By Sunil Bhardwaj

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It is difficult to apply the collision theory to a unimolecular reaction because only one molecule takes part in unimolecular reaction, for example, the decomposition of nitrogen pentoxide. $$ N_2O_5 \longrightarrow N_2O_4 + \frac12O_2 $$ Lindemann suggested that the two different molecules of N2O5 colloids and during this collision one of the molecule acquires activation energy for the reaction. $$ N_2O_5 + N_2O_5 \longleftrightarrow \left[ { N_2O_5 } \right]^{ * } \longrightarrow N_2O_4 + \frac12O_2 $$

Consider a unimolecular reaction $$ A \longrightarrow Product$$ This reaction is given by Lindemann in stages as $$ A + A \overset { { k }_{ 1 } }{ \underset { { k }_{ 2 } }{ \leftrightharpoons } } { A }^{ * } + A \qquad ...(1) $$ $${ A }^{ * } \overset { { k }_{ 3 } }{ \longrightarrow } Product \qquad ...(2)$$ In the first stage one molecule gets activated giving \({ A }^{ * }\) and the rate constant is \({ k }_{ 1 }\). If there is some time-lag, then this activated molecule gets deactivated and reverse process takes place and that rate constant is given by \({ k }_{ 2 }\). If there is comparatively small or no time-lag then the activated molecule decomposes as usual and the rate constant is given by \({ k }_{ 3 }\). Thus, $$ \left( { \text { Rate of formation of activated molecules } } \right) = { k }_{ 1 }{ \left[ A \right] }^{ 2 } \qquad ...(3)$$ $$ \left( { \text { Rate of disappearance of activated molecules } } \right) = { k }_{ 2 }\left[ { A }^{ * } \right] \left[ A \right] + { k }_{ 3 }\left[ { A }^{ * } \right] \qquad ...(4)$$ According to steady state principle, when a short-lived intermediate exists in a system, the rate of formation of intermediate is same as the rate of disappearance. $$ \therefore \left( { \text { Rate of formation of activated molecules } } \right) = \left( { \text { Rate of disappearance of activated molecules } } \right) $$ Lets put the values from equation (3) and (4) $$ \therefore { k }_{ 1 }{ \left[ A \right] }^{ 2 } = { k }_{ 2 }\left[ { A }^{ * } \right] \left[ A \right] + { k }_{ 3 }\left[ { A }^{ * } \right] $$ $$ { k }_{ 1 }{ \left[ A \right] }^{ 2 } = \left[ { A }^{ * } \right] \left\{ { k }_{ 2 }\left[ A \right] + { k }_{ 3 } \right\} $$ $$ \therefore \boxed { \left[ { A }^{ * } \right] = \frac { { k }_{ 1 }{ \left[ A \right] }^{ 2 } }{ { k }_{ 2 }\left[ A \right] + { k }_{ 3 } } } \qquad ...(5)$$ Since decomposition of activated molecule is rate determining step. The rate of reaction is given by, $$ \text { Rate of reaction } = \frac { -d\left[ A \right] }{ dt } = { k }_{ 3 }\left[ { A }^{ * } \right] $$ Lets put the value of \(\left[ { A }^{ * } \right] \) from equation (5) $$ \boxed { \frac { -d\left[ A \right] }{ dt } = { k }_{ 3 }\frac { { k }_{ 1 }{ \left[ A \right] }^{ 2 } }{ { k }_{ 2 }\left[ A \right] + { k }_{ 3 } } } \qquad ...(6)$$ Now there are two possibilities,

(1) When \({ k }_{ 3 } << { k }_{ 2 }[A] \), then \(\left( { k }_{ 2 }\left[ A \right] + { k }_{ 3 } \right) \approx { k }_{ 2 }\left[ A \right] \)

so equation (6) becomes, $$ \frac { -d\left[ A \right] }{ dt } = \frac { { k }_{ 1 }{ k }_{ 3 }{ \left[ A \right] }^{ 2 } }{ { k }_{ 2 }\left[ A \right] } $$ $$ \frac { -d\left[ A \right] }{ dt } = \frac { { k }_{ 1 }{ k }_{ 3 } }{ { k }_{ 2 } } \left[ A \right] $$ $$ \therefore \boxed { \frac { -d\left[ A \right] }{ dt } = k’ \left[ A \right] } \qquad ...(7)$$ where \(k’ =\frac { { k }_{ 1 }k_{ 3 } }{ { k }_{ 2 } } \) is a new constant,

(2) When \({ k }_{ 2 }\left[ A \right] << { k }_{ 3 }\), then \(\left( { k }_{ 2 }\left[ A \right] + { k }_{ 3 } \right) \approx { k }_{ 3 }\) so equation (6) becomes $$ \frac { -d\left[ A \right] }{ dt } = { k }_{ 3 }\frac { k_{ 1 }{ \left[ A \right] }^{ 2 } }{ { k }_{ 3 } } $$ $$ \therefore \boxed { \frac { -d\left[ A \right] }{ dt } = { k }_{ 1 }{ \left[ A \right] }^{ 2 } } \qquad ...(8) $$ The equation (7) and (8) represent unimolecular and bimolecular reactions respectively.