IIT-JEE 2009 · Paper 2 · Chemistry Q1

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ChemistrySingle correct+3 / -1Question 1 of 57

For a first order reaction A → P, the temperature (T) dependent rate constant ( $k$ ) was found to follow the equation $\log k=-(2000)\dfrac1T+6.0$ . The pre-exponential factor A and the activation energy $E_a$ , respectively, are

1.0\times10^6\ \mathrm{s^{-1}}

and

9.2\ \mathrm{kJ\,mol^{-1}}

6.0\ \mathrm{s^{-1}}

and

16.6\ \mathrm{kJ\,mol^{-1}}

1.0\times10^6\ \mathrm{s^{-1}}

and

16.6\ \mathrm{kJ\,mol^{-1}}

1.0\times10^6\ \mathrm{s^{-1}}

and

38.3\ \mathrm{kJ\,mol^{-1}}

Answer and worked solution

Official answer: D

Arrhenius in log form. $\log k=\log A-\dfrac{E_a}{2.303RT}$ .

Compare. $\log A=6.0$ , so $A=1.0\times10^6\ \mathrm{s^{-1}}$ , and
$\frac{E_a}{2.303R}=2000\implies E_a=2000\times2.303\times8.314=3.83\times10^4\ \text{J mol}^{-1}=38.3\ \text{kJ mol}^{-1}.$

(D)

Topics:Arrhenius equation

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IIT-JEE 2009 · Paper 2, Chemistry question 1