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NEET CHEMISTRYEasy

Activation energy EaE_a and rate constant (k1k_1 and k2k_2) of a chemical reaction at two different temperatures (T1T_1 and T2T_2) are related by:

A

lnk2k1=EaR(1T21T1)\ln \frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

B

lnk2k1=EaR(1T2+1T1)\ln \frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2} + \frac{1}{T_1}\right)

C

lnk2k1=EaR(1T21T1)\ln \frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

D

lnk2k1=EaR(1T11T2)\ln \frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Step-by-Step Solution

According to the Arrhenius equation, the rate constant kk is given by k=AeEa/RTk = A e^{-E_a/RT}. Taking the natural logarithm on both sides yields lnk=EaRT+lnA\ln k = -\frac{E_a}{RT} + \ln A. For two different temperatures T1T_1 and T2T_2, the equations are: lnk1=EaRT1+lnA\ln k_1 = -\frac{E_a}{RT_1} + \ln A lnk2=EaRT2+lnA\ln k_2 = -\frac{E_a}{RT_2} + \ln A Subtracting the first equation from the second gives: lnk2lnk1=EaRT2(EaRT1)\ln k_2 - \ln k_1 = -\frac{E_a}{RT_2} - \left(-\frac{E_a}{RT_1}\right) lnk2k1=EaR(1T21T1)\ln \frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) This is also commonly written as lnk2k1=EaR(1T11T2)\ln \frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right).

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