Chapter 6, Problem 6.13P

(a)

Interpretation Introduction

Interpretation:

The activation Gibbs energy for the given decomposition reaction of urea has to be estimated.

Concept introduction:

• Activation energy: It is defined as the minimum energy required by the reacting species in order to undergo chemical reaction.
• Activation energy can be represented as $E_a$. The unit of activation energy is $\text{kJ/mol}$.
• Catalyst: The catalyst is a chemical substance that increases the rate of the reaction without participating in the reaction by reducing the activation energy of the reaction.

Arrhenius equation:

• Arrhenius equation is a formula that represents the temperature dependence of reaction rates
• The Arrhenius equation has to be represented as follows $f=e^{-E_a/RT}$
• $E_a$ represents the activation energy and it’s unit is kJ/mol
• R represents the universal gas constant and it has the value of $\text{8}\text{.314 J/K}\text{.mol}$
• T represents the absolute temperature
• f  represents the frequency factor or collision frequency

The activation energy from the given information has to be calculated from the modified Arrhenius equation.

  $\text{ln}\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{2}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

Explanation of Solution

Given pseudo first-order decomposition reaction is,

$CO{\left(N{H}_3\right)}_{\left(aq\right)}+2H_2{O}_{\left(l\right)}\underset{}{\to }2N{H}_4^{+}{}_{\left(aq\right)}+C{O}_3^{2-}{}_{\left(aq\right)}$

Record the given data’s

$\begin{array}{l}{k}_1=1.2\times {10}^{-7}{s}^{-1};T_1={60}^{o}C=60+273=333K\\ k_2=4.6\times {10}^{-7}{s}^{-1};T_2={70}^{o}C=70+273=343K\end{array}$

The activation energy equation is,

$\begin{array}{l}\text{ln}\frac{{\text{k}}_2}{{\text{k}}_1}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_1}\text{-}\frac{\text{1}}{{\text{T}}_2}\right]\\ \left(\text{or}\right)\\ \text{ln}\frac{{\text{k}}_2}{{\text{k}}_1}\text{=}\frac{\Delta \text{H}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_1}\text{-}\frac{\text{1}}{{\text{T}}_2}\right]\end{array}$

$\begin{array}{l}\text{ln}\left(\frac{4.6\times {10}^{-7}}{1.2\times {10}^{-7}}\right)\text{=}\frac{\Delta \text{H}}{8.314}\left[\frac{\text{1}}{333}\text{-}\frac{\text{1}}{343}\right]\\ \Delta \text{H=}\text{127582}\text{.6}\text{J/mol}\\ \Delta \text{H=}\text{127}\text{.582k}\text{J/mol}\\ \Delta \text{H=}1.3\times {10}^2\text{kJ/mol}\end{array}$

Next we calculated the entropy $\Delta S$

Activation complex theory is,

$A=\left(\frac{K_{B}T}{h}\right)e^{\left(\frac{\Delta S}{R}\right)}$

$\begin{array}{c}{k}_B=1.38\times {10}^{-23}J/K,h=6.624\times {10}^{-34}J/S\\ T=333K,R=8.314J/mol\\ A=2.9\times {10}^{13}{s}^{-1}\end{array}$

$\begin{array}{l}2.9\times {10}^{13}=\left(\frac{1.38\times {10}^{-23}\times 333}{6.62\times {10}^{-34}}\right)e\left(\frac{\Delta S}{8.314}\right)\\ \Delta S=1.2\times {10}^{1}J/K\end{array}$

According to standard free energy equation is,

$\Delta G=\Delta H-T\Delta S$

Calculated values are

$\begin{array}{l}\Delta H=1.3\times {10}^{2}KJ/mol\\ \Delta S=1.2\times {10}^{1}J/K\\ =1.2\times {10}^{-2}kJ/K\left[\because 1J={10}^{-3}KJ\right]\\ T=333\end{array}$

Hence,

$\begin{array}{l}\Delta G=1.3\times {10}^2-333\times \left(1.2\times {10}^{-2}\right)\\ \Delta G=1.3\times {10}^{2}KJ/mol\end{array}$

Therefore, the Gibbs energy is $\Delta G=1.3\times {10}^{2}KJ/mol$

(b)

Interpretation Introduction

Interpretation:

The entropy of activation energy for the given decomposition reaction of urea has to be calculated.

Concept introduction:

• Activation energy: It is defined as the minimum energy required by the reacting species in order to undergo chemical reaction.
• Activation energy can be represented as $E_a$. The unit of activation energy is $\text{kJ/mol}$.
• Catalyst: The catalyst is a chemical substance that increases the rate of the reaction without participating in the reaction by reducing the activation energy of the reaction.

Arrhenius equation:

• Arrhenius equation is a formula that represents the temperature dependence of reaction rates
• The Arrhenius equation has to be represented as follows

  $K=A{e}^{-E_a/RT}$

• $E_a$ represents the activation energy and it’s unit is kJ/mol
• R represents the universal gas constant and it has the value of $\text{8}\text{.314 J/K}\text{.mol}$
• T represents the absolute temperature
• f  represents the frequency factor or collision frequency

Explanation of Solution

Given pseudo first-order decomposition reaction is,

$CO{\left(N{H}_3\right)}_{\left(aq\right)}+2H_2{O}_{\left(l\right)}\underset{}{\to }2N{H}_4^{+}{}_{\left(aq\right)}+C{O}_3^{2-}{}_{\left(aq\right)}$

The activation energy equation is,

$K=A{e}^{-E_a/RT}$

Record the given data’s

$\begin{array}{l}{k}_1=1.2\times {10}^{-7}{s}^{-1};T={60}^{o}C=60+273=333K\\ \Delta H=1.3\times {10}^{2}kJ/mol;R=8.314\times {10}^{-3}KJ/mol-k\end{array}$

These values are plugging above activation energy equation.

$\begin{array}{l}1.2\times {10}^{-7}=A\times e^{-\frac{1.3\times {10}^{-3}}{8.314\times {10}^{-3}\times 333}}\\ A=2.9\times {10}^{13}{s}^{-1}\end{array}$

Calculation for entropy of activation,

$A=\left(\frac{K_{B}T}{h}\right)e^{\left(\frac{\Delta S}{R}\right)}$

$\begin{array}{c}{k}_B=1.38\times {10}^{-23}J/K,h=6.624\times {10}^{-34}J/S\\ T=333K,R=8.314J/mol\\ A=2.9\times {10}^{13}{s}^{-1}\end{array}$

$\begin{array}{l}2.9\times {10}^{13}=\left(\frac{1.38\times {10}^{-23}\times 333}{6.62\times {10}^{-34}}\right)e\left(\frac{\Delta S}{8.314}\right)\\ \Delta S=1.2\times {10}^{1}J/K\end{array}$

Therefore, the entropy of activation is $\Delta S=\underset{\_}{1.2\times {10}^{1}J/K}$