Q14 / 90 · JEE Main 2017 (08 Apr Online) · Chemistry

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.

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

Solution

log 2=EaR1300-1310 .......(i)

log  2=2EaR 1300-1T ......(ii)

2EaR 1300-1T=EaR 1300-1310

1300+1310=2T

⇒     T=300×310610×2

=304.92

Hence, the temperature of reaction B should be increased from 300 K by 304.92−300=4.92 K.

Hence, the correct answer is option: 'A'.

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