In a pseudo first order hydrolysis of ester in water the following result6s were obtained:
`{:(t//s,0,30,60,90),(["Ester"],0.55,0.31,0.17,0.085):}`
(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.

Options

A. `1.91 xx 10^(-2) s^(-1)`
B. `4.67 xx 10^(-3) mol L^(-1) s^(-1)`
C. `1.98 xx 10^(-3) s^(-1)`
D. `2.07 xx 10^(-2) s^(1)`

Explanation

To solve the given problem step by step, we will break it down into two parts as indicated in the question.

### Part (i): Calculate the average rate of reaction between the time interval 30 to 60 seconds.

1. **Identify the concentrations at the given times:**
- At `t = 30 seconds`, the concentration of ester, \([E]\) = 0.31 M
- At `t = 60 seconds`, the concentration of ester, \([E]\) = 0.17 M

2. **Calculate the change in concentration:**
\[ \Delta [E] = [E]_{30s} - [E]_{60s} = 0.31 - 0.17 = 0.14 \, \text{M} \]

3. **Calculate the change in time:**
\[ \Delta t = 60s - 30s = 30s \]

4. **Calculate the average rate of reaction:**
\[ \text{Average Rate} = \frac{\Delta [E]}{\Delta t} = \frac{0.14 \, \text{M}}{30 \, \text{s}} = 4.67 \times 10^{-3} \, \text{M/s} \]

### Part (ii): Calculate the pseudo first order rate constant for the hydrolysis of ester.

1. **Use the first-order rate constant formula:**
The formula for the pseudo first-order rate constant \(k\) is given by:
\[ k = \frac{2.303}{t} \log\left(\frac{[E]_0}{[E]}\right) \]
where \([E]_0\) is the initial concentration and \([E]\) is the concentration at time \(t\).

2. **Calculate \(k\) at different time intervals:**
- **At \(t = 30\) seconds:**
\[ k_{30} = \frac{2.303}{30} \log\left(\frac{0.55}{0.31}\right) \]
\[ k_{30} = \frac{2.303}{30} \log(1.774) \approx 1.91 \times 10^{-2} \, \text{s}^{-1} \]

- **At \(t = 60\) seconds:**
\[ k_{60} = \frac{2.303}{60} \log\left(\frac{0.55}{0.17}\right) \]
\[ k_{60} = \frac{2.303}{60} \log(3.235) \approx 1.96 \times 10^{-2} \, \text{s}^{-1} \]

- **At \(t = 90\) seconds:**
\[ k_{90} = \frac{2.303}{90} \log\left(\frac{0.55}{0.085}\right) \]
\[ k_{90} = \frac{2.303}{90} \log(6.471) \approx 2.07 \times 10^{-2} \, \text{s}^{-1} \]

3. **Calculate the average rate constant \(k\):**
\[ k = \frac{k_{30} + k_{60} + k_{90}}{3} = \frac{1.91 \times 10^{-2} + 1.96 \times 10^{-2} + 2.07 \times 10^{-2}}{3} \]
\[ k \approx 1.98 \times 10^{-2} \, \text{s}^{-1} \]

### Final Answers:
- (i) The average rate of reaction between 30 to 60 seconds is \(4.67 \times 10^{-3} \, \text{M/s}\).
- (ii) The pseudo first order rate constant for the hydrolysis of ester is \(1.98 \times 10^{-2} \, \text{s}^{-1}\).