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 $$\boxed { c = \frac { 1 }{ a } }$$ 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 $$\boxed { c = \log { \frac { a }{ b } } }$$ 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 $$\boxed { k = \frac { slope \times 2.303 }{ \left( a - b \right) } }$$

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