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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 - CBSE (Science) Class 12 - Chemistry

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Question

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)

Solution

Given:

k1 = 4 × 10−2

k2 = 8 × 10−2

T1 = 300 K

T2 = 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`

Ea = 53598.5 J

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Solution 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 Concept: Temperature Dependence of the Rate of a Reaction.
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