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Chemical Kinetics

Question

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The decomposition of hydrocarbon follows the equation
k = (4.5 x 10¹¹ s^–1) e^–28000K/TCalculate E_a.

Solution

The given equation is

K = (4.5 \times 10^{11} S^{- 1}) e^{- 28000 k / T} \log K = \log A - \frac{E_{a}}{RT} or \frac{E_{a}}{R} = 28000 K or E_{a} = 28000 \times R = 28000 \times 8.314 = 232792 J {mol}^{- 1} = 232.79 kJ {mol}^{- 1}

Some More Questions From Chemical Kinetics Chapter

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.

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