Question

The first order rate constant for a certain reaction increases from $1.667\times {10}^{-6}$ $s^{-1}$ at ${727}^{o}C$ to $1.667\times {10}^{-4}$ $s^{-1}$ at ${1571}^{o}C$. The rate constant at ${1150}^{o}C$, assuming constancy of activation energy over the given temperature range is
[Given:$\log 19.9-1.299$]

• A. $3.911\times {10}^{-5}{s}^{-1}$
• B. $1.139\times {10}^{-5}{s}^{-1}$
• C. $3.318\times {10}^{-5}{s}^{-1}$
• D. $1.193\times {10}^{-5}{s}^{-1}$

Answer 1

Answer: (C)

$3.318\times {10}^{-5}{s}^{-1}$

According to Arrehnius equation,

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
$2.303\log \frac{k_2}{k_1}=\frac{E_a}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
$2.303\log \left[\frac{1.667\times {10}^{-4}}{1.667\times {10}^{-6}}\right]=-\frac{E_a}{R}[\frac{1}{1844}-\frac{1}{1000}]$
$2.303\times 2=\frac{E_a}{R}\times \frac{844}{1844\times 1000}.......\left(1\right)$
$\therefore \frac{E_a}{R}=\frac{4.606\times 1844\times 1000}{844}$
$2.303\log \left[\frac{k_3}{1.667\times {10}^{-6}}\right]=\frac{E_a}{R}\times \frac{1423-1000}{1423\times 1000}$
$=\frac{E_a}{R}\times \frac{423}{1423\times 1000}........\left(2\right)$
Dividing equation $\left(2\right)$ by equation $\left(1\right)$
$\log \left[\frac{k_3}{1.667\times {10}^{-6}}\right]=\frac{423}{1423\times 1000}\times \frac{1844\times 100}{844}$
$\therefore \log \left[\frac{k_3}{1.667\times {10}^{-6}}\right]=2\times \frac{423\times 1844}{1423\times 844}=1.299$
On taking Antilog, $k_3=19.9$
$\therefore k_3=19.9\times 1.667\times {10}^{-6}=3.318\times {10}^{-5}{s}^{-1}$