Chapter 25, Problem 25.36AP

Interpretation Introduction

(a)

Interpretation:

The $∆\mathrm{G}°\mathrm{\text{'}}$ for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for $∆\mathrm{G}°\mathrm{\text{'}}$ is negative.

Answer to Problem 25.36AP

$∆\mathrm{G}{°}^{\mathrm{\text{'}}}=-14.0\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

Explanation of Solution

The given reaction is:

$\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{y}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{y}\mathrm{l}-\mathrm{C}\mathrm{o}\mathrm{A}+\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{e}\leftrightharpoons \mathrm{ }\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{y}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{y}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{e}+\mathrm{ }\mathrm{H}\mathrm{S}\mathrm{C}\mathrm{o}\mathrm{A}$

This reaction is the addition of two reactions that are shown below:

$\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{y}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{y}\mathrm{l}-\mathrm{C}\mathrm{o}\mathrm{A}+{\mathrm{H}}_2\mathrm{O}\mathrm{ }\leftrightharpoons \mathrm{R}\mathrm{C}{\mathrm{O}}_2^{-}\mathrm{ }+\mathrm{ }\mathrm{ }\mathrm{H}\mathrm{S}\mathrm{C}\mathrm{o}\mathrm{A}\mathrm{ }+{\mathrm{H}}^{+}\mathrm{ }\mathrm{ }∆\mathrm{G}{°}^{\mathrm{\text{'}}}=-31.4\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$ (1)

$\mathrm{R}\mathrm{C}{\mathrm{O}}_2^{-}+\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{e}+{\mathrm{H}}^{+}\leftrightharpoons \mathrm{ }\mathrm{a}\mathrm{ }\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{y}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{y}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{e}+{\mathrm{H}}_2\mathrm{O}\mathrm{ }∆\mathrm{G}{°}^{\mathrm{\text{'}}}=+17.4\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$ (2)

Thus, the $∆\mathrm{G}°\mathrm{\text{'}}$ for the given reaction is calculated as below:

$\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{y}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{y}\mathrm{l}-\mathrm{C}\mathrm{o}\mathrm{A}+{\mathrm{H}}_2\mathrm{O}\mathrm{ }\leftrightharpoons \mathrm{R}\mathrm{C}{\mathrm{O}}_2^{-}\mathrm{ }+\mathrm{ }\mathrm{ }\mathrm{H}\mathrm{S}\mathrm{C}\mathrm{o}\mathrm{A}\mathrm{ }+{\mathrm{H}}^{+}∆\mathrm{G}{°}^{\mathrm{\text{'}}}=-31.4\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$ (1)

+

$\mathrm{R}\mathrm{C}{\mathrm{O}}_2^{-}+\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{e}+{\mathrm{H}}^{+}\leftrightharpoons \mathrm{ }\mathrm{a}\mathrm{ }\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{y}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{y}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{e}+{\mathrm{H}}_2\mathrm{O}\mathrm{ }∆\mathrm{G}{°}^{\mathrm{\text{'}}}=+17.4\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$ (2)

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$\mathrm{a}\mathrm{ }\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{y}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{y}\mathrm{l}-\mathrm{C}\mathrm{o}\mathrm{A}+\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{e}\leftrightharpoons \mathrm{a}\mathrm{ }\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{y}\mathrm{ }\mathrm{a}\mathrm{c}\mathrm{y}\mathrm{l}\mathrm{ }\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{e}+\mathrm{ }\mathrm{H}\mathrm{S}\mathrm{C}\mathrm{o}\mathrm{A}∆\mathrm{G}{°}^{\mathrm{\text{'}}}=-14.0\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

Conclusion

$∆\mathrm{G}{°}^{\mathrm{\text{'}}}=-14.0\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant (at $pH=7$) for this reaction at $37℃ \left(310 K\right)$.

Concept Introduction:

The relation between $∆\mathrm{G}{°}^{\mathrm{\text{'}}}$ and the equilibrium constant is:

$∆\mathrm{G}{°}^{\mathrm{\text{'}}}=-2.3\mathrm{ }\mathrm{R}\mathrm{T}\log K_{eq}$

Answer to Problem 25.36AP

$K=230 at 37℃ \left(310 K\right)$

Explanation of Solution

The relation between $∆\mathrm{G}{°}^{\mathrm{\text{'}}}$ and the equilibrium constant is:

$∆\mathrm{G}{°}^{\mathrm{\text{'}}}=-2.3\mathrm{ }\mathrm{R}\mathrm{T}\log K_{eq}$

By substituting the value of $T=37℃ \left(310 K\right)$ and $∆\mathrm{G}{°}^{\mathrm{\text{'}}}=-14.0\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$, the value for the equilibrium constant is calculated as below:

$∆\mathrm{G}{°}^{\mathrm{\text{'}}}=-2.3\mathrm{ }\mathrm{R}\mathrm{T}\log K_{eq}$

$-14000\frac{\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}=-2.3\mathrm{ }\times 8.314\frac{\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}·\mathrm{K}}\mathrm{ }\times 310\mathrm{ }\mathrm{K}\times \log K_{eq}$

$\log K_{eq}=\frac{-14000 J/mol}{-2.3\times 8.314 J/mol·K\times 310 K}$

$\log K_{eq}=2.362$

${\mathrm{K}}_{\mathrm{e}\mathrm{q}}=230$

Thus, the equilibrium constant for the temperature $T=37℃=310 K$ is $230$.

Conclusion

${\mathrm{K}}_{\mathrm{e}\mathrm{q}}=230$

Interpretation Introduction

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.

Answer to Problem 25.36AP

This reaction is favorable because the overall conversion is from a less stable thioester to a more stable oxoester.

Explanation of Solution

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Conclusion

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.