Question
For a first order reaction $$A → P,$$ the temperature $$(T)$$ dependent rate constant $$(k)$$ was found to follow the equation $$\log k = - \left( {2000} \right)\frac{1}{T} + 6.0.$$ The pre-exponential factor $$A$$ and the activation energy $${E_a}$$ ,respectively, are
A.
$$1.0 \times {10^6}{s^{ - 1}}\,{\text{and}}\,9.2\,kJ\,mo{l^{ - 1}}$$
B.
$$6.0{s^{ - 1}}\,{\text{and}}\,16.6\,kJ\,mo{l^{ - 1}}$$
C.
$$1.0 \times {10^6}{s^{ - 1}}\,{\text{and}}\,16.6\,kJ\,mo{l^{ - 1}}$$
D.
$$1.0 \times {10^6}{s^{ - 1}}\,{\text{and}}\,38.3\,kJ\,mo{l^{ - 1}}$$
Answer :
$$1.0 \times {10^6}{s^{ - 1}}\,{\text{and}}\,38.3\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& \log k = \log A - \frac{{{E_a}}}{{2.303RT}}\,\,...(1) \cr
& {\text{Also}}\,{\text{given}}\,{\text{log}}k = 6.0 - \left( {2000} \right)\frac{1}{T}\,\,\,...(2) \cr
& {\text{On comparing equations, (1) and (2)}} \cr
& \log A = 6.0 \Rightarrow \,A = {10^6}{s^{ - 1}} \cr
& and\,\frac{{{E_a}}}{{2.303\,R}} = 2000\,; \cr
& \Rightarrow {E_a} = 2000 \times 2.303 \times 8.314 = 38.29\,mo{l^{ - 1}} \cr} $$