Verified Solution • 2025–26 Session moderate

QJEEM17Ch51c6: JEE Main 2017 Chemistry - Class 12 JEEM17Ch51c6 Solution: The rate of a reaction A doubles on increasing the temperature…

The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be Increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.

Multiple Choice Options

  • A 9.84 K
  • B 4.92 K
  • C 2.45 K
  • D 19.67 K

Step-by-Step Verified Solution

For reaction A, T1 = 300 K, T2 = 310 K, k2 = 2 k1

log⁡k2k1=Ea12.303R[1T1−1T2]\log {{{k_2}} \over {{k_1}}} = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]

∴\therefore log⁡2k1k1=log⁡2=Ea12.303R[1300−1310]\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right] ...(1)

For reaction B, T1 = 300 K, T2 = ?, k2 = 2k1, Ea2 = 2Ea1

log⁡2k1k1=log⁡2=2Ea12.303R[1300−1T2]\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right]

From eq. (i) and (ii), we get

2Ea12.303R[1300−1T2]=Ea12.303R[1300−1310]{{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]

⇒\Rightarrow 2[1300−1T2]=[1300−1310]2\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = \left[ {{1 \over {300}} - {1 \over {310}}} \right]

⇒\Rightarrow T2 = 300×310610{{{300 \times 310} \over {610}}} ×\times 2 = 304.92 K

∴\therefore Increased temperature = (304.92 – 300) = 4.92 K

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring