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 \(\log 5=0.6989]\)
- A \(8\)
- B \(5\)
- C \(1\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
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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