Question

The rate of a reaction doubles when temperature changes from 27oc to 37oc. The activation energy is

Answer 1

Answer:

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The rate of a particular reaction doubles when temperature changes from 270 C to 370 C. Calculate the energy of activation of the reaction.

We are given that:

When T1 = 27 + 273 = 300 K

Let k1 = k

When T2 = 37 + 273 = 310 K

k2 = 2 k

Substituting these values the equation:

Log (k2 / k1)

= Ea / 2.303 R {(T2 – T1) / T1 T2}

We will get:

Log (2 k / k) = Ea / 2.303 x 8.314 {(310 – 300) / 300 x 310}

Log 2 = Ea / 2.303 x 8.314 x (10 / 300 x 310)

Ea = 53598.6 J mol-1

Ea = 53.6 kJ mol-1

Explanation: