Q5P The following reaction has a val... [FREE SOLUTION]

91影视

Chapter 4: Q5P (page 199)

The following reaction has a value of ${螖骋}^{0} =鈥�-2.1鈥塳闯/尘辞濒鈥夆赌�(-0.50鈥夆赌塳肠补濒/尘辞濒)$
${CH}_{3} {叠谤鈥夆赌�+鈥夆赌塇}_{2} 厂鈥夆赌� 鈬� {鈥夆赌塁贬}_{3} 厂贬鈥夆赌�+鈥夆赌塇叠谤$
(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. ${[CH}_{3} 叠谤闭鈥�=鈥�0.4鈥夆赌塎$

${[H}_{2} 厂闭鈥�=鈥�0.4鈥夆赌塎$

${[CH}_{3} 厂贬闭鈥�=鈥夆赌�0.6鈥塎$

$摆贬叠谤闭鈥�=鈥夆赌�0.6鈥塎$

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, $补础鈥夆赌�+鈥夆赌塨叠鈥夆赌� 鈬� 鈥夆赌塩颁鈥夆赌�+鈥夆赌塪顿$ , 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鈥檚 free energy is ${螖骋}^{0} = 鈭� RTln K_{e q}$ . This expression can also be written as data-custom-editor="chemistry" ${螖骋}^{0} = 鈭� 2.303 RT(log K_{e q})$ .

As per given data,

data-custom-editor="chemistry" ${螖骋}^{0} =鈥�-2.1鈥塳闯/尘辞濒鈥夆赌�(-0.50鈥夆赌塳肠补濒/尘辞濒)$

data-custom-editor="chemistry" ${T=25}^{0} 颁鈥�=(25+273)碍鈥夆赌�=鈥�298碍$

data-custom-editor="chemistry" $搁=鈥�8.314鈥塉/碍.尘辞濒鈥夆赌�(耻苍颈惫别谤蝉补濒鈥夆赌塯补蝉鈥夆赌塩辞苍蝉迟补苍迟)$

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

Now,

data-custom-editor="chemistry" $\begin{matrix} {螖骋}^{0} = 鈭� 2.303 RTlog K_{e q} \\ 鈬� log K_{e q} = \frac{{螖骋}^{0}}{鈭� 2.303 RT} \\ 鈬� log K_{e q} = \frac{鈭� 2.1鈥塳闯/ mol 鈥�}{鈭� 2.303 脳鈥�8.314鈥塉/ K . mol 鈥� 脳 298 K} \\ 鈬� log K_{e q} = 鈥� \frac{2100 鈥� J}{5705.848 鈥� J} \\ 鈬� log K_{e q} = 鈥� 0.368 \\ 鈬� K_{e q} = 2.33 \end{matrix}$

b) Set up an ICE table as:

data-custom-editor="chemistry" $\begin{array}{l} {________鈥夆赌夆€夆赌塁贬}_{3} {叠谤鈥夆赌�+鈥夆赌塇}_{2} 厂鈥夆赌� 鈬� {鈥夆赌塁贬}_{3} 厂贬鈥夆赌�+鈥夆赌塇叠谤 \\ 滨苍颈迟颈补濒冲冲冲冲鈥夆赌夆赌夆赌夆赌夆赌夆赌�1惭冲冲冲1惭冲冲冲冲冲0冲冲冲冲冲0 \\ 颁丑补苍驳别冲冲冲冲鈥�-虫冲冲冲冲-虫冲冲冲冲冲冲+虫冲冲冲冲+虫 \\ Equilibrium__(1-x)__(1-x)____x_____x \end{array}$

Now,

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

Here, $补鈥�=1.33鈥�,鈥塨鈥�=鈥�-4.66鈥�,鈥塩鈥�=鈥�2.33$

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

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

Either,

$\begin{matrix} x = \frac{4.66鈥� + 3.05}{2.66} \\ = \frac{7.71}{2.66} \\ = 2.89 鈥�(rejected鈥夆赌塻ince鈥夆赌塿alue鈥夆赌塱s鈥夆赌塯reater鈥夆赌塼han鈥夆赌�1M) \end{matrix}$

$\begin{matrix} or,鈥夆赌墄 = \frac{4.66鈥� 鈭� 3.05}{2.66} \\ = \frac{1.61}{2.66} \\ = 0.6 鈥�(accepted鈥夆赌塻ince鈥夆赌塿alue鈥夆赌塱s鈥夆赌塻maller鈥夆赌塼han鈥夆赌�1M) \end{matrix}$