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MHT CET · Chemistry · Chemical Kinetics

For the reaction \(2 \mathrm{~A}+2 \mathrm{~B} \rightarrow 2 \mathrm{C}+\mathrm{D}\), the rate law is expressed as rate \(=k[A]^2[B]\). Calculate the rate constant if rate of reaction is \(0.24 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\).
\(([\mathrm{A}]=0.5 \mathrm{M}\) and \([\mathrm{B}]=0.2 \mathrm{M}\) )

  1. A \(4.8 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
  2. B \(9.6 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
  3. C \(12.1 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
  4. D \(14.4 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(4.8 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)

Step-by-step Solution

Detailed explanation

Rate \(=k[A]^2[B]\)
\(\therefore \quad \mathrm{k}=\frac{\text { Rate }}{[\mathrm{A}]^2[\mathrm{~B}]}\)
\(=\frac{0.24 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}}{\left(0.5 \mathrm{~mol} \mathrm{dm}^{-3}\right)^2\left(0.2 \mathrm{~mol} \mathrm{dm}^{-3}\right)}\)
\(=4.8 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)