Search any question & find its solution
Question:
Answered & Verified by Expert
In a pseudo first order hydrolysis of ester in water, the following results were obtained:
\(\begin{array}{ccccc}t / s & 0 & 30 & 60 & 90 \\ \text { [Ester] } \text { mol L }^{-1} & 0.55 & 0.31 & 0.17 & 0.085\end{array}\)
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
\(\begin{array}{ccccc}t / s & 0 & 30 & 60 & 90 \\ \text { [Ester] } \text { mol L }^{-1} & 0.55 & 0.31 & 0.17 & 0.085\end{array}\)
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Solution:
2366 Upvotes
Verified Answer
(i) Average rate during the interval \(30-60 \mathrm{sec}\)
\(\begin{aligned}
&=\frac{C_2-C_1}{t_2-t_1} \\
&=\frac{0 \cdot 31-0 \cdot 17}{60-30}=\frac{0 \cdot 14}{30}=4.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} .
\end{aligned}\)
(ii) \(k^{\prime}=\frac{2 \cdot 303}{t} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\) where initial concentration of \(\mathrm{A},\left[\mathrm{A}_0\right]=0.55 \mathrm{M}\) when \(t=30 \mathrm{sec} \quad k_1^{\prime}=\frac{2.303}{30} \log \frac{0.55}{0.31}\) \(=1.92 \times 10^{-2} \mathrm{~s}^{-1}\)
when \(t=60 \sec k_2^{\prime}=\frac{2 \cdot 303}{60} \log \frac{0 \cdot 55}{0 \cdot 17}\)
\(=1.96 \times 10^{-2} \mathrm{~s}^{-1}\)
when \(t=90 \sec \quad k_3^{\prime}=\frac{2 \cdot 303}{90} \log \frac{0 \cdot 55}{0.085}\)
\(=2.07 \times 10^{-2} \mathrm{~s}^{-1}\)
\(\therefore\) Average
\(\begin{aligned}
k^{\prime} &=\frac{k_1^{\prime}+k_2^{\prime}+k_3^{\prime}}{3}=\frac{(1.91+1.96+2.07) \times 10^{-2}}{3} \\
&=1.98 \times 10^{-2} \mathrm{~s}^{-1} .
\end{aligned}\)
\(\begin{aligned}
&=\frac{C_2-C_1}{t_2-t_1} \\
&=\frac{0 \cdot 31-0 \cdot 17}{60-30}=\frac{0 \cdot 14}{30}=4.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} .
\end{aligned}\)
(ii) \(k^{\prime}=\frac{2 \cdot 303}{t} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\) where initial concentration of \(\mathrm{A},\left[\mathrm{A}_0\right]=0.55 \mathrm{M}\) when \(t=30 \mathrm{sec} \quad k_1^{\prime}=\frac{2.303}{30} \log \frac{0.55}{0.31}\) \(=1.92 \times 10^{-2} \mathrm{~s}^{-1}\)
when \(t=60 \sec k_2^{\prime}=\frac{2 \cdot 303}{60} \log \frac{0 \cdot 55}{0 \cdot 17}\)
\(=1.96 \times 10^{-2} \mathrm{~s}^{-1}\)
when \(t=90 \sec \quad k_3^{\prime}=\frac{2 \cdot 303}{90} \log \frac{0 \cdot 55}{0.085}\)
\(=2.07 \times 10^{-2} \mathrm{~s}^{-1}\)
\(\therefore\) Average
\(\begin{aligned}
k^{\prime} &=\frac{k_1^{\prime}+k_2^{\prime}+k_3^{\prime}}{3}=\frac{(1.91+1.96+2.07) \times 10^{-2}}{3} \\
&=1.98 \times 10^{-2} \mathrm{~s}^{-1} .
\end{aligned}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.