Chapter 3, Problem 3.49AP

Interpretation Introduction

(a)

Interpretation:

The standard free-energy change at $\text{25}°\text{C}$ for ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+\text{H}-\text{CN}\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}+{}^{-}\text{CN}$ reaction is to be predicted.

Concept introduction:

The relation between change in standard free energy of ionization $\text{Δ}G°$ and equilibrium constant, $K_{\text{eq}}$ is given by an expression shown below.

$\text{Δ}G°=-2.3RT\log K_{\text{eq}}$

Where,

• $R$ is the Gas constant.

• $T$ is the absolute temperature in Kelvin.

• $K_{\text{eq}}$ is the equilibrium constant.

Answer to Problem 3.49AP

The standard free-energy change at $\text{25}°\text{C}$ for ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+\text{H}-\text{CN}\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}+{}^{-}\text{CN}$ reaction is $62.17\text{kJ}{\text{mol}}^{-1}$.

Explanation of Solution

The standard free-energy change is given by the formula shown below.

$\text{Δ}G°=-2.3RT\log K_{\text{eq}}$…(1)

Where,

• $R$ is the Gas constant. The value of $R$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

• $T$ is the absolute temperature in Kelvin.

• $K_{\text{eq}}$ is the equilibrium constant.

The calculated value of $K_{\text{eq}}$ for ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+\text{H}-\text{CN}\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}+{}^{-}\text{CN}$ reaction is $10.91$.

The temperature is given as $\text{25}°\text{C}$. Conversion of temperature in Celsius to Kelvin is shown below.

$T\text{(K)}=T\text{(}°\text{C)}+273\text{K}$…(2)

Substitute the temperature $\left(\text{25}\text{.0}°\text{C}\right)$ in the equation (2).

$\begin{array}{l}T=\text{25}\text{.0}°\text{C}+273\\ T=298\text{K}\end{array}$

Substitute $K_{\text{eq}}$ value, Gas constant and temperature in the equation (1).

$\begin{array}{rll}\text{Δ}G°& =& 2.3\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(298\text{K}\right)\left(10.91\right)\\ & =& 62169.71\text{J}{\text{mol}}^{-1}\times \frac{1\text{J}}{1000\text{kJ}}\\ & =& 62.17\text{kJ}{\text{mol}}^{-1}\end{array}$

Therefore, the standard free-energy change at $\text{25}°\text{C}$ ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+\text{H}-\text{CN}\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}+{}^{-}\text{CN}$ reaction 3.48 is $62.17\text{kJ}{\text{mol}}^{-1}$.

Conclusion

The standard free-energy change at $\text{25}°\text{C}$ for ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+\text{H}-\text{CN}\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}+{}^{-}\text{CN}$ reaction is $62.17\text{kJ}{\text{mol}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The standard free-energy change at $\text{25}°\text{C}$ for ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{S}-\text{H}+{}^{-}\text{OH}⥂{\text{CH}}_{\text{3}}{\text{CH}}_2{\text{S}}^{-}+{\text{H}}_{\text{2}}\text{O}$ reaction is to be predicted.

Concept introduction:

The relation between change in standard free energy of ionization $\text{Δ}G°$ and equilibrium constant, $K_{\text{eq}}$ is given by an expression shown below.

$\text{Δ}G°=-2.3RT\log K_{\text{eq}}$

Where,

• $R$ is the Gas constant.

• $T$ is the absolute temperature in Kelvin.

• $K_{\text{eq}}$ is the equilibrium constant.

Answer to Problem 3.49AP

The standard free-energy change at $\text{25}°\text{C}$ for ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{S}-\text{H}+{}^{-}\text{OH}⥂{\text{CH}}_{\text{3}}{\text{CH}}_2{\text{S}}^{-}+{\text{H}}_{\text{2}}\text{O}$ reaction is $178.132\text{kJ}{\text{mol}}^{-1}$.

Explanation of Solution

The standard free-energy change is given by the formula shown below.

$\text{Δ}G°=-2.3RT\log K_{\text{eq}}$…(1)

Where,

• $R$ is the Gas constant. The value of $R$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

• $T$ is the absolute temperature in Kelvin.

• $K_{\text{eq}}$ is the equilibrium constant.

The calculated value of $K_{\text{eq}}$ for ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{S}-\text{H}+{}^{-}\text{OH}⥂{\text{CH}}_{\text{3}}{\text{CH}}_2{\text{S}}^{-}+{\text{H}}_{\text{2}}\text{O}$ reaction is $31.26$.

The temperature is given as $\text{25}°\text{C}$. Conversion of temperature in Celsius to Kelvin is shown below.

$T\text{(K)}=T\text{(}°\text{C)}+273\text{K}$…(2)

Substitute the temperature $\left(\text{25}\text{.0}°\text{C}\right)$ in the equation (2).

$\begin{array}{l}T=\text{25}\text{.0}°\text{C}+273\\ T=298\text{K}\end{array}$

Substitute $K_{\text{eq}}$ value, Gas constant and temperature in the equation (1).

$\begin{array}{rll}\text{Δ}G°& =& 2.3\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(298\text{K}\right)\left(31.26\right)\\ & =& 178132.472\text{J}{\text{mol}}^{-1}\times \frac{1\text{J}}{1000\text{kJ}}\\ & =& 178.132\text{kJ}{\text{mol}}^{-1}\end{array}$

Therefore, the standard free-energy change at $\text{25}°\text{C}$ for ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{S}-\text{H}+{}^{-}\text{OH}⥂{\text{CH}}_{\text{3}}{\text{CH}}_2{\text{S}}^{-}+{\text{H}}_{\text{2}}\text{O}$ reaction is $178.132\text{kJ}{\text{mol}}^{-1}$.

Conclusion

The standard free-energy change at $\text{25}°\text{C}$ for ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{S}-\text{H}+{}^{-}\text{OH}⥂{\text{CH}}_{\text{3}}{\text{CH}}_2{\text{S}}^{-}+{\text{H}}_{\text{2}}\text{O}$ reaction is $178.132\text{kJ}{\text{mol}}^{-1}$.

Interpretation Introduction

(c)

Interpretation:

The concentration of each species at equilibrium for ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+\text{H}-\text{CN}\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}+{}^{-}\text{CN}$ reaction is to be predicted.

Concept introduction:

Equilibrium constant of a reaction is expressed by the ratio of concentration of product species raised to the power of their stoichiometric coefficients to the concentration of reactant species raised to the power of their stoichiometric coefficients.

Answer to Problem 3.49AP

The concentration of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}$ and ${}^{-}\text{CN}$ at equilibrium is $0.9768M$ and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ and $\text{HCN}$ is $0.0232M$ respectively.

Explanation of Solution

The reaction (a) in Problem 3.48 is given as shown below.

${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+\text{H}-\text{CN}\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}+{}^{-}\text{CN}$

In the above reaction, ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ is a base while the ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}$ is the conjugate acid. $\text{H}-\text{CN}$ is an acid while ${}^{-}\text{CN}$ is the conjugate base.

The initial concentration of ${\left({\text{CH}}_{\text{3}}\right)}_3\text{N}$ and $\text{HCN}$ are given as zero and ${\left({\text{CH}}_{\text{3}}\right)}_3\stackrel{+}{\text{N}}\text{H}$ and ${}^{-}\text{CN}$ are given as $0.1M$.

Let the concentration consumed by each reactant be $x$. The reaction will go on backward direction; the concentration of each species for reversible reaction is shown below.

$\begin{array}{l}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}+\text{H}-\text{CN}\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}+{}^{-}\text{CN}\\ \text{Initial conc}\text{.(mol) }000.10.\text{1}\\ \text{Change in conc}\text{.(mol) }+x+x-x-x\\ \text{Conc}\text{. at equilibrium(mol) }+x+x0.1-x0.1-x\end{array}$

The equilibrium constant for the reversible reaction is written as shown below.

$K_{\text{eq}}=\frac{\left[{\left({\text{CH}}_{\text{3}}\right)}_3\stackrel{+}{\text{N}}\text{H}\right]\left[{}^{-}\text{CN}\right]}{\left[{\left({\text{CH}}_{\text{3}}\right)}_3\text{N}\right]\left[\text{HCN}\right]}$…(1)

The calculated value of $K_{\text{eq}}$ for reaction (a) in Problem 3.48 is $10.91$.

Substitute the concentration of each species at equilibrium in equation (1).

$\begin{array}{rll}10.91& =& \frac{\left(0.1-x\right)\left(0.1-x\right)}{\left(x\right)\left(x\right)}\\ 10.91x^2& =& 0.01-0.2x+x^2\\ 10.91x^2-x^2+0.2x-0.01& =& 0\\ 9.91x^2+0.2x-0.01& =& 0\end{array}$

Using the quadratic formula, the value of $x$ is calculated as follows.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute the value of $a=9.91$, $b=0.2$ and $c=-0.01$ in the above formula.

$\begin{array}{rll}x& =& \frac{-0.2\pm \sqrt{{\left(0.2\right)}^2-4\times 9.91\times \left(-0.01\right)}}{2\times 9.91}\\ & =& \frac{-0.2\pm 0.66}{19.82}\end{array}$

Therefore, the two possible values of $x$ are shown below.

$\begin{array}{l}x=\frac{-0.2+0.66}{19.82}=0.0232\\ x=\frac{-0.2-0.66}{19.82}=-0.0434\end{array}$

Therefore, the value of $x$ is $0.0232$.

The concentration of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}$ at equilibrium is calculated as shown below.

$\begin{array}{rll}\left[{\left({\text{CH}}_{\text{3}}\right)}_3\stackrel{+}{\text{N}}\text{H}\right]& =& 1-x\\ & =& 1-0.0232\\ & =& 0.9768M\end{array}$

The concentration of ${}^{-}\text{CN}$ at equilibrium is calculated as shown below.

$\begin{array}{rll}\left[{}^{-}\text{CN}\right]& =& 1-x\\ & =& 1-0.0232\\ & =& 0.9768M\end{array}$

The concentration of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ at equilibrium is calculated as shown below.

$\begin{array}{rll}\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}\right]& =& x\\ & =& 0.0232M\end{array}$

The concentration of $\text{HCN}$ at equilibrium is calculated as shown below.

$\begin{array}{rll}\left[\text{HCN}\right]& =& x\\ & =& 0.0232M\end{array}$

Therefore, the concentration of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}$ and ${}^{-}\text{CN}$ at equilibrium is $0.9768M$ and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ and $\text{HCN}$ is $0.0232M$ respectively.

Conclusion

The concentration of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\stackrel{+}{\text{N}}\text{H}$ and ${}^{-}\text{CN}$ at equilibrium is $0.9768M$ and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ and $\text{HCN}$ is $0.0232M$ respectively.