Question

In Arrhenius's equation for a certain reaction, the value of A and E (activation energy) are $4\times {10}^{13}{\text{s}}^{-1}$ and $92.12{\text{kJ mol}}^{-1}$ respectively. If the reaction is of first order, at what temperature will its half-life period be $11.55$ minutes?$\text{Take}R=8{\text{J mol}}^{-1}{\text{K}}^{-1}\text{and}\log 2=0.3$)

• A. 301.20
• B. 301.21
• C. 301.2

Answer 1

Answer: (A), (B), (C)

Arrhenius’s equation represented as:
$k=A{e}^{-E/RT}$
$lnk=lnA-\frac{E}{RT}$
$2.303\log k=2.303\log A-\frac{E}{RT}$
$\Rightarrow \log k=\log A-\frac{E}{2.303RT}$..........(1)
Given that $A=4\times {10}^{13}{\text{s}}^{-1};E=92.12{\text{kJ mol}}^{-1};t_{1/2}=11.55\times 60\text{s};R=8\times {10}^{-3}{\text{kJ mol}}^{-1}{\text{K}}^{-1}$
For first-order reaction $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{693}{\text{s}}^{-1}$
$\Rightarrow k={10}^{-3}{\text{s}}^{-1}$
From equation (1)
$T=\frac{E}{2.303\times R(\log A-\log k)}$
$T=\frac{92.12}{2.303\times 8\times {10}^{-3}\left[\log \right(4\times {10}^{13})-\log ({10}^{-3}\left)\right]}$
$T=\frac{92.12}{2.303\times 8\times {10}^{-3}\left[2\log \right(2)+13\log 10)+3\log \left(10\right)]}$
$\therefore T=301.2\text{K}$.