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Chemical Kinetics

Question
CBSEENCH12010280
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The rate constant of a first order reaction increases from 2 — 10-2 to 4 — 10-2when the temperature changes from 300 K to 310 K. Calculate the energy of activation (Ea).

(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)

Solution

According to the Arrhenius equation.

K = Ae(-Ea/RT)

From this, we get
log space K subscript 2 over K subscript 1 space equals fraction numerator E subscript a over denominator 2.303 end fraction space fraction numerator T subscript 2 minus T subscript 1 over denominator T subscript 1 T subscript 2 end fraction

 

We are given that

initial temperature T1=300K

Final temperature T2=310 K

Rate constant at initial temperature, k 1 = 2 x 10-2

Rate constant at final temperature, k2 = 4 x 10-2

Gas constant, R = 8.314 J K-1

Substituting the value, we get

  log space fraction numerator 4 space x space 10 to the power of negative 2 end exponent over denominator 2 space x space 10 to the power of negative 2 end exponent end fraction space equals space fraction numerator E subscript a over denominator 2.303 space x space 8.314 space x space end fraction open parentheses fraction numerator 310 minus 300 over denominator 300 space x 310 end fraction close parentheses

Therefore activation energy of the reaction, Ea =  straight E subscript straight a space equals fraction numerator log space straight x space 2 space straight x space 2.303 space straight x space 8.314 space straight x space 300 space straight x space 310 over denominator 10 end fraction

= 535985.94 J mol-

= 535.98 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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