By Sunil Bhardwaj

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A reaction is said to be of the second order if the rate of the reaction depends upon two concentration term only.$$2O_3(g) \overset { 100C }{ \longrightarrow } 3O_2(g) \qquad \text{ (Second Order reaction) }$$ $$CH_3COOC_2H_5 + NaOH \longrightarrow CH_3COONa + C_2H_5OH $$

$$ \boxed { A + B \longrightarrow Product } $$ The half life is the time taken for the initial concentration of the reactant to be reduced to half its value. The rate equation for second order reaction is $$ \boxed { k = \frac { 1 }{ at } \frac { x }{ \left( a - x \right) } } $$ So at half life time \({ t }_{ 1/2 }\), then \(x = \frac { a }{ 2 }\) $$ k = \frac { 1 }{ a{ t }_{ 1/2 } } \frac { \frac { a }{ 2 } }{ \left( a - \frac { a }{ 2 } \right) } $$ $$ \therefore { t }_{ 1/2 } = \frac { 1 }{ ak } \frac { \frac { a }{ 2 } }{ \left( \frac { \left( 2a - a \right) }{ 2 } \right) } $$ $$ { t }_{ 1/2 } = \frac { 1 }{ ak } \frac { \frac { a }{ 2 } }{ \left( \frac { a }{ 2 } \right) } = \frac { 1 }{ ak } $$ $$ \therefore \boxed { { t }_{ 1/2 } = \frac { 1 }{ ak } } $$ This indicates that the half life is inversely proportional to the initial concentration.