Question

For the reaction, $aA+bB\to cC+dD$ , the plot of $logk$ vs $\frac{1}{T}$ is given below:

The temperature at which the rate constant of the reaction is $10^{-4}{s}^{-1}$ is [Rounded off to the nearest integer) [Given: The rate constant of the reaction is $10^{-5}{s}^{-1}$ at $500K$ ]

Answer 1

Answer: 526

$lo{g}_{10}K=lo{g}_{10}A-\frac{E_a}{2.303RT}$
Slope $=\frac{E_a}{2.303R}=-10000$
$lo{g}_{10}\frac{K_2}{K_1}=\frac{E_a}{2.303R}\times \left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$lo{g}_{10}\frac{10^{-4}}{10^{-5}}=10000\times \left[\frac{1}{500}-\frac{1}{T}\right]$
$1=10000\times \left[\frac{1}{500}-\frac{1}{T}\right]$
$\frac{1}{10000}=\frac{1}{500}-\frac{1}{T}$
$\frac{1}{T}=\frac{1}{500}-\frac{1}{10000}$
$\frac{1}{T}=\frac{20-1}{10000}=\frac{19}{10000}$
$T=\frac{10,000}{19}\Rightarrow 526K$