By Sunil Bhardwaj

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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 }$$ on a tugsten wire at 857C from the followring data.  Total pressure (torr) Time (sec) 228 250 273 318 200 400 600 1000 Solution: The rate of reaction will be $$= \frac { dx }{ dt }$$ in this case change in concentration will be equal to change in pressure. $$\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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