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Question: In a pseudo first order hydrolysis of ester in water, the following results were obtained. | t/s ...

In a pseudo first order hydrolysis of ester in water, the following results were obtained.

t/s0306090
Ester/molL−1molL^{- 1}0.550.550.310.310.170.170.0850.085

What will be the average rate of reaction between the time intervals 30 to 60 seconds?

A

1.91×10−2s−11.91 \times 10^{- 2}s^{- 1}

B

4.67×10−3molL−1s−14.67 \times 10^{- 3}molL^{- 1}s^{- 1}

C

1.98×10−3s−11.98 \times 10^{- 3}s^{- 1}

D

2.07×10−2s−12.07 \times 10^{- 2}s^{- 1}

Answer

4.67×10−3molL−1s−14.67 \times 10^{- 3}molL^{- 1}s^{- 1}

Explanation

Solution

Average rate during the time interval 30−60s30 - 60s

Rate=C2−C1t2−t1=(0.17−0.31)60−30=0.1430Rate = \frac{C_{2} - C_{1}}{t_{2} - t_{1}} = \frac{(0.17 - 0.31)}{60 - 30} = \frac{0.14}{30}

=4.67×10−3molL−1s−1= 4.67 \times 10^{- 3}molL^{- 1}s^{- 1}