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Q.

For a reaction $A \to 2 B + C$ the half-lives are 100s and 50s when the concentration of reactant A is 0.5 and $1.0 m o l L^{- 1}$ respectively. The order of reaction is

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a

2

b

1

c

Zero

d

3

answer is C.

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Detailed Solution

$t_{1 / 2} α \frac{1}{a^{n - 1}}$

$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{a_{2}}{a_{1}})}^{n - 1}$

$\frac{100}{50} = {(\frac{1.0}{0.5})}^{n - 1}$

$2^{1} = 2^{n - 1}$

$n - 1 = 1$

$n = 2$

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