KCET · Chemistry · Chemical Kinetics
The rate of a gaseous reaction is given by the expression \(k[A][B]^{2}\). If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is
- A \(\frac{1}{16}\)
- B \(\frac{1}{8}\)
- C 8
- D 16
Answer & Solution
Correct Answer
(C) 8
Step-by-step Solution
Detailed explanation
Given, rate law expression \((r)=k[A][B]^{2}\)
Final volume \(\left(V_{2}\right)\) is reduced to half of initial volume \(\left(V_{1}\right)\) i.e. \(V_{2}=1 / 2 V_{1}\)
\(\therefore\) Final concentration \([A]\) and \([B]=[2 A]\) and \(2[B]\)
\(\therefore\) Final rate \(r^{\prime}=k[A]^{0}[B]^{2}=k[2 A][2 B]^{2}\)
\(\begin{aligned}
& r^{\prime}=k \times 2 \times 4[A][B]^{2} \Rightarrow r^{\prime}=8 k[A][B]^{2} \\
\therefore \quad & r^{\prime}=8 r
\end{aligned}\)
i.e. rate increases by 8 times.
Final volume \(\left(V_{2}\right)\) is reduced to half of initial volume \(\left(V_{1}\right)\) i.e. \(V_{2}=1 / 2 V_{1}\)
\(\therefore\) Final concentration \([A]\) and \([B]=[2 A]\) and \(2[B]\)
\(\therefore\) Final rate \(r^{\prime}=k[A]^{0}[B]^{2}=k[2 A][2 B]^{2}\)
\(\begin{aligned}
& r^{\prime}=k \times 2 \times 4[A][B]^{2} \Rightarrow r^{\prime}=8 k[A][B]^{2} \\
\therefore \quad & r^{\prime}=8 r
\end{aligned}\)
i.e. rate increases by 8 times.
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