1.

If the rate constant of a first order reaction at a certain temperature is 1.5xx10^(-1)s^(-1) and t_(1) and t_(2) are the respective times for 50% and 75% completion of the reaction, determine the ratio of t_(2)" to "t_(1).

Answer»


Solution :`k=(2.303)/(k)LOG""(a)/(a-x)`
For `3//4` of the REACTION to OCCUR, `t=t_(3//4),(a-x)=a-3a//4=a//4:.t_(3//4)=(2.303)/(k)log4`
For HALF of a reaction to occur, `t=t_(1//2),(a-x)=a-a//2=a//2.:.t_(1//2)=(2.303)/(k)log""(a)/(a//2)=(2.303)/(k)LOG2`
Hence, `(t_(3//4))/(t_(1//4))=(log4)/(log2)=(0.6021)/(0.3010)=2.`


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