For the reaction, $\mathrm{CH}_3 \mathrm{Br}_{(\mathrm{aq})}+\mathrm{OH}_{(\mathrm{sq})}^{-} \rightarrow…

For the reaction, $\mathrm{CH}_3 \mathrm{Br}_{(\mathrm{aq})}+\mathrm{OH}_{(\mathrm{sq})}^{-} \rightarrow \mathrm{CH}_3 \mathrm{OH}_{(\mathrm{aq})}+\mathrm{Br}_{(\mathrm{aq})}^{-}$ the rate law is rate $=\mathrm{k}\left[\mathrm{CH}_3 \mathrm{Br}\right][\mathrm{OH}]$. What is change in rate of reaction if concentration of both reactants is doubled?
  1. Rate increases by factor 2
  2. Rate increases by factor 4
  3. Rate remains same
  4. Rate decreases by factor 2

Solution

Rate $=\mathrm{k}\left[\mathrm{CH}_3 \mathrm{Br}\right][\mathrm{OH}]$ $(\text { Rate })_1=\mathrm{k} \times 2\left[\mathrm{CH}_3 \mathrm{Br}\right] \times 2\left[\mathrm{OH}^{-}\right]$ $=4 \mathrm{k}\left[\mathrm{CH}_3 \mathrm{Br}\right][\mathrm{OH}]$ $\therefore \quad \frac{\text { (Rate })_1}{\text { Rate }}=\frac{4 \mathrm{k}\left[\mathrm{CH}_3 \mathrm{Br}\right]\left[\mathrm{OH}^{-}\right]}{\mathrm{k}\left[\mathrm{CH}_3 \mathrm{Br}\right]\left[\mathrm{OH}^{-}\right]}=4$ $\therefore \quad(\text { Rate })_1=4 \times$ Rate

Asked in: MHT CET 2023 (13 May Shift 2)

Practice more Chemical Kinetics questions on Aicharya