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.

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