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MHT CET · Chemistry · Chemical Kinetics

The rate law for the reaction \(\mathrm{A}+\mathrm{B} \rightarrow\) product is given by rate \(=k[A][B]\) Calculate \([A]\) if rate of reaction and rate constant are \(0.25 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\) and \(6.25 \mathrm{~mol}^{-1} \mathrm{dm}^3 \mathrm{~s}^{-1}\) respectively \(\left[[\mathrm{B}]=0.25 \mathrm{~mol} \mathrm{dm}^{-3}\right]\)

  1. A \(0.22 \mathrm{~mol} \mathrm{dm}^3\)
  2. B \(0.16 \mathrm{~mol} \mathrm{dm}^3\)
  3. C \(0.30 \mathrm{~mol} \mathrm{dm}^3\)
  4. D \(0.25 \mathrm{~mol} \mathrm{dm}^3\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.16 \mathrm{~mol} \mathrm{dm}^3\)

Step-by-step Solution

Detailed explanation

\(\text {rate }=\mathrm{k}[\mathrm{A}][\mathrm{B}]\)
\(\therefore [\mathrm{A}]=\frac{\text { rate }}{\mathrm{k}[\mathrm{B}]} =\) \(\frac{0.25 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}}{6.25 \mathrm{~mol}^{-1} \mathrm{dm}^3 \mathrm{~s}^{-1} \times 0.25 \mathrm{~mol} \mathrm{dm}^{-3}}\)
\(=0.16 \mathrm{~mol} \mathrm{dm}^3\)