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A reaction is said to be of zero order if the rate of the reaction is independent of the concentration of all the reactants. $$ { H }_{ 2 }(g) + Cl_2 (g) \overset { h\nu }{ \longrightarrow } 2HCl (g) \text{ (Photochemical reaction) } $$
Rate equation for Zero order Reaction. $$ $$ If a is the initial concentration, the expression for the rate of reaction of zero order may be written as: $$ \frac { dx }{ dt } \propto { a }^{ 0 }$$ or $$ \frac { dx }{ dt } = k{ a }^{ 0 } = k \qquad ...(as \ { a }^{ 0 } = 1) $$ or rate of reaction remains constant throughout. where \(k\) is the rate constant for the zero order reaction. The above expression may be written as $$ dx = kdt \qquad ...(1)$$ Integrating this equation we get,$$ \int { dx } = k\int { dt } $$ $$ x = kt + I \qquad ...(2)$$ where \(I\) is constant of integration.
When \(t = 0, x = 0\) (at the start of reaction),$$ \therefore 0 = k\left( 0 \right) + I$$ $$ or \ I = 0 ...(3)$$ Lets put the value of \(I\) again in eq (2) $$ x = kt + 0$$ Hence $$ x = kt \qquad ...(4) $$ or $$ x \propto t $$ i.e., the amount of substance reacted \(\propto \) time.
The expression for the half-life period: Half life period is the time taken for half the reaction to complete, it may be obtained as: At \(t = { t }_{ 1/2 }, x = \frac { a }{ 2 } \) Substituting these values in rate equation,$$ \boxed { x = kt } $$ we get, $$ \frac { a }{ 2 } = k{ t }_{ 1/2 } $$ $$ \therefore \boxed { { t }_{ 1/2 } = \frac { a }{ 2k } } $$ i.e., the half-life period \propto initial concentration.$$
Numerical: Determine the order of reaction and the rate constsnt of the decomposition of \({ NH }_{ 3 }\) on a tugsten wire at 857C from the followring data.
Total pressure (torr) | 228 | 250 | 273 | 318 |
---|---|---|---|---|
Time (sec) | 200 | 400 | 600 | 1000 |
\( \therefore \) Rate of reaction \(= \frac { dp }{ dt } \)
at time t = 400sec $$ \frac { dp }{ dt } = \frac { 250 - 228 }{ 400 - 200 } = \frac { 22 }{ 200 } = 0.11 $$
at time t = 600sec $$ \frac { dp }{ dt } = \frac { 273 - 228 }{ 600 - 200 } = \frac { 45 }{ 400 } = 0.1125 $$
at time t = 1000sec $$ \frac { dp }{ dt } = \frac { 318 - 228 }{ 1000 - 200 } = \frac { 90 }{ 800 } = 0.1125 $$ It is clear from the pressure is changing but rate of reaction is not changing. Or rate of recation does not depend on concentration of any reactant.
\( \therefore \) It is zero order reaction.
and specific reaction rate is \(= 0.1117 torr.{ sec }^{ -1 } \)
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