The rate constant of a reaction is 1.2 x 10^–3 sec^–3 at 30°C and 2.1 x 10^–3 sec^–1 at 40°C. Calculate the energy of activation of the reaction.

Solution

We have given that

$k_{1} = 1.2 \times 10^{- 3} \sec^{- 1} T_{1} = 30 + 273 = 303 K k_{2} = 2.1 \times 10^{- 3} \sec^{- 1} T_{2} = 40 + 273 = 313$

Substituting these values in the equation

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

we get
$\log \frac{2.1 \times 10^{- 3}}{1.2 \times 10^{- 3}} = \frac{E_{a}}{2.303 \times 8.314} (\frac{313 - 303}{303 \times 313})$

or

$\log \frac{2.1}{1.2} = \frac{E_{a}}{2.303 \times 8.314} \times \frac{10}{303 \times 313}$

$E_{a} = 44.13 kJ {mol}^{- 1} .$

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Chemical Kinetics

The rate constant of a reaction is 1.2 x 10^–3 sec^–3 at 30°C and 2.1 x 10^–3 sec^–1 at 40°C. Calculate the energy of activation of the reaction.

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Chemical Kinetics

The rate constant of a reaction is 1.2 x 10–3 sec–3 at 30°C and 2.1 x 10–3 sec–1 at 40°C. Calculate the energy of activation of the reaction.

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What will be the effect of temperature on rate constant?

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Mock Test Series

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The rate constant of a reaction is 1.2 x 10^–3 sec^–3 at 30°C and 2.1 x 10^–3 sec^–1 at 40°C. Calculate the energy of activation of the reaction.

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