1.

The rate constants of a reaction at 500 K and 700 K are 0.02s^(-1)" and "0.07s^(-1) respectively. Calculate the values of E_(a) and A.

Answer»

SOLUTION :`log""(k_(2))/(k_(1))=(E_(a))/(2.303"R")((T_(2)-T_(1))/(T_(1)T_(2)))`
`log""(0.07)/(0.02)=(E_(a))/(2.303xx8.314" JK"^(-1)MOL^(-1))xx(700"K"-500"K")/(700"K"xx500"K")`,
`log3.5=(E_(a))/(2.303xx8.314" JK"^(-1)mol^(-1))xx(200"K")/(700"K"xx500"K")`
or `""E_(a)=(0.5441xx2.303xx8.314xx700xx500)/(200)"J mol"^(-1)=18231.4" J mok"^(-1)=18.23" kJ mol"^(-1)`
Further, `logk=-(E_(a))/(2.303"R")+LOGA" or"logA=logk+(E_(a))/(2.303"RT")`
Substituting `T=500 K, k=0.02s^(-1)`, we get
`logA=log0.02+(18231.4" J mol"^(-1))/(2.303xx8.314"JK"^(-1)mol^(-1)xx500"K")=-log50+12.4095=-1.699+1.904=0.205`
`A=" Antilog "(0.205)=1.603`


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