Question

For a reaction consider the plot of ln k versus $1/T$ given in the figure. If the rate constant of this reaction at $400$K is ${10}^{-5}$ $s^{-1}$, then the rate constant at $500$K is:

• A. $2\times {10}^{-4}{s}^{-1}$
• B. ${10}^{-4}{s}^{-1}$
• C. ${10}^{-6}{s}^{-1}$
• D. $4\times {10}^{-4}{s}^{-1}$

Answer 1

Answer: (B)

${10}^{-4}{s}^{-1}$

We know,

$ln\frac{K_2}{K_1}=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

Given that,

$K_1={10}^{-5};T_1=400K;T_2=500K;-\frac{E_a}{R}=-4606K$

Now,

$2.303log\frac{K_2}{{10}^{-5}}=4606[\frac{1}{400}-\frac{1}{500}]$

$\Rightarrow K_2={10}^{-4}{s}^{-1}$.