1. Tardigrade
  2. Question
  3. Chemistry
  4. In a reaction A arrow Products, when start is made from 8.0 × 10-2 M of A, half-life is found to be 120 minute. For the initial concentration 4.0 × 10-2 M, the half-life of the reaction becomes 240 minute. The order of the reaction is:

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Q. 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:

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Solution:

From the half life reaction,
$\frac{\left( t _{1 / 2}\right)_{1}}{\left( t _{1 / 2}\right)_{2}}=\left(\frac{ a _{2}}{ a _{1}}\right)^{ n -1}$
$\frac{120}{140}=\left(\frac{4 \times 10^{-2}}{8 \times 10^{-2}}\right)^{ n -1}$
$n =2$