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MHT CET · Chemistry · Chemical Kinetics

For the reaction,
\(\mathrm{CH}_3 \mathrm{Br}_{(\mathrm{aq})}+\mathrm{OH}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{CH}_3 \mathrm{OH}_{(\mathrm{aq})}+\mathrm{Br}_{(\mathrm{aq})}^{-}\)
rate of consumption of \(\mathrm{OH}_{(\mathrm{aq})}^{-}\)is \(\mathrm{x} \mathrm{mol} \mathrm{dm}{ }^{-3} \mathrm{~s}^{-1}\) What is the rate of formation of \(\mathrm{Br}_{(\mathrm{aq})}^{-}\)?

  1. A \(0.5 \mathrm{x} \mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
  2. B \(\mathrm{x} \mathrm{mol} \mathrm{dm}{ }^{-3} \mathrm{~s}^{-1}\)
  3. C \(2 \mathrm{x} \mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
  4. D \(\frac{3}{2} \mathrm{x} \mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\mathrm{x} \mathrm{mol} \mathrm{dm}{ }^{-3} \mathrm{~s}^{-1}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \text {Rate of consumption of }\left[\mathrm{OH}^{-}\right]=-\frac{\mathrm{d}\left[\mathrm{OH}^{-}\right]}{\mathrm{dt}} \\ & \text { Rate of formation of }\left[\mathrm{Br}^{-}\right] \\ & =\frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{OH}^{-}\right]}{\mathrm{dt}}=\mathrm{x} \mathrm{mold} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\end{aligned}\)