MHT CET · Chemistry · Chemical Kinetics
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?
- A Rate increases by factor 2
- B Rate increases by factor 4
- C Rate remains same
- D Rate decreases by factor 2
Answer & Solution
Correct Answer
(B) Rate increases by factor 4
Step-by-step Solution
Detailed explanation
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
\((\text { Rate })_1=\mathrm{k} \times 2\left[\mathrm{CH}_3 \mathrm{Br}\right] \times 2\left[\mathrm{OH}^{-}\right]\)
\(=4 \mathrm{k}\left[\mathrm{CH}_3 \mathrm{Br}\right][\mathrm{OH}]\)
\(\therefore \quad \frac{\text { (Rate })_1}{\text { Rate }}=\frac{4 \mathrm{k}\left[\mathrm{CH}_3 \mathrm{Br}\right]\left[\mathrm{OH}^{-}\right]}{\mathrm{k}\left[\mathrm{CH}_3 \mathrm{Br}\right]\left[\mathrm{OH}^{-}\right]}=4\)
\(\therefore \quad(\text { Rate })_1=4 \times\) Rate
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