First slide

Kinetics of chemical reactions

Question

The rate of a reaction triple when temperature changes from 20°C to 50°C. Calculate energy of activation for the reaction. $(R = 8.314 J K^{- 1} m o l^{- 1})$

Moderate

Solution

The Arrhenius equation is,
$\log_{10} \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R \times 2.303} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
$Given: \frac{k_{2}}{k_{1}} = 3; R = 8.314 {JK}^{- 1} {mol}^{- 1}; T_{1} = 20 + 273 = 293 K$
and $T_{2} = 50 + 273 = 323 K$
Substituting the given values in the Arrhenius equation,
$\log_{10} 3 = \frac{E_{a}}{8.314 \times 2.303} [\frac{323 - 293}{323 \times 293}]$
$E_{a} = \frac{2.303 \times 8.314 \times 323 \times 293 \times 0.477}{30} = 28811.8 J m o l^{- 1}$
$E_{a} = 28.81 k J m o l^{- 1}$

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