# NEET Chemistry Chemical Kinetics Questions Solved

BOARD

The rate of  a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature. [R=8.314 J K-1 mol -1, log =0.6021]

We know, log $\frac{{K}_{2}}{{K}_{1}}=\frac{{E}_{a}}{2.303R}\left(\frac{{T}_{2}-{T}_{1}}{{T}_{1}{T}_{2}}\right)$

Given : T1 = 293 K,     T2 = 313 K,

Let ${K}_{1}$ = K(consider)  ${K}_{2}$ = 4K,

R = 8.314,     E= ?

As,    log$\frac{4K}{K}=\frac{{E}_{a}}{2.303×8.314}\left(\frac{313-293}{293×313}\right)$

or  log

or  0.6020 $=\frac{{E}_{a}}{19.1471}×\left(\frac{20}{91709}\right)$

$\therefore$ Ea$\frac{0.6020×19.1471×91709}{20}$

= 52854.43 = 52.8 K J mol-1

Difficulty Level:

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