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.$$2O_3(g) \overset { 100C }{ \longrightarrow } 3O_2(g) \qquad \text{ (Second Order reaction) }$$ $$CH_3COOC_2H_5 + NaOH \longrightarrow CH_3COONa + C_2H_5OH $$

Case 1: The rate equation for second order reaction with equal initial concentration is $$ \boxed { k = \frac { 1 }{ at } \frac { x }{ \left( a - x \right) } } $$ $$ kt = \frac { 1 }{ a } \frac { x }{ \left( a - x \right) } $$ From method of partial fractions $$ kt = \frac { 1 }{ \left( a - x \right) } - \frac { 1 }{ a } $$ $$ or \boxed { \frac { 1 }{ \left( a - x \right) } = kt + \frac { 1 }{ a } } $$ which is similar to straight line graph, $$ \boxed { y = mx + c } $$ where slope \(\boxed { m = k }\) and intercept \(\boxed { c = \frac { 1 }{ a } }\)

Second Order Reaction

Thus from the plot of \(\frac { 1 }{ \left( a - x \right) } \) Vs time t we can calculate the value of rate constant k, as $$ \boxed { k = slope } $$

Case 2: If the reaction is of second order with unequal concentrations. The rate equation for this will be, $$ \boxed { k = \frac { 2.303 }{ t\left( a - b \right) } \log { \frac { b\left( a - x \right) }{ a\left( b - x \right) } } } $$ $$ \frac { kt\left( a - b \right) }{ 2.303 } = \log { \frac { b\left( a - x \right) }{ a\left( b - x \right) } } $$ $$ \frac { kt\left( a - b \right) }{ 2.303 } = \log { \frac { b }{ a } } + \log { \frac { \left( a - x \right) }{ \left( b - x \right) } } $$ $$ \log { \frac { \left( a - x \right) }{ \left( b - x \right) } } = \frac { kt\left( a - b \right) }{ 2.303 } - \log { \frac { b }{ a } } $$ or $$ \boxed { \log { \frac { \left( a - x \right) }{ \left( b - x \right) } } = \frac { k\left( a - b \right) }{ 2.303 } t + \log { \frac { a }{ b } } } $$ which is similar to straight line graph, $$ \boxed { y = mx + c } $$ where slope \(\boxed { m = \frac { k\left( a - b \right) }{ 2.303 } } \) and intercept \(\boxed { c = \log { \frac { a }{ b } } } \)

Second Order Reaction

Thus from the plot of \(\log { \frac { \left( a - x \right) }{ \left( b - x \right) } } \) Vs time t we can calculate the value of rate constant k, as $$ \boxed { k = \frac { slope \times 2.303 }{ \left( a - b \right) } } $$