Q32 / 120 · VITEEE 2016 · Chemistry

In a reaction $A \rightarrow$ Products, when start is made from $8.0 \times 10^{-2} M$ of $A$, half-life is found to be 120 minute. For the initial concentration $4.0 \times 10^{-2}$ $M$, the half-life of the reaction becomes 240 minute. The order of the reaction is :

  • A
    zero
  • B
    one
  • Correct answer
    C
    two
  • D
    $0.5$

Solution

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

Hence, the correct answer is option: 'C'.

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