Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The rate constant for the first order decomposition of $H_{2}O_{2}$ is given by the following equation :

$log\; k=14.2-\frac{1.0\times 10^{4}K}{T}$

Calculate $E_{a}$ for this reaction and rate constant $k$ if its half -life period be 200 minutes. (Given : $R=8.314\; JK^{-1}mol^{-1}$ )

Answers (1)

$E_{a}=?\; \; \; \; \; \; k=?$

$R=8.314\; JK^{-1}mol^{-1}$

$t_{\frac{1}{2}}=200\; min$

$log\; k=log \; A-\frac{E_{a}}{2.303\; RT}$

$\frac{E_{a}}{2.303\; RT}=\frac{1\times 10^{4}}{T}$

$E_{a}=1\times 10^{4}\times 2.303\; R$

$=2.303\times 10^{4}\times 8.314$

$=191471.42\; J\; mol^{-1}$

$t_{\frac{1}{2}}=200\; min$

$K=\frac{0.693}{T_{\frac{1}{2}}}=\frac{0.693}{200}$

$=0.003465\; min^{-1}$

$=0.2079\; sec^{-1}$

Posted by

Sumit Saini

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE revaluation payment system faces malicious attack; IIT experts, PSU banks step in

anam-khan

From next year, CBSE Class 12 answer sheets on Digilocker: Education ministry

anam-khan

CBSE Class 12 students can apply for verification and re-evaluation from June 1

Latest Question