KCET · Chemistry · Chemical Kinetics
For the reaction, \(\mathrm{PCl}_5 \longrightarrow \mathrm{PCl}_3+\mathrm{Cl}_2\), rate and rate constant are \(1.02 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) and \(3.4 \times 10^{-5} \mathrm{~s}^{-1}\) respectively at a given instant. The molar concentration of \(\mathrm{PCl}_5\) at that instant is
- A \(8.0 \mathrm{~mol} \mathrm{~L}^{-1}\)
- B \(3.0 \mathrm{~mol} \mathrm{~L}^{-1}\)
- C \(0.2 \mathrm{~mol} \mathrm{~L}^{-1}\)
- D \(2.0 \mathrm{~mol} \mathrm{~L}^{-1}\)
Answer & Solution
Correct Answer
(B) \(3.0 \mathrm{~mol} \mathrm{~L}^{-1}\)
Step-by-step Solution
Detailed explanation
Given,
\(\mathrm{PCl}_5 \longrightarrow \mathrm{PCl}_3+\mathrm{Cl}_2\)
Rate \(=1.02 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
\(k=3.4 \times 10^{-5} \mathrm{~s}^{-1}\)
\(\left[\mathrm{PCl}_5\right]=?\)
For the given equation, rate will be given by
Rate \(=k\left[\mathrm{PCl}_5\right]\)
\(1.02 \times 10^{-4}=3.4 \times 10^{-5}\left[\mathrm{PCl}_5\right]\)
\(\left[\mathrm{PCl}_5\right]=\frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}\)
\(\left[\mathrm{PCl}_5\right]=3 \mathrm{~mol} \mathrm{~L}^{-1}\)
\(\mathrm{PCl}_5 \longrightarrow \mathrm{PCl}_3+\mathrm{Cl}_2\)
Rate \(=1.02 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
\(k=3.4 \times 10^{-5} \mathrm{~s}^{-1}\)
\(\left[\mathrm{PCl}_5\right]=?\)
For the given equation, rate will be given by
Rate \(=k\left[\mathrm{PCl}_5\right]\)
\(1.02 \times 10^{-4}=3.4 \times 10^{-5}\left[\mathrm{PCl}_5\right]\)
\(\left[\mathrm{PCl}_5\right]=\frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}\)
\(\left[\mathrm{PCl}_5\right]=3 \mathrm{~mol} \mathrm{~L}^{-1}\)
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