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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

\(\mathrm{PCl}_{5}(g) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) In the above first order reaction the concentration of \(\mathrm{PCl}_{5}\) reduces from initial concentration \(50\, mol\,\mathrm{L}^{-1}\) to \(10\, \mathrm{~mol} \,\mathrm{~L}^{-1}\) in \(120\, minutes\) at \(300\, \mathrm{~K}\). The rate constant for the reaction at \(300\, \mathrm{~K}\) is \(\mathrm{X}\) \(\times 10^{-2} \mathrm{~min}^{-1}\). The value of \(x\) is \(......\) \([\) Given \(\log 5=0.6989]\)

  1. A \(8\)
  2. B \(5\)
  3. C \(1\)
  4. D \(4\)
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Answer & Solution

Correct Answer

(C) \(1\)

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Detailed explanation

\(\mathrm{PCl}_{5}(\mathrm{~g}) \frac{1 \text { order }}{300 \mathrm{~K}} \mathrm{PCl}_{3(g)}+\mathrm{Cl}_{2(g)}\) \(\mathrm{t}=0 \quad\quad\quad\quad 50 \mathrm{M}\) \(\mathrm{t}=120 \min \quad10 \mathrm{M}\)…
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