The rate constant of a first order reaction increases from 4 × 10^{−2} to 8 × 10^{−2} when the temperature changes from 27°C to 37°C. Calculate the energy of activation (Ea). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)

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#### Solution

Given:

k_{1} = 4 × 10^{−2}

k_{2} = 8 × 10^{−2}

T_{1} = 300 K

T_{2} = 310 K

Solution:

`log(k_2/k_1)=E_a/(2.303R)[(T_2-T_1)/(T_1T_2)]`

`log((8xx10^(-2)|)/(4xx10^(-2)))=E_a/(2.303R)[(T_2-T_1)/(T_1T_2)]`

`0.301= E_a/(2.303xx 8.314JK^(-1)mol^(-1))[(310-300)/(310xx300)]`

`E_a=(0.301 × 2.303 × 8.314 × 93000)/10`

E_{a} = 53598.5 J

Concept: Temperature Dependence of the Rate of a Reaction

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