Chapter 4, Problem 4.68AP

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for interconversion of given alkenes is to be calculated. The alkene that is favored at equilibrium is to be stated.

Concept introduction:

The free energy diagram of a reaction is the plot of standard free energy versus reaction coordinate or reaction progress. The products and reactants are placed at their respective free energy. The difference in the free energy of products and reactants is the standard free energy of the reaction.

Answer to Problem 4.68AP

The equilibrium constant for interconversion of given alkenes is $3.50$.

The alkene that is more favorable is shown below.

Explanation of Solution

The given alkenes undergoing interconversion along with their free energy of formation are shown below.

$\begin{array}{l}\Delta G_{\text{f}}^{°}79.0\text{kJ}{\text{mol}}^{-1}75.9\text{kJ}{\text{mol}}^{-1}\\ 18.9\text{kcal}{\text{mol}}^{-1}18.1\text{kcal}{\text{mol}}^{-1}\end{array}$

Figure 1

The free energy change for the interconversion of alkenes is equal to the free energy of formation of the product minus the free energy of formation of the reactant.

$\Delta G^{°}={\left(\Delta G_{\text{f}}^{°}\right)}_{\text{products}}-{\left(\Delta G_{\text{f}}^{°}\right)}_{\text{reactants}}$ …(1)

Substitute the free energy of formation of product alkene and reactant alkene in the equation (1) as shown below.

$\begin{array}{rll}\Delta G^{°}& =& 75.9\text{kJ}{\text{mol}}^{-1}-79.0\text{kJ}{\text{mol}}^{-1}\\ & =& -3.1\text{kJ}{\text{mol}}^{-1}\end{array}$

The Gibbs free energy of the reaction is related to its equilibrium constant by the relation shown below.

$\Delta G^{\circ }=-RT\ln K_{\text{eq}}$ …(2)

Where,

• $R$ is the gas constant.

• $T$ is the temperature.

The value of $R$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$. The value of temperature is $298\text{K}$.

Substitute the value of $R$, $T$, and $\Delta G^{\circ }$ in the equation (2).

$\begin{array}{rll}-3.1\text{kJ}{\text{mol}}^{-1}& =& -\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(298\text{K}\right)\ln K_{\text{eq}}\\ 3.1\text{kJ}{\text{mol}}^{-1}& =& \left(2477.57\text{J}{\text{mol}}^{-1}\right)\ln K_{\text{eq}}\end{array}$

Rearrange above equation to calculate the $K_{\text{eq}}$ as shown below.

$\begin{array}{rll}\ln K_{\text{eq}}& =& \frac{3.1\text{kJ}{\text{mol}}^{-1}}{\left(2477.57\text{J}{\text{mol}}^{-1}\right)}\\ & =& \frac{3.1\times {10}^3\text{J}{\text{mol}}^{-1}}{\left(2477.57\text{J}{\text{mol}}^{-1}\right)}\\ & =& 1.251\end{array}$

Take the antilog on both sides of the equation as shown below.

$\begin{array}{rll}{K}_{\text{eq}}& =& e^{1.251}\\ & =& 3.494\\ & \approx & 3.50\end{array}$

The equilibrium constant for the interconversion of alkenes is $3.50$.

The value of equilibrium constant is high, therefore, the alkene on the product is more favored. This can also be understood from the negative value of Gibbs free energy of the reaction which indicates that the reaction is spontaneous. Therefore, alkene on the product side is more favorable which is shown below.

Figure 2

Conclusion

The equilibrium constant for interconversion of given alkenes is $3.50$.

The alkene that is more favorable is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The information of the rate at which the interconversion is taking place from the equilibrium constant value is to be stated.

Concept introduction:

The equilibrium constant of the reaction gives information about the rate of reaction. The equilibrium constant is the ratio of forward reaction rate constant and backward reaction rate constant.

Answer to Problem 4.68AP

The rate of interconversion of alkene is moderate. The rate of forward reaction is $3.5$ times the rate of backward reaction.

Explanation of Solution

The equilibrium constant in terms of the concentration of reactant alkene and product alkene is shown below.

$K_{\text{eq}}=\frac{\left[\text{Product}\text{alkene}\right]}{\left[\text{Reactant}\text{alkene}\right]}$

The value of the equilibrium constant is $3.50$.

Substitute the value of the equilibrium constant in the above expression.

$\begin{array}{rll}3.50& =& \frac{\left[\text{Product}\text{alkene}\right]}{\left[\text{Reactant}\text{alkene}\right]}\\ 3.50\left[\text{Reactant}\text{alkene}\right]& =& \left[\text{Product}\text{alkene}\right]\\ \left[\text{Product}\text{alkene}\right]& =& 3.50\left[\text{Reactant}\text{alkene}\right]\end{array}$

The concentration of product alkene at equilibrium is only $3.5$ times that of reactant alkene. This is the indication that the rate of interconversion of alkene is moderate or not very much high. The rate of reaction is very high for the equilibrium of the order ${10}^5$ and more.

The equilibrium constant of the reaction is the ratio of rate constant of forward reaction and backward reaction.

Conclusion

The rate of interconversion of alkene is moderate from the value of the equilibrium constant.