Chapter 23, Problem 23.81QA

Interpretation Introduction

To calculate:

The equilibrium constant of the reaction

$Hg{\left(methionine\right)}^{+2}+penicillamine \rightleftharpoons Hg{\left(penicillamine\right)}^{+2}+methionine$

Answer to Problem 23.81QA

Solution:

The equilibrium constant of the reaction is $126$.

Explanation of Solution

1) Concept:

The logarithm of formation constant of $Hg{\left(methionine\right)}^{+2}$ is given, and the logarithm of formation constant of $Hg{\left(penicillamine\right)}^{+2}$ is also given. If we reverse the first reaction and add to the second reaction, we get the desired reaction:

$Hg{\left(methionine\right)}^{+2}+penicillamine \rightleftharpoons Hg{\left(penicillamine\right)}^{+2}+methionine$

When we reverse any reaction, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.

If we add two reactions, then the equilibrium constant for the resultant reaction is the product of the equilibrium constants of those two reactions.

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2) Given:

i) $\log K_{f_{Hg{\left(methionine\right)}^{+2}}}=14.2$

ii) $\log K_{f_{Hg{\left(penicillamine\right)}^{+2}}}=16.3$

3) Calculations:

The formation constants are given in logarithm form. Therefore, we have to remove the logarithm to calculate the equilibrium constant.

For $Hg{\left(methionine\right)}^{+2}$, the formation constant is

$\mathrm{ }{K}_{f_{Hg{\left(methionine\right)}^{+2}}}={10}^{14.2}$

$\mathrm{ }{K}_{f_{Hg{\left(methionine\right)}^{+2}}}=1.58489 \times {10}^{14}$

For $Hg{\left(penicillamine\right)}^{+2}$, the formation constant is

$\mathrm{ }{K}_{f_{Hg{\left(penicillamine\right)}^{+2}}}= {10}^{16.3}$

$\mathrm{ }{K}_{f_{Hg{\left(penicillamine\right)}^{+2}}}=1.995 \times {10}^{16}$

Reversing the first reaction, we get

$Hg{\left(methionine\right)}^{+2} \rightleftharpoons H{g}^{+2}+methionine$

$K_{reverse}= \frac{1}{K_{f_{Hg{\left(methionine\right)}^{+2}}}}= \frac{1}{1.58489 \times {10}^{14}}=6.3095\times {10}^{-15}$

Adding this to the second reaction, we get

$Hg{\left(methionine\right)}^{+2} \rightleftharpoons H{g}^{+2}+methionine K= 6.3095\times {10}^{-15}$

$H{g}^{+2}+penicillamine \rightleftharpoons Hg{\left(penicillamine\right)}^{+2} K= 1.995 \times {10}^{16}$

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$Hg{\left(methionine\right)}^{+2}+penicillamine \rightleftharpoons Hg{\left(penicillamine\right)}^{+2}+methionine$

The equilibrium constant of the resultant reaction is the product of the two equilibrium constants:

$K_{eq}=\left(6.3095\times {10}^{-15}\right) \times \left(1.995 \times {10}^{16}\right)$

$K_{eq}=125.87$

Thus, the equilibrium constant of the resultant reaction is $126$.

Conclusion:

The equilibrium constant of the resultant reaction is determined by using the $K_f$ values for the given reactions.