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. $$ N_2O_2(g) \longrightarrow N_2O_4(g) + \frac12{ O }_{ 2 }(g) \text{ (First Order reaction) } $$

Half life period is defined as the time at which half of the reactant is converted into product. \( \therefore \) at half life period \(t = { t }_{ 1/2 }\), $$ \text{ conc. of reactant } = \frac { \text{ initial conc. of reactant } }{ 2 } $$ $$ or \ x = \frac { a }{ 2 } \qquad ...(1)$$ Equation for the rate constant for first order reaction is, $$ \boxed { k = \frac { 2.303 }{ t } \log { \frac { a }{ a-x } } } \qquad ...(2)$$ Substituting the value of x from eq (1), $$ k = \frac { 2.303 }{ { t }_{ 1/2 } } \log { \frac { a }{ a-\frac { a }{ 2 } } } $$ $$ k = \frac { 2.303 }{ { t }_{ 1/2 } } \log { \frac { a }{ \left( \frac { 2a-a }{ 2 } \right) } } $$ $$ k = \frac { 2.303 }{ { t }_{ 1/2 } } \log { \frac { a }{ \left( \frac { a }{ 2 } \right) } } $$ $$ k = \frac { 2.303 }{ { t }_{ 1/2 } } \log { \left( 2 \right) } $$ $$ k = \frac { 2.303 }{ { t }_{ 1/2 } } \times \left( 0.3010 \right) $$ $$ \boxed { { t }_{ 1/2 } = \frac { 0.693 }{ k } } $$ Thus the half life for a first order reaction is a constant and independent of the initial concentration.

Numerical: A substance decomposes according to first order rate law with a half time of 1 hr. What fraction of a given sample of the substance will remain after 2 hrs?

Solution: Given that,

\({ t }_{ 1/2 } = 1 hr\)

\(t = 2 hr\)

Fraction of sample remaining \(= \frac { C_{ t } }{ C_{ 0 } } = ?\)

For the half life equation, $$ \boxed { k = \frac { 0.693 }{ { t }_{ 1/2 } } } \\ k = \frac { 0.693 }{ 1 hr } = 0.693 { hr }^{ -1 }\qquad ...(1)$$ also, rate constant for first order reaction is, $$ \boxed { k =\frac { 2.303 }{ t } \log { \frac { C_{ 0 } }{ C_{ t } } } } $$ Lets substitute the values, $$ 0.693 = \frac { 2.303 }{ 2 hr } \log { \frac { C_{ 0 } }{ C_{ t } } } $$ $$ \therefore \log { \frac { C_{ 0 } }{ C_{ t } } } = \frac { 0.693 \times 2 }{ 2.303 } = 0.602$$ $$ \therefore \frac { C_{ 0 } }{ C_{ t } } = Anti.\log { \left( 0.602 \right) } = 3.998$$ $$ \therefore \frac { C_{ t } }{ C_{ 0 } } = \frac { 1 }{ 3.998 } = 0.25$$ i.e. fraction remaining after 2 hrs is 0.25