By Sunil Bhardwaj

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Case 2: When concentration of reactants are not equal. If a is the initial concentration of A is a and B is b, At time t the same amount of A and B will react to produce product let it be x. $$ \underset { \left( a - x \right) }{ A } + \underset { \left( b - x \right) }{ B } \longrightarrow \underset { \left( x \right) }{ Product } $$ Applying the law of mass action, $$ \frac { dx }{ dt } \propto \left[ a - x \right] \left[ b - x \right] $$ $$ \frac { dx }{ dt } = k\left[ a - x \right] \left[ b - x \right] $$ $$ \frac { dx }{ \left[ a - x \right] \left[ b - x \right] } = k.dt \qquad ...(1)$$ Resolving the left side into partial fractions $$ \frac { 1 }{ \left( a - b \right) } \left[ \frac { 1 }{ \left[ b - x \right] } - \frac { 1 }{ \left[ a - x \right] } \right] dx = k.dt$$ $$ \frac { 1 }{ \left( a - b \right) } \left[ \frac { dx }{ \left[ b - x \right] } - \frac { dx }{ \left[ a - x \right] } \right] = k.dt$$ On integrating this equation $$ \frac { 1 }{ \left( a - b \right) } \left[ \int { \frac { 1 }{ \left[ b - x \right] } dx } - \int { \frac { 1 }{ \left[ a - x \right] } dx } \right] = k\int { dt } $$ $$ \frac { 1 }{ \left( a - b \right) } \left\{ \left[ -\ln { \left( b - x \right) } \right] - \left[ -\ln { \left( a - x \right) } \right] \right\} = kt + k’$$ where k’ is the constant of integration. $$ \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \left( a - x \right) } - \ln { \left( b - x \right) } \right\} = kt + k’$$ $$ \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \frac { \left( a - x \right) }{ \left( b - x \right) } } \right\} = kt + k’ \qquad ...(2)$$ When t = 0, x = 0 equation (2) beacomes, $$ \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \frac { \left( a - 0 \right) }{ \left( b - 0 \right) } } \right\} = k\left( 0 \right) + k’$$ $$ \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \frac { a }{ b } } \right\} = k’$$ on substituting value of k’ in equation (2) we get, $$ \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \frac { \left( a - x \right) }{ \left( b - x \right) } } \right\} = kt + \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \frac { a }{ b } } \right\} $$ $$ \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \frac { \left( a - x \right) }{ \left( b - x \right) } } \right\} - \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \frac { a }{ b } } \right\} = kt $$ $$ kt = \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \frac { \left( a - x \right) }{ \left( b - x \right) } } - \ln { \frac { a }{ b } } \right\} $$ $$ kt = \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \left( \frac { \left( a - x \right) }{ \left( b - x \right) } \times \frac { b }{ a } \right) } \right\} $$ $$ kt = \frac { 1 }{ \left( a - b \right) } \left\{ \ln { \frac { b\left( a - x \right) }{ a\left( b - x \right) } } \right\} $$ $$ k = \frac { 1 }{ t\left( a - b \right) } \ln { \frac { b\left( a - x \right) }{ a\left( b - x \right) } } $$ $$ \boxed { k = \frac { 2.303 }{ t\left( a - b \right) } \log { \frac { b\left( a - x \right) }{ a\left( b - x \right) } } } $$ This is the integrated rate law for second order reaction with different initial concentration.