Q5 / 120 · VITEEE 2010 · Chemistry

A chemical reaction wascarried out at $320 \mathrm{~K}$ and $300 \mathrm{~K}$. The rate constants were found to be $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$ respectively. Then

  • A
    $\bar{k}_{2}=4 \mathrm{k}_{1}$
  • B
    $\mathrm{k}_{2}=2 \mathrm{k}_{1}$
  • Correct answer
    C
    $\mathrm{k}_{2}=0.25 \mathrm{k}_{1}$
  • D
    $k_{2}=0.5 \mathrm{k}_{1}$

Solution

As we know that for every $10^{\circ}$ rise in temperature, rate constant, $k$ becomes doubled. Hence, on rising the temperature $20^{\circ}$, the rate constant will be four times, i.e., $k_{1}=4 k_{2} \Rightarrow k_{2}=\frac{1}{4} k_{1}=0.25 k_{1}$

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

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