Chapter 4, Problem 4.28P

Interpretation Introduction

(a)

Interpretation:

The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

$\text{rate}={10}^{-\Delta G{°}^{\ast }/2.3RT}$

Where,

• $\Delta G{°}^{\ast }$ represents the activation energy for reaction.

• $R$ is the gas constant.

• $T$ is the temperature.

Answer to Problem 4.28P

The reaction B is faster than the reaction A by a factor of $429.18$.

Explanation of Solution

It is given that standard free energy of activation of reaction A is $90\text{kJ}{\text{mol}}^{-1}\left(21.5\text{kcal}{\text{mol}}^{-1}\right)$ and for reaction B is $75\text{kJ}{\text{mol}}^{-1}\left(17.9\text{kcal}{\text{mol}}^{-1}\right)$.

The relative rates of two reactions are expressed as,

$\log \left(\frac{{\text{rate}}_A}{{\text{rate}}_B}\right)=\frac{\Delta G{°}_B^{\ast }-\Delta G{°}_A^{\ast }}{2.3RT}$

Where,

• $\Delta G{°}_B^{\ast }$ and $\Delta G{°}_A^{\ast }$ represents the activation energy for reaction A and B.

• $R$ is the gas constant.

• $T$ is the temperature.

Substitute the activation energy for reaction A and B, gas constant and temperature in the given formula.

$\begin{array}{l}\log \left(\frac{{\text{rate}}_A}{{\text{rate}}_B}\right)=\frac{75\text{kJ}{\text{mol}}^{-1}-90\text{kJ}{\text{mol}}^{-1}}{2.3\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(298\text{K}\right)}\\ \log \left(\frac{{\text{rate}}_A}{{\text{rate}}_B}\right)=\frac{-15\text{kJ}{\text{mol}}^{-1}}{5698.41\text{J}}\\ \log \left(\frac{{\text{rate}}_A}{{\text{rate}}_B}\right)=\frac{-15\times {10}^3\text{J}{\text{mol}}^{-1}}{5698.41\text{J}{\text{mol}}^{-1}}\\ \log \left(\frac{{\text{rate}}_A}{{\text{rate}}_B}\right)=-2.632\end{array}$

Inverse of the logarithm is taken on both the sides of the equation.

$\begin{array}{l}\left(\frac{{\text{rate}}_A}{{\text{rate}}_B}\right)=0.00233\\ \left(\frac{{\text{rate}}_B}{{\text{rate}}_A}\right)=429.18\end{array}$

${\text{rate}}_B=429.18\left({\text{rate}}_A\right)$ …(1)

Thus, from the equation (1), the reaction B is faster than the reaction A by a factor of $429.18$.

Conclusion

The reaction B is faster than the reaction A by a factor of $429.18$.

Interpretation Introduction

(b)

Interpretation:

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

$\text{rate}={10}^{-\Delta G{°}^{\ast }/2.3RT}$

Where,

• $\Delta G{°}^{\ast }$ represents the activation energy for reaction.

• $R$ is the gas constant.

• $T$ is the temperature.

Answer to Problem 4.28P

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is $357.6\text{K}$.

Explanation of Solution

It is given that standard free energy of activation of reaction A is $90\text{kJ}{\text{mol}}^{-1}\left(21.5\text{kcal}{\text{mol}}^{-1}\right)$ and for reaction B is $75\text{kJ}{\text{mol}}^{-1}\left(17.9\text{kcal}{\text{mol}}^{-1}\right)$.

The rate of the reaction is expressed as,

$\text{rate}={10}^{-\Delta G{°}^{\ast }/2.3RT}$

Where,

• $\Delta G{°}^{\ast }$ represents the activation energy for reaction.

• $R$ is the gas constant.

• $T$ is the temperature.

The temperature for reaction A at which rate of reaction of A and B are equal is given by $T\text{'}$. The calculation for the $T\text{'}$ is as follows,

$\begin{array}{l}{\text{rate}}_A={\text{rate}}_A\\ {10}^{-\Delta G{°}_A^{\ast }/2.3RT\text{'}}={10}^{-\Delta G{°}_B^{\ast }/2.3RT}\end{array}$

Take log on both sides of equation.

$-\frac{\Delta G{°}_A^{\ast }}{2.3RT\text{'}}=-\frac{\Delta G{°}_B^{\ast }}{2.3RT}$

$\frac{\Delta G{°}_A^{\ast }}{T\text{'}}=\frac{\Delta G{°}_B^{\ast }}{T}$ …(1)

Substitute the activation energy for reaction A and B, and temperature for reaction B in the given equation (2).

$\begin{array}{rll}\frac{90\text{kJ}{\text{mol}}^{-1}}{T\text{'}}& =& \frac{75\text{kJ}{\text{mol}}^{-1}}{298\text{K}}\\ T\text{'}& =& \frac{90\text{kJ}{\text{mol}}^{-1}\times 298\text{K}}{75\text{kJ}{\text{mol}}^{-1}}\\ T\text{'}& =& 357.6\text{K}\end{array}$

Thus, the increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is $357.6\text{K}$.

Conclusion

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is $357.6\text{K}$.