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Chemistry · Ch 6 — Chemical Kinetics

Determination of activation energy

6.7.3

Determination of activation energy

For two different temperatures T1T_1 and T2T_2,

log⁡10k1=log⁡10A−Ea2.303 R T1...(6.21)\log_{10} k_1 = \log_{10} A - \frac{E_a}{2.303\,R\,T_1} \qquad \text{...(6.21)}

log⁡10k2=log⁡10A−Ea2.303 R T2...(6.22)\log_{10} k_2 = \log_{10} A - \frac{E_a}{2.303\,R\,T_2} \qquad \text{...(6.22)}

where k1k_1 and k2k_2 are the rate constants at temperatures T1T_1 and T2T_2 respectively. Subtracting Eq. (6.21) from Eq. (6.22),

log⁡10k2−log⁡10k1=−Ea2.303 R 1T2+Ea2.303 R 1T1\log_{10} k_2 - \log_{10} k_1 = -\frac{E_a}{2.303\,R}\,\frac{1}{T_2} + \frac{E_a}{2.303\,R}\,\frac{1}{T_1}

Hence, log⁡10k2k1=Ea2.303 R(1T1−1T2)=Ea2.303 R(T2−T1T1T2)...(6.23)\text{Hence, } \log_{10}\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right) \qquad \text{...(6.23)} …

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