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Chemical Kinetics

Question

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The rate of reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction that it does not change with temperature.

Solution

According to Arrhenius equation,

\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) T_{1} = 293 K, T_{2} = 313 K \log \frac{4}{1} = \frac{E_{a}}{2.303 \times 8.314 J {mol}^{- 1} K^{- 1}} [\frac{1}{293 K} - \frac{1}{313 K}] \log 4 = \frac{E_{a}}{2.303 \times 8.314 J {mol}^{- 1}} \times \frac{20}{293 \times 313} E_{a} = \frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20} = 5.2863 \times 10^{4} J {mol}^{- 1} = 52.863 kJ {mol}^{- 1}

Some More Questions From Chemical Kinetics Chapter

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

What will be the effect of temperature on rate constant?

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