JEE Advance - Chemistry (2009 - Paper 2 Offline - No. 1)
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
$$1.0\times10^6~\mathrm{s^{-1}}$$ and 9.2 kJ mol$$^{-1}$$
$$6.0~\mathrm{s^{-1}}$$ and 16.6 kJ mol$$^{-1}$$
$$1.0\times10^6~\mathrm{s^{-1}}$$ and 16.6 kJ mol$$^{-1}$$
$$1.0\times10^6~\mathrm{s^{-1}}$$ and 38.3 kJ mol$$^{-1}$$
Explanation
Given that $$\log k = 6 - {{2000} \over T}$$
According to Arrhenius equation
$$k = A{e^{ - {E_a}/RT}}$$
$$\log K = \log A - {E \over {2.303RT}}$$
Comparing with the given equation and solving, we get
$$A = 1.0 \times {10^6}\,{s^{ - 1}}$$
$${E_a} = 38.3$$ kJ/mol
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