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Chemical Kinetics

Question

The rate constant of reaction is 1.5 x 10⁷ sec^–1 at 50°C and 4.5 x10⁷ sec^–1 at 100°C. Calculate the value of activation energy, E_a for the reaction (R =8.314 JK^–1 mol^–1).

Solution

by using of arrhenius of equation

$\frac{\log k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{2} T_{1}}]$

or

$\frac{\log 4.5 \times 10^{7}}{1.5 \times 10^{7}} = \frac{E_{a}}{2.303 \times 8.314} [\frac{373 - 323}{373 \times 323}]$

or

$\log 3 = \frac{E_{a}}{2.303 \times 8.314} [\frac{50}{373 \times 323}]$

or

$E_{a} = \frac{0.4771 \times 2.303 \times 8.314 \times 373 \times 323}{50} = 22011.75 J {mol}^{- 1} = 22.01 kJ {mol}^{- 1} .$

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Chemical Kinetics

The rate constant of reaction is 1.5 x 10⁷ sec^–1 at 50°C and 4.5 x10⁷ sec^–1 at 100°C. Calculate the value of activation energy, E_a for the reaction (R =8.314 JK^–1 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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Chemical Kinetics

The rate constant of reaction is 1.5 x 107 sec–1 at 50°C and 4.5 x107 sec–1 at 100°C. Calculate the value of activation energy, Ea for the reaction (R =8.314 JK–1 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.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The rate constant of reaction is 1.5 x 10⁷ sec^–1 at 50°C and 4.5 x10⁷ sec^–1 at 100°C. Calculate the value of activation energy, E_a for the reaction (R =8.314 JK^–1 mol^–1).

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