Chapter 9, Problem 9.7P

Interpretation Introduction

(a)

Interpretation:

The ratio of rate constants $k_{\text{A}}/k_{\text{B}}$ at $25\text{°C}$ for two reactions A and B is to be stated.

Concept introduction:

Relative rate is defined as the ratio of rate of a reaction and rate of a standard reaction. The rate constant is numerically equal to the rate under standard condition under which concentration is $1\text{mol}{\text{L}}^{-1}$ for all reactants.

The relationship between relative rate of two reactions A and B under standard condition is shown below.

$\log \left(\frac{k_{\text{A}}}{k_{\text{B}}}\right)=\frac{\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}}{2.303RT}$

Answer to Problem 9.7P

The ratio of rate constants $k_{\text{A}}/k_{\text{B}}$ at $25\text{°C}$ for two reactions A and B is $288.4$.

Explanation of Solution

The ratio of rate constants $k_{\text{A}}/k_{\text{B}}$ at $25\text{°C}$ for two reactions A and B is calculated by the formula given below.

$\log \left(\frac{k_{\text{A}}}{k_{\text{B}}}\right)=\frac{\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}}{2.303RT}$…(1)

Where,

• $\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}$ is difference in standard free energies of activation for reaction B and reaction A.

• $R$ is the gas constant $\left(8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\right)$.

• $T$ is temperature $\left(25\text{°C}\right)$.

The values of $\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}$, $R$ and $T$ are $14\text{kJ}{\text{mol}}^{-1}$, $\left(8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\right)$ and $298\text{K}$ respectively.

Substitute the values of $\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}$, $R$ and $T$ in equation (1).

$\begin{array}{rll}\log \left(\frac{k_{\text{A}}}{k_{\text{B}}}\right)& =& \frac{\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}}{2.303RT}\\ & =& \frac{14\text{kJ}{\text{mol}}^{-1}}{2.303\times 8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 298\text{K}}\\ & =& \frac{14\text{kJ}{\text{mol}}^{-1}}{5705.84\text{J}{\text{mol}}^{-1}}\end{array}$

The conversion of $\text{J}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is shown below.

$\text{1}\text{J}{\text{mol}}^{-1}={\text{10}}^{-3}\text{kJ}{\text{mol}}^{-1}$

Therefore, the conversion of $5705.84\text{J}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is shown below.

$\begin{array}{rll}5705.84\text{J}{\text{mol}}^{-1}& =& 5705.84\times {\text{10}}^{-3}\text{kJ}{\text{mol}}^{-1}\\ & =& 5.70\text{kJ}{\text{mol}}^{-1}\end{array}$

Solve the above equation.

$\begin{array}{rll}\log \left(\frac{k_{\text{A}}}{k_{\text{B}}}\right)& =& \frac{14\text{kJ}{\text{mol}}^{-1}}{5.70\text{k}{\text{Jmol}}^{-1}}\\ \log \left(\frac{k_{\text{A}}}{k_{\text{B}}}\right)& =& 2.46\\ \frac{k_{\text{A}}}{k_{\text{B}}}& =& \text{antilog}\left(2.46\right)\\ & =& 288.4\end{array}$

Therefore, the ratio of rate constants $k_{\text{A}}/k_{\text{B}}$ at $25\text{°C}$ for two reactions A and B is $288.4$.

Conclusion

The ratio of rate constants $k_{\text{A}}/k_{\text{B}}$ at $25\text{°C}$ for two reactions A and B is $288.4$.

Interpretation Introduction

(b)

Interpretation:

The difference in the standard free energies of activation at $25\text{°C}$ of two reactions A and B if reaction B is $450$ times faster than reaction, is to be calculated.

Concept introduction:

Relative rate is defined as the ratio of rate of a reaction and rate of a standard reaction. The rate constant is numerically equal to the rate under standard condition under which concentration is $1\text{mol}{\text{L}}^{-1}$ for all reactants.

The relationship between relative rate of two reactions A and B under standard condition is shown below.

$\log \left(\frac{k_{\text{A}}}{k_{\text{B}}}\right)=\frac{\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}}{2.303RT}$

Answer to Problem 9.7P

The difference in the standard free energies of activation at $25\text{°C}$ of two reactions A and B is $-15.16{\text{kJmol}}^{-1}$ and $\text{Δ}G{°}_{\text{A}}^{+}$ is greater than $\text{Δ}G{°}_{\text{B}}^{+}$.

Explanation of Solution

The formula to calculate relative rate is written below.

$\log \left(\frac{k_{\text{A}}}{k_{\text{B}}}\right)=\frac{\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}}{2.303RT}$…(1)

Rearrange the equation (1) to calculate the value of difference in the standard free energies of activation as shown below.

$\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}=\log \left(\frac{k_{\text{A}}}{k_{\text{B}}}\right)\times 2.303RT$…(2)

Where,

• $k_{\text{B}}$ and $k_{\text{A}}$ is rate constant of B and A.

• $R$ is the gas constant $\left(8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\right)$.

• $T$ is temperature $\left(25\text{°C}\right)$.

The reaction B is $450$ times faster than reaction A. Therefore the value of $\frac{k_{\text{A}}}{k_{\text{B}}}$ is $\frac{1}{450}$.

The values of $R$ and $T$ are $\left(8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\right)$ and $298\text{K}$ respectively.

Substitute the value of $R$ and $T$ in equation (2).

$\begin{array}{rll}\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}& =& \log \left(\frac{k_{\text{A}}}{k_{\text{B}}}\right)\times 2.303RT\\ & =& \log \left(\frac{1}{450}\right)\times 2.303\times \left(8.314{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\times 298\text{K}\\ & =& \log \left(0.0022\right)\times 5705.84{\text{Jmol}}^{-1}\\ & =& \left(-2.657\right)\times 5705.84{\text{Jmol}}^{-1}\end{array}$

$\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}=\left(-15160.42{\text{Jmol}}^{-1}\right)$

The conversion of $\text{J}{\text{mol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is shown below.

$\text{1}\text{J}{\text{mol}}^{-1}={\text{10}}^{-3}\text{kJ}{\text{mol}}^{-1}$

Therefore, the conversion of $-15160.42{\text{Jmol}}^{-1}$ to $\text{kJ}{\text{mol}}^{-1}$ is shown below.

$\begin{array}{rll}-15160.42\text{J}{\text{mol}}^{-1}& =& -15160.42\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}\\ & =& -15.16\text{kJ}{\text{mol}}^{-1}\end{array}$

Solve the above equation.

$\begin{array}{rll}\text{Δ}G{°}_{\text{B}}^{+}-\text{Δ}G{°}_{\text{A}}^{+}& =& \left(-15160.42\text{J}{\text{mol}}^{-1}\right)\\ & =& -15.16\text{kJ}{\text{mol}}^{-1}\end{array}$

The difference in the standard free energies of activation at $25\text{°C}$ of two reactions A and B is $-15.16\text{kJ}{\text{mol}}^{-1}$. The negative sign indicates that $\text{Δ}G{°}_{\text{A}}^{+}$ is greater than $\text{Δ}G{°}_{\text{B}}^{+}$.

Conclusion

The difference in the standard free energies of activation at $25\text{°C}$ of two reactions A and B is $-15.16\text{kJ}{\text{mol}}^{-1}$ and $\text{Δ}G{°}_{\text{A}}^{+}$ is greater than $\text{Δ}G{°}_{\text{B}}^{+}$.