Question
The decomposition of a hydrocarbon follows the equation $$k = \left( {4.5 \times {{10}^{11}}\,{s^{ - 1}}} \right){e^{ - \frac{{28000\,K}}{T}}}.$$ What will be the value of activation energy?
A.
$$669\,kJ\,mo{l^{ - 1}}$$
B.
$$232.79\,kJ\,mo{l^{ - 1}}$$
C.
$$4.5 \times {10^{11}}\,kJ\,mo{l^{ - 1}}$$
D.
$$28000\,kJ\,mo{l^{ - 1}}$$
Answer :
$$232.79\,kJ\,mo{l^{ - 1}}$$
Solution :
Arrhenius equation, $$k = A{e^{ - \frac{{{E_a}}}{{RT}}}}$$
Given equation is $$k = \left( {4.5 \times {{10}^{11}}\,{s^{ - 1}}} \right){e^{ - \frac{{28000\,K}}{T}}}$$
Comparing both the equations, we get $$ - \frac{{{E_a}}}{{RT}} = - \frac{{28000\,K}}{T}$$
$$\eqalign{
& {E_a} = 28000\,K \times R \cr
& \,\,\,\,\,\,\,\, = 28000\,K \times 8.314\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr
& \,\,\,\,\,\,\,\, = 232.79\,kJ\,mo{l^{ - 1}} \cr} $$