The rate of a particular reaction doubles when temperature changes from 27°C to 37°C. Calculate the energy of activation of such a reaction.

Solution

When $T_{1} = 27 ° C = 300 k k_{1} = k (say)$

when, $T_{2} = 37 ° C = 310 K k_{2} = 2 k$

Substituting these values in the equation

$\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
We get

$\log \frac{2 k}{k} = \frac{E_{a}}{2.303 \times 8.314} \times \frac{310 - 300}{300 \times 310}$

or

$\log 2 = \frac{E_{a}}{2.303 \times 8.314} \times \frac{10}{300 \times 310}$

or $E_{a} = 53.6 kJ {mol}^{- 1}$

Some More Questions From Chemical Kinetics Chapter

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Chemical Kinetics

The rate of a particular reaction doubles when temperature changes from 27°C to 37°C. Calculate the energy of activation of such a reaction.

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Chemical Kinetics

The rate of a particular reaction doubles when temperature changes from 27°C to 37°C. Calculate the energy of activation of such a reaction.

Some More Questions From Chemical Kinetics Chapter

What will be the effect of temperature on rate constant?

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction.

The rate law for the decomposition of N2O5 is rate = k[N2O5]. What is the significance of k in this equation?

Define activation energy of a reaction.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The rate law for the decomposition of N₂O₅ is rate = k[N₂O₅]. What is the significance of k in this equation?