Medium INTEGER +4 / -1 PYQ · JEE Mains 2021

The rate constant of a reaction increases by five times on increase in temperature from 27$^\circ$C to 52$^\circ$C. The value of activation energy in kJ mol$-$1 is _________. (Rounded off to the nearest integer)

[R = 8.314 J K$-$1mol$-$1]

Answer (integer) 52

Solution

T<sub>1</sub> = (273 + 27) = 300 K, T<sub>2</sub> = (273 + 52) = 325 K<br/><br/>Given, temperature coefficient of the reaction,<br/><br/>${\alpha _T} = {{{K_{325}}} \over {{K_{300}}}} = 5$<br/><br/>$$\log {{{K_{{T_2}}}} \over {{K_{{T_1}}}}} = {{{E_a}} \over {2.303R}} \times \left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$$<br/><br/>$$\log {{{K_{325}}} \over {{K_{300}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{{325 - 300} \over {300 \times 325}}} \right)$$<br/><br/>$$\log 5 = {{{E_a}} \over {2.303 \times 8.319}} \times {{25} \over {300 \times 325}}$$<br/><br/>E<sub>a</sub> = 52194.78 J mol<sup>$-$</sup><br/><br/>= 52.194 kJ mol<sup>$-$1</sup><br/><br/>$\simeq$ 52 kJ mol<sup>$-$1</sup>

About this question

Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction

This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Chemical Kinetics are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →