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.

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