Chapter 4, Problem 4.29P

Interpretation Introduction

(a)

Interpretation:

The standard free energy of activation of reaction B is to be calculated.

Concept introduction:

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

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

Where,

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

• $R$ is the gas constant.

• $T$ is the temperature.

Answer to Problem 4.29P

The standard free energy of activation for reaction B is $55.8\text{kJ}{\text{mol}}^{-1}$.

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 reaction B is one million times faster than the reaction A at the same temperature.

The relative rates of two reactions are expressed as,

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

Where,

• $\text{Δ}G{°}_B^{\ast }$ and $\text{Δ}G{°}_A^{\ast }$ 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, the relative rate of A and B, gas constant and temperature in the given formula.

$\begin{array}{rll}\log \left(\frac{1}{{10}^6}\right)& =& \frac{\text{Δ}G{°}_B^{\ast }-90\text{kJ}{\text{mol}}^{-1}}{2.3\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(298\text{K}\right)}\\ \log \left({10}^{-6}\right)& =& \frac{\text{Δ}G{°}_B^{\ast }-90\text{kJ}{\text{mol}}^{-1}}{5698.41\text{J}{\text{mol}}^{-1}}\\ -6& =& \frac{\text{Δ}G{°}_B^{\ast }-90\text{kJ}{\text{mol}}^{-1}}{5698.41\text{J}{\text{mol}}^{-1}}\end{array}$

Rearrange the above equation for the calculation of activation energy of B as shown below.

$\begin{array}{l}\text{Δ}G{°}_B^{\ast }=-6\times 5698.41\text{J}{\text{mol}}^{-1}+90\text{kJ}{\text{mol}}^{-1}\\ \text{Δ}G{°}_B^{\ast }=-6\times 5.7\text{kJ}{\text{mol}}^{-1}+90\text{kJ}{\text{mol}}^{-1}\\ \text{Δ}G{°}_B^{\ast }=55.8\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, standard free energy of activation for reaction B is $55.8\text{kJ}{\text{mol}}^{-1}$.

Conclusion

The standard free energy of activation for reaction B is $55.8\text{kJ}{\text{mol}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The reaction free energy diagram for the two reactions showing the two values of $\text{Δ}G{°}^{\ast }$ to scale is to be drawn.

Concept introduction:

The transition state is formed during the conversion of reactants into products in the chemical reaction. In an energy level diagram, it corresponds to the high potential energy along the y-axis. In this state, the dashed bond implies that bonds are partially broken and partially formed in the reaction.

Answer to Problem 4.29P

The reaction free-energy diagram for the reaction A showing the two values of $\text{Δ}G{°}^{\ast }$ to scale is,

The reaction free-energy diagram for the reaction B showing the two values of $\text{Δ}G{°}^{\ast }$ to scale is,

Explanation of Solution

It is given that standard free energy of activation of reaction A is $90\text{kJ}{\text{mol}}^{-1}$ and for reaction B is $55.8\text{kJ}{\text{mol}}^{-1}$. The products of each reaction are $10\text{kJ}{\text{mol}}^{-1}$ more stable than the reactants. This indicates that the energy of the product is $10\text{kJ}{\text{mol}}^{-1}$ less than that of the reactants.

The reaction free-energy diagram for the reaction A showing the two values of $\text{Δ}G{°}^{\ast }$ to scale is,

Figure 1

This diagram represents the plot between standard free energy and the reaction coordinates. The point at which the energy is maximum represents the transition state of the reaction.

The reaction free energy diagram for the reaction B showing the two values of $\text{Δ}G{°}^{\ast }$ to scale is,

Figure 2

Conclusion

The reaction free-energy diagram for the two reactions showing the two values of $\text{Δ}G{°}^{\ast }$ to scale is shown in Figure 1 and Figure 2.

Interpretation Introduction

(c)

Interpretation:

The standard free energy of activation of the reverse reaction in both reactions A and B is to be calculated.

Concept introduction:

The free energy diagram represents the plot between standard free energy and the reaction coordinates. The point at which the energy is maximum represents the transition state of the reaction.

Answer to Problem 4.29P

The standard free energy of activation of the reverse reaction in both reactions A and B is $100\text{kJ}{\text{mol}}^{-1}$ and $65.8\text{kJ}{\text{mol}}^{-1}$ respectively.

Explanation of Solution

It is given that standard free energy of activation of reaction A is $90\text{kJ}{\text{mol}}^{-1}$ and for reaction B is $55.8\text{kJ}{\text{mol}}^{-1}$.

The products of each reaction are $10\text{kJ}{\text{mol}}^{-1}$ more stable than the reactants. This indicates that the energy of the product is $10\text{kJ}{\text{mol}}^{-1}$ less than that of the reactants. Thus, standard free energy of activation of the reverse reaction is $10\text{kJ}{\text{mol}}^{-1}$ more than that for the forward reaction.

The standard free energy of activation of the reverse reaction A is,

$\begin{array}{l}\text{Δ}G{°}_A^{\ast }\left(\text{reverse}\right)=90\text{kJ}{\text{mol}}^{-1}+10\text{kJ}{\text{mol}}^{-1}\\ \text{Δ}G{°}_A^{\ast }\left(\text{reverse}\right)=100\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the standard free energy of activation of the reverse reaction A is $100\text{kJ}{\text{mol}}^{-1}$.

The standard free energy of activation of the reverse reaction B is,

$\begin{array}{l}\text{Δ}G{°}_A^{\ast }\left(\text{reverse}\right)=55.8\text{kJ}{\text{mol}}^{-1}+10\text{kJ}{\text{mol}}^{-1}\\ \text{Δ}G{°}_A^{\ast }\left(\text{reverse}\right)=65.8\text{kJ}{\text{mol}}^{-1}\end{array}$

Thus, the standard free energy of activation of the reverse reaction A is $65.8\text{kJ}{\text{mol}}^{-1}$.

Conclusion

The standard free energy of activation of the reverse reaction in both reactions A and B is $100\text{kJ}{\text{mol}}^{-1}$ and $65.8\text{kJ}{\text{mol}}^{-1}$ respectively.