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

$$ \boxed { A + B \longrightarrow Product } $$ Case 1: When concentration of reactants are equal If a is the initial concentration of reactants A and B, and x amount of A and B are reacted at time t, so at time t the concentrations will be,$$ \underset { \left( a - x \right) }{ A } + \underset { \left( a - x \right) }{ B } \longrightarrow \underset { \left( x \right) }{ Product } $$ Applying the law of mass action,$$ \frac { dx }{ dt } \propto { \left[ a - x \right] }^{ 2 }$$ $$ \frac { dx }{ dt } = k{ \left[ a - x \right] }^{ 2 }$$ $$ \frac { dx }{ { \left[ a - x \right] }^{ 2 } } = k.dt \qquad ...(1)$$ On integrating this equation $$ \int { \frac { 1 }{ { \left[ a - x \right] }^{ 2 } } dx } = k\int { dt } $$ $$ \frac { 1 }{ { \left( a - x \right) } } = kt + k’ \qquad ...(2)$$ where k’ is the constant of integration.

When t = 0, x = 0 equation (2) beacomes, $$ \frac { 1 }{ { \left( a - 0 \right) } } = k\left( 0 \right) + k’$$ $$ \therefore \frac { 1 }{ { a } } = k’$$ on substituting value of k’ in equation (2) we get, $$ \frac { 1 }{ { \left( a - x \right) } } = kt + \frac { 1 }{ { a } } $$ $$ kt = \frac { 1 }{ { \left( a - x \right) } } - \frac { 1 }{ { a } } $$ $$ kt = \frac { a - \left( a - x \right) }{ { a\left( a - x \right) } } $$ $$ kt = \frac { x }{ { a\left( a - x \right) } } $$ $$ \boxed { k = \frac { 1 }{ { at } } \frac { x }{ { \left( a - x \right) } } } $$ This is the integrated rate law for second order reaction with equal initial concentration.

Numerical: A second order reduction with equal initial concentration of the reactants is 75% complete in 1 hrs. Calculate how much will be left unreacted at the end of 2 hrs.

Solution: In 1 hr 75% of reactants are reacted.$$ \therefore x = 0.75a \ and \ \left( a - x \right) = 0.25a$$ For a second order reaction with equal initial concentrations rate constant is given by,$$ \boxed { k = \frac { 1 }{ { at } } \frac { x }{ { \left( a - x \right) } } } $$ Lets substitute the values $$ \therefore k = \frac { 1 }{ a \times 1hrs } \frac { 0.75a }{ 0.25a } = \frac { 3 }{ a } $$ After 2nd hr the equation will be $$ k = \frac { 1 }{ a \times 2hrs } \frac { x }{ \left( a - x \right) } $$ $$ \frac { 3 }{ a } = \frac { 1 }{ 2a } \frac { x }{ \left( a - x \right) } $$ $$ \frac { 3 \times 2a }{ a } = \frac { x }{ \left( a - x \right) } $$ $$ x = 6\left( a - x \right) = 6a - 6x$$ $$ 7x = 6a$$ $$ \therefore x = \frac { 6 }{ 7 } a = 0.857a$$ This is amount reacted after 2 hrs so amount remaining will be, 1 - 0.857 = 0.143 or 14.3%

Numerical: Decomposition of a gas is of second order. It takes 40 minutes for 40% of a gas to be decomposed when its initial concentration is \(4 \times { 10 }^{ -2 } \)mole/litre. Calculate the specific reaction rate.

Ans: K = 0.4167 moles/liter