations of all four species at equilibrium.

Short answer

a.$K_{eq}=2.33$

b.${\text{[CH}}_{\text{3}}\text{Br]\hspace{0.17em}=\hspace{0.17em}0.4\hspace{0.17em}\hspace{0.17em}M}$

${\text{[H}}_2\text{S]\hspace{0.17em}=\hspace{0.17em}0.4\hspace{0.17em}\hspace{0.17em}M}$

${\text{[CH}}_{\text{3}}\text{SH]\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0.6\hspace{0.17em}M}$

$\text{[HBr]\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0.6\hspace{0.17em}M}$

Step-by-step solution

Equilibrium constant  (Keq)

Equilibrium constant $\left(K_{eq}\right)$of the reaction governs the equilibrium concentration of the reactants and products. For a general reaction of the type, $\text{aA\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}bB\hspace{0.17em}\hspace{0.17em}}\rightleftarrows \text{\hspace{0.17em}\hspace{0.17em}cC\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}dD}$, the equilibrium constant $\left(K_{eq}\right)$expression can be written as:

$\begin{array}{c}{K}_{eq}=\frac{\text{[products]}}{\text{[reactants]}}\\ \text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}\end{array}$

Position of equilibrium

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

Explanation

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

As per given data,

data-custom-editor="chemistry" ${\text{ΔG}}^{\text{0}}\text{=\hspace{0.17em}-2.1\hspace{0.17em}kJ/mol\hspace{0.17em}\hspace{0.17em}(-0.50\hspace{0.17em}\hspace{0.17em}kcal/mol)}$

data-custom-editor="chemistry" ${\text{T=25}}^{\text{0}}\text{C\hspace{0.17em}=(25+273)K\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}298K}$

data-custom-editor="chemistry" $\text{R=\hspace{0.17em}8.314\hspace{0.17em}J/K.mol\hspace{0.17em}\hspace{0.17em}(universal\hspace{0.17em}\hspace{0.17em}gas\hspace{0.17em}\hspace{0.17em}constant)}$

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

Now,

data-custom-editor="chemistry"

b) Set up an ICE table as:

data-custom-editor="chemistry" $\begin{array}{l}{\text{\_\_\_\_\_\_\_\_\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}CH}}_{\text{3}}{\text{Br\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}\text{S\hspace{0.17em}\hspace{0.17em}}\rightleftarrows {\text{\hspace{0.17em}\hspace{0.17em}CH}}_{\text{3}}\text{SH\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}HBr}\\ \text{Initial\_\_\_\_\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1M\_\_\_1M\_\_\_\_\_0\_\_\_\_\_0}\\ \text{Change\_\_\_\_\hspace{0.17em}-x\_\_\_\_-x\_\_\_\_\_\_+x\_\_\_\_+x}\\ \text{Equilibrium\_\_(1-x)\_\_(1-x)\_\_\_\_x\_\_\_\_\_x}\end{array}$

Now,

$\begin{array}{l}{\mathrm{K}}_{\mathrm{eq}}=\frac{\text{[products]}}{\text{[reactants]}}\\ \Rightarrow 2.33\text{\hspace{0.17em}\hspace{0.17em}=}\frac{{{\text{[CH}}_{\text{3}}\text{SH]}}^1{\text{[HBr]}}^1}{{{\text{[CH}}_{\text{3}}\text{Br]}}^1{{\text{[H}}_{\text{2}}\text{S]}}^1}\\ \Rightarrow 2.33\text{\hspace{0.17em}\hspace{0.17em}=}\frac{\text{x*x}}{\text{(1-x)*(1-x)}}\\ \Rightarrow 2.33=\frac{{\text{x}}^{\text{2}}}{{\text{(1-x)}}^{\text{2}}}\\ \Rightarrow 2.33=\frac{{\text{x}}^{\text{2}}}{{\text{(1}}^2{\text{+x}}^{\text{2}}-2\text{x})}\\ \Rightarrow 2.33*({\text{1}}^2{\text{+x}}^{\text{2}}-2\text{x})={\text{\hspace{0.17em}x}}^{\text{2}}\\ \Rightarrow 2.33+2.33{\text{x}}^{\text{2}}-4.66\text{x}-{\text{x}}^{\text{2}}=0\\ \Rightarrow 1.33{\text{x}}^{\text{2}}-4.66\text{x}+2.33=0\end{array}$

Here, $\text{a\hspace{0.17em}=1.33\hspace{0.17em},\hspace{0.17em}b\hspace{0.17em}=\hspace{0.17em}-4.66\hspace{0.17em},\hspace{0.17em}c\hspace{0.17em}=\hspace{0.17em}2.33}$

Use quadratic formula $\frac{\text{-b\hspace{0.17em}±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}$ to solve for the value of $\mathrm{x}$.

Either,

$\begin{array}{c}\text{x}=\frac{\text{4.66\hspace{0.17em}}+\text{3.05}}{\text{2.66}}\\ =\frac{7.71}{2.66}\\ =2.89\text{\hspace{0.17em}(rejected\hspace{0.17em}\hspace{0.17em}since\hspace{0.17em}\hspace{0.17em}value\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}greater\hspace{0.17em}\hspace{0.17em}than\hspace{0.17em}\hspace{0.17em}1M)}\end{array}$

$\begin{array}{c}\text{or,\hspace{0.17em}\hspace{0.17em}x}=\frac{\text{4.66\hspace{0.17em}}-\text{3.05}}{\text{2.66}}\\ =\frac{1.61}{2.66}\\ =0.6\text{\hspace{0.17em}(accepted\hspace{0.17em}\hspace{0.17em}since\hspace{0.17em}\hspace{0.17em}value\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}smaller\hspace{0.17em}\hspace{0.17em}than\hspace{0.17em}\hspace{0.17em}1M)}\end{array}$

Finally,

${\text{[CH}}_{\text{3}}\text{Br]\hspace{0.17em}=\hspace{0.17em}1-x\hspace{0.17em}=\hspace{0.17em}1-0.6\hspace{0.17em}=\hspace{0.17em}0.4\hspace{0.17em}\hspace{0.17em}M}$

${\text{[H}}_2\text{S]\hspace{0.17em}=\hspace{0.17em}1-x\hspace{0.17em}=\hspace{0.17em}1-0.6\hspace{0.17em}=\hspace{0.17em}0.4\hspace{0.17em}\hspace{0.17em}M}$

${\text{[CH}}_{\text{3}}\text{SH]\hspace{0.17em}=\hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}0.6\hspace{0.17em}M}$

$\text{[HBr]\hspace{0.17em}=\hspace{0.17em}x\hspace{0.17em}=\hspace{0.17em}0.6\hspace{0.17em}M}$