Rate constant k of a reaction varies with temperature according to the equation:

log k constant $- \frac{E_{a}}{2.303} \frac{1}{T}$

where E_a is the energy of activation for the reaction. When a graph is plotted for log k versus 1/T a straight line with a slope - 6670 k is obtained. Calculate energy of activation for this reaction. State the units (R = 8.314 J K^–1 mol^–1)

Solution

We have given the slope value ,
Slope = -6670 k

$E_{a} = 2 R = 8.314 J K^{- 1} {mol}^{- 1}$

Thus using the equation

$Slope = - \frac{E_{a}}{2.303 R}$

$\Rightarrow E_{a} = - slope \times 2.303 \times R = - (- 6670) \times 2.303 \times 8.314 J = + 127710.49 J = + 127.71049 kJ .$

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?

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Chemical Kinetics

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?

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction.

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