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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
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Question : 8 of 39
Marks: +1, -0
In a pseudo first order hydrolysis of ester in water, the following results were obtained :
  t / s  0  30  60  90
  [ Ester] /mol L−1^{-1}  0.55  0.31  0.17  0.085
(i) Calculate the average rate of reaction between the time interval 30 to 60seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis ofester.
Solution:  
(i) Average rate during the interval 30-60 sec
Rate =−C2−C1t2−t1= -\frac{C_{2}-C_{1}}{t_{2}-t_{1}}
=−0.17−0.3160−30= -\frac{0.17-0.31}{60-30}
=0.1430 mol L−1 s−1= \frac{0.14}{30} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}
=4.67×10−3 mol L−1 s−1= 4.67 \times 10^{-3} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}
(ii) Using formula, k=2.303tlog⁡[R0][R]k = \frac{2.303}{t} \log \frac{[R_{0}]}{[R]} in which [R0]=0.55 M[R_{0}] = 0.55 \, \text{M}
At t=30 sec,[R]=0.31 mol L−1t = 30 \, \text{sec}, [R] = 0.31 \, \text{mol} \, \text{L}^{-1}
∴ k′=2.30330log⁡0.550.31\therefore\, k^{\prime} = \frac{2.303}{30} \log \frac{0.55}{0.31}
=1.91×10−2 s−1= 1.91 \times 10^{-2} \, \text{s}^{-1}
At t=60 sec,[R]=0.17 mol L−1t = 60 \, \text{sec}, [R] = 0.17 \, \text{mol} \, \text{L}^{-1}
∴ k′=2.30360log⁡0.550.17\therefore\, k^{\prime} = \frac{2.303}{60} \log \frac{0.55}{0.17}
=1.96×10−2 s−1= 1.96 \times 10^{-2} \, \text{s}^{-1}
At t=90 sec,[R]=0.085 mol L−1t = 90 \, \text{sec}, [R] = 0.085 \, \text{mol} \, \text{L}^{-1}
∴ k′=2.30390log⁡0.550.085\therefore\, k^{\prime} = \frac{2.303}{90} \log \frac{0.55}{0.085}
=2.07×10−2 s−1= 2.07 \times 10^{-2} \, \text{s}^{-1}
Hence, average K=1.91+1.96+2.073×10−2K = \frac{1.91+1.96+2.07}{3} \times 10^{-2}
=1.98×10−2 s−1= 1.98 \times 10^{-2} \, \text{s}^{-1}
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