InterviewSolution
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InterviewSolution
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In a pseudo first order hydrolysis of ester in water the following results 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. |
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Answer» (i) Average rate of reaction between the time interval , 30 to 60 seconds , = `(d ["Ester"])/("dt")` = `(0.31 - 0.17)/(60 - 30)` `= (0.14)/(30)` `= 4. 67 xx 10^(-3) "mol" L^(-1) s^(-1)` (ii) For a pseudo first order reaction , `k = (2.303)/(t) "log" ([R]_(0))/([R])` For t = 30 s , `k_(2) = (2.303)/(30) "log" (0.55)/(0.31)` `= 1.911 xx 10^(-2) s^(-1)` For t = 60 s `k_(2) = (2.303)/(60) "log" (0.55)/(0.17)` `= 1.957 xx 10^(-2) s^(-1)` For t = 90 s , `k_(3) = (2.303)/(90) "log" (0.55)/(0.085)` =`2.075 xx 10^(-2) s^(-1)` Then , average rate constant , `k = (k_(1) + k_(2) + k_(3))/(3)` = `((1.911 xx 10^(-2)) + (1.957 xx 10^(-2)) + (2.075 xx 10^(-2)))/(3)` `= 1.98 xx 10^(-2) s^(-1)` |
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