By Sunil Bhardwaj

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A reaction is said to be of the first order if the rate of the reaction depends upon one concentration term only. E.g. $$N_2O_2(g) \longrightarrow N_2O_4(g) + \frac12{ O }_{ 2 }(g) \text{ (First Order reaction) }$$

Suppose we start with a moles per litre of the reactant A $$\underset { a \ moles/liter }{ A } \longrightarrow Products \qquad ... (at \ start)$$ After time t, suppose x moles per litre of it have reacted. Therefore the concentration of A remaining after time $$t = (a - x)$$ moles per litre. $$\underset { (a-x) \ moles/liter }{ A } \longrightarrow Products ... (after \ time \ t)$$ Then according to law of mass action, $$\text{Rate of reaction } \propto \text{ Conc of A remaining after time t.}$$ $$i.e., \frac { dx }{ dt } \propto (a-x)$$ or $$\frac { dx }{ dt } = k(a-x) ... (1)$$ where k is called the Rate constant. $$\therefore \frac { dx }{ (a-x) } = kdt$$ Integrating equation this equation, we get,$$\int { \frac { dx }{ (a-x) } } = k\int { dt }$$ $$-\ln { (a-x) } = kt + I \qquad ... (2)$$ where $$I$$ is a constant of integration.

In the beginning, when t = 0, x = 0 (as no substance has reacted at the start of the reaction). Putting these values in equation (2), we get,$$-\ln { (a-0) } = k\left( 0 \right) + I\\ -\ln { (a) } = I \qquad ... (3)$$ Substituting the value of I in equation (2), we get, $$-\ln { (a-x) } = kt + \left[ -\ln { (a) } \right]$$ $$-\ln { (a-x) } + \ln { (a) } = kt$$ $$kt = \ln { (a) } - \ln { (a-x) }$$ $$kt = \ln { \frac { a }{ a-x } }$$ $$k = \frac { 1 }{ t } \ln { \frac { a }{ a-x } }$$ $$\boxed { k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } } \qquad ... (4)$$ This equation can also be written as, $$\boxed { k = \frac { 2.303 }{ t } \log { \frac { { C }_{ 0 } }{ { C }_{ t } } } } ... (5)$$ Where $${ C }_{ 0 }$$ is initial concentration and $${ C }_{ t }$$ is Concentration at time t.

Unit of Rate Constant (k): Equation for the rate constant for first order reaction is, $$\boxed { k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } }$$ where $$\frac { a }{ a-x }$$ is ratio of concentration thus no unit.$$\therefore unit \ of \ k = \frac { 1 }{ time } = \frac { 1 }{ sec } = { sec }^{ -1 } \ or \ { min }^{ -1 }$$

Numerical: The catalytic decomosition of hydrogen peroxide in aqueous solution was followed at 298K by titrating aliquot volumes of the reaction solution against standard $$KMn{ O }_{ 4 }$$ solution. The following results were obtained.

Time in min051015
Volm of KMnO4 (cm3)37.533.529.926.7
Calculate the specific reaction rate assuming the reaction to be first order.

Solution: Given that Volume of $$KMn{ O }_{ 4 }$$ consumed will be equal to the conc of reactant remaining after that time. Thus the rate constant, $$\boxed { k =\frac { 2.303 }{ t } \log { \frac { { C }_{ 0 } }{ { C }_{ t } } } }$$ when t = 5min, $$k = \frac { 2.303 }{ 5 } \log { \frac { 37.5 }{ 33.5 } } =0.02256 { min }^{ -1 }$$ when t = 10min, $$k = \frac { 2.303 }{ 10 } \log { \frac { 37.5 }{ 29.9 } } =0.02264 { min }^{ -1 }$$ when t = 15min, $$k = \frac { 2.303 }{ 15 } \log { \frac { 37.5 }{ 26.7 } } =0.02264 { min }^{ -1 }$$ The average value of specific reaction rate is $$0.02261 { min }^{ -1 }$$

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