Chapter 6, Problem 6D.2E

(a)

Interpretation Introduction

Interpretation:

The half-life time of cyclobutane at $20{}^{o}C$ has to be found.

Concept introduction:

Half-life period:

Half-life is defined as the time taken for amount of the sample to reduce to half of its starting value.  The symbol is ${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$.

$t_{\frac{1}{2}}=\frac{\ln \left(2\right)}{k}$

Half-life for a first order reaction is, $t_{\frac{1}{2}}=\frac{0.693}{k}$

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 $8.314J/K\times mol$
• T represents the absolute temperature
• A represents the frequency factor or collision frequency

Explanation of Solution

Given gas-phase decomposition reaction is,

$C_4{H}_8(g)\underset{}{\to }2C_2{H}_4(g)$

$\begin{array}{l}logA=15.6\\ A=anti\log \left(15.6\right)=3.98\times {10}^{15}\\ E_a=261kJmo{l}^{-1}=2.61\times {10}^{-5}Jmo{l}^{-1}\end{array}$

Arrhenius equation is,

$\begin{array}{l}k=A{e}^{-E_a/RT}\\ T={20}^{o}C=293K\\ k=3.98\times {10}^{15}{e}^{-\left(2.61\times {10}^5/8.314\times 298\right)}\\ k=3.98\times {10}^{15}\times 2.30\times {10}^{-18}\\ k=9.154\times {10}^{-3}{s}^{-1}\end{array}$

Therefore, the half-life is,

$\begin{array}{l}{t}_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{1.17\times {10}^{-31}{s}^{-1}}\\ t_{\frac{1}{2}}=5.923\times {10}^{-32}s\end{array}$

Hence, the cyclobutane half-life time value is $t_{\frac{1}{2}}=5.923\times {10}^{-32}s$.

(b)

Interpretation Introduction

Interpretation:

The half-life time of cyclobutane at $500{}^{o}C$ has to be found.

Concept introduction:

Half-life period:

Half-life is defined as the time taken for amount of the sample to reduce to half of its starting value.  The symbol is ${\text{t}}_{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.}$.

$t_{\frac{1}{2}}=\frac{\ln \left(2\right)}{k}$

Half-life for a first order reaction is, $t_{\frac{1}{2}}=\frac{0.693}{k}$

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 $8.314J/K\times mol$
• T represents the absolute temperature
• A represents the frequency factor or collision frequency

Explanation of Solution

Given gas-phase decomposition reaction is,

$C_4{H}_8(g)\underset{}{\to }2C_2{H}_4(g)$

$\begin{array}{l}logA=15.6\\ A=anti\log \left(15.6\right)=3.98\times {10}^{15}\\ E_a=261kJmo{l}^{-1}=2.61\times {10}^{-5}Jmo{l}^{-1}\end{array}$

Arrhenius equation is,

$\begin{array}{l}k=A{e}^{-E_a/RT}\\ T={500}^{o}C=773K\\ k=3.98\times {10}^{15}{e}^{-\left(2.61\times {10}^5/8.314\times 773\right)}\\ k=2.30\times {10}^{-18}{s}^{-1}\end{array}$

Therefore, the half-life is,

$\begin{array}{l}{t}_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{2.30\times {10}^{-18}{s}^{-1}}\\ t_{\frac{1}{2}}=3.01\times {10}^{-19}s\end{array}$

Hence, the cyclobutane half-life time value is $t_{\frac{1}{2}}=3.01\times {10}^{-19}s$ at $500{}^{o}C$.