For a first-order reaction A \to P, the temperature (T) dependent rate constant (k) was found to follow the equation \log k = - (2000)1 \over T + 6.0 . The pre-exponential factor A and activation energy E_a , respectively, are

1.0\times10^6~s^-1 and 38.3 kJ mol ^-1

1.0\times10^6~s^-1 and 9.2 kJ mol ^-1

1.0\times10^6~s^-1 and 16.6 kJ mol ^-1

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Advanced 2009 · Shift 2 · Q1 with Solution

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Correct answer: D

Use the Arrhenius equation in logarithmic form

For a reaction,

k = A e^{-E_a/(RT)}

Taking common logarithm,

\log k = \log A - \frac{E_a}{2.303RT}

Compare with the given equation

Given:

\log k = -\frac{2000}{T} + 6.0

Comparing with

\log k = \log A - \frac{E_a}{2.303RT}

we get:

Intercept: $\log A = 6.0$
Slope term: $\frac{E_a}{2.303R} = 2000$

Calculate the pre-exponential factor $A$

Since

\log A = 6.0

therefore,

A = 10^6~\mathrm{s^{-1}}

Calculate the activation energy $E_a$

E_a = 2000 \times 2.303 \times R

Using $R = 8.314~\mathrm{J~mol^{-1}~K^{-1}}$ ,

E_a = 2000 \times 2.303 \times 8.314

E_a \approx 38288~\mathrm{J~mol^{-1}}

E_a \approx 38.3~\mathrm{kJ~mol^{-1}}

Match with the options

Thus,

A = 1.0\times10^6~\mathrm{s^{-1}}, \qquad E_a = 38.3~\mathrm{kJ~mol^{-1}}

This corresponds to Option D.

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