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In a reaction A — Products, when start is made from $8 .0 \times 10^{- 2}$ M f A, half-life is found to be 120 minute. For the initial concentration 4.0 x 10^-2 M, the half-life of the reaction becomes 240 minute. The order of the reaction is :

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zero

one

two

0.5

answer is C.

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Detailed Solution

$\frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} = {(\frac{a_{2}}{a_{1}})}^{n - 1}; \frac{120}{240} = {(\frac{4 \times 10^{- 2}}{8 \times 10^{- 2}})}^{n - 1}; n = 2$

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