Chapter 20, Problem 20.68E

Interpretation Introduction

(a)

Interpretation:

The activation energy for the isomerization reaction is to be predicted.

Concept introduction:

A small packet of energy is known as quanta. Light is emitted in the form of quanta or photons. The Planck’s law gives the relation between the energy and wavelength, frequency and wavenumber.

Answer to Problem 20.68E

The activation energy for the isomerization reaction is $159.6\text{kJ}\cdot {\text{mol}}^{-1}$.

Explanation of Solution

The given isomerization reaction is,

$hv+cis\text{-retinal}\to trans\text{-retinal}$

It is given that the wavelength of least energetic photon is $750\text{nm}$.

To calculate the activation energy of given isomerization reaction the formula used is,

$E=\frac{hc}{\lambda }$

Where,

• $h$ is the Planck’s constant

• $E$ is the energy

• $c$ is the speed of light

• $\lambda$ is the wavelength

Substitute the values of Planck’s constant, speed of light and wavelength of photon in the given formula.

$\begin{array}{l}E=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times \text{3}\times {\text{10}}^8{\text{ms}}^{-1}}{750\times {10}^{-9}\text{m}}\\ E=2.65\times {10}^{-19}\text{J}\end{array}$

Thus, the energy for theleast energetic photon is $2.65\times {10}^{-19}\text{J}$. The activation energy for the isomerization reaction is calculated for one mole of photons as shown below.

$\begin{array}{rll}{E}_A& =& 2.65\times {10}^{-19}\text{J}\times 6.022\times {10}^{23}{\text{mol}}^{-1}\\ E_A& =& 159.6\times {10}^3\text{J}\cdot {\text{mol}}^{-1}\\ E_A& =& 159.6\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Thus, the activation energy for the isomerization reaction is $159.6\text{kJ}\cdot {\text{mol}}^{-1}$.

Conclusion

The activation energy for the isomerization reaction is $159.6\text{kJ}\cdot {\text{mol}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The value of pre-exponential factor for the given isomerization reaction is to be calculated.

Concept introduction:

The Arrhenius equation gives the temperature dependence of reaction rates.

$k=A{e}^{-E_A/RT}$

Where,

• $k$ is the rate constant.

• $A$ is the pre-exponential factor.

• $E_A$ is the activation energy.

• $R$ is the gas constant and its value is $8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

• $T$ is the temperature.

The pre-exponential factor is also known as the frequency factor or the steric factor.

Answer to Problem 20.68E

The value of pre-exponential factor is $2.347\times {10}^{38}{\text{s}}^{-1}$.

Explanation of Solution

The rate constant for the given isomerization reaction is $3\times {10}^{11}{\text{s}}^{-1}$ at $37°\text{C}$. The activation energy for the given reaction is $159.6\text{kJ}\cdot {\text{mol}}^{-1}$.

The Arrhenius equation can be used for the calculation of activation energy. The Arrhenius equation is,

$k=A{e}^{-E_A/RT}$

Where,

• $k$ is the rate constant.

• $A$ is the pre-exponential factor.

• $E_A$ is the activation energy.

• $R$ is the gas constant and its value is $8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

• $T$ is the temperature.

Conversion of temperature in Celsius to Kelvin is done by the formula,

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

Substitute the temperature $\left(37°\text{C}\right)$ in the formula.

$\begin{array}{l}T=37°\text{C}+273\\ T=310\text{K}\end{array}$

Thus, the given temperature in Kelvin is $310\text{K}$.

Substitute the values of activation energy, gas constant, rate constant and temperature.

$\begin{array}{rll}3\times {10}^{11}{\text{s}}^{-1}& =& A{e}^{-159.6\text{kJ}/\text{mol}/8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\times 310\text{K}}\\ 3\times {10}^{11}{\text{s}}^{-1}& =& A{e}^{-159.6\times {10}^3\text{J}/\text{mol}/8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\times 310\text{K}}\\ 3\times {10}^{11}{\text{s}}^{-1}& =& A{e}^{-61.92}\end{array}$

The above equation if further solved to obtain the value of pre-exponential factor as shown below.

$\begin{array}{rll}3\times {10}^{11}{\text{s}}^{-1}& =& A\left(1.278\times {10}^{-27}\right)\\ A& =& \frac{3\times {10}^{11}{\text{s}}^{-1}}{1.278\times {10}^{-27}}\\ A& =& 2.347\times {10}^{38}{\text{s}}^{-1}\end{array}$

The value of pre-exponential factor is $2.347\times {10}^{38}{\text{s}}^{-1}$.

Conclusion

The value of pre-exponential factor is $2.347\times {10}^{38}{\text{s}}^{-1}$.

Interpretation Introduction

(c)

Interpretation:

The value of the rate constant at $-2°\text{C}$ for the given isomerization reaction taking place in ice fish eye is to be predicted.

Concept introduction:

The Arrhenius equation gives the temperature dependence of reaction rates.

$k=A{e}^{-E_A/RT}$

Where,

• $k$ is the rate constant.

• $A$ is the pre-exponential factor.

• $E_A$ is the activation energy.

• $R$ is the gas constant and its value is $8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

• $T$ is the temperature.

The pre-exponential factor is also known as the frequency factor or the steric factor.

Answer to Problem 20.68E

The value of the rate constant at $-2°\text{C}$ for the given isomerization reaction taking place in ice fish eye is $4.04\times {10}^7{\text{s}}^{-1}$.

Explanation of Solution

It is given that the rate constant at $37°\text{C}$$\left(T_1\right)$ is $3\times {10}^{11}{\text{s}}^{-1}$ and value of activation energy is $159.6\text{kJ}\cdot {\text{mol}}^{-1}$.

The form of Arrhenius equation used to calculate the rate constant at different temperature is,

$\ln \frac{k_1}{k_2}=\left(-\frac{E_A}{R}\right)\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$ …(1)

Where,

• $k_1$ and $k_2$ is the rate constant at two different temperatures.

• $E_A$ is the activation energy.

• $R$ is the gas constant and its value is $8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

• $T_1$ and $T_2$ is the are different temperatures.

Conversion of temperature in Celsius to Kelvin is done by the formula,

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

Substitute the temperature $\left(-2°\text{C}\right)$ in the formula.

$\begin{array}{l}T=-2°\text{C}+273\\ T=271\text{K}\end{array}$

Thus, the temperature $\left(T_2\right)$ is $271\text{K}$.

Substitute the values of activation energy, temperatures, the rate constant at $310\text{K}$ and gas constant in equation (1).

$\begin{array}{rll}\ln \frac{3\times {10}^{11}{\text{s}}^{-1}}{k_2}& =& \left(-\frac{159.6\times {10}^3\text{J}/\text{mol}}{8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}}\right)\left(\frac{1}{310\text{K}}-\frac{1}{271\text{K}}\right)\\ \ln \frac{3\times {10}^{11}{\text{s}}^{-1}}{k_2}& =& -19196.5\text{K}\left(\frac{39\text{K}}{271\text{K}\times 310\text{K}}\right)\\ \ln \frac{3\times {10}^{11}{\text{s}}^{-1}}{k_2}& =& 8.912\end{array}$

Take inverse of logarithm on both sides of equation to solve for the value of rate constant at $271\text{K}$ as shown below.

$\begin{array}{rll}\frac{3\times {10}^{11}{\text{s}}^{-1}}{k_2}& =& e^{8.912}\\ \frac{3\times {10}^{11}{\text{s}}^{-1}}{k_2}& =& 7420.487\\ k_2& =& \frac{3\times {10}^{11}{\text{s}}^{-1}}{7420.487}\\ k_2& =& 4.04\times {10}^7{\text{s}}^{-1}\end{array}$

Thus, the value of rate constant at $271\text{K}$ is $4.04\times {10}^7{\text{s}}^{-1}$.

Conclusion

The value of the rate constant at $-2°\text{C}$ for the given isomerization reaction taking place in ice fish eye is $4.04\times {10}^7{\text{s}}^{-1}$.