The specific reaction rate of a reaction quadruples when temperature changes from 30°C to 50°C. Calculate the energy of activation of the reaction [Given : R = 8.314 "JK"^(-1) "mol"^(-1) ]

The specific reaction rate of a reaction quadruples when temperature c

1.	The specific reaction rate of a reaction quadruples when temperature changes from 30°C to 50°C. Calculate the energy of activation of the reaction [Given : R = 8.314 "JK"^(-1) "mol"^(-1) ]
Answer» <P> Solution :(a) In first order reaction the rate of the reaction is proportional to first power of the concentration of the reactants. Let us CONSIDER the FOLLOWING first order reaction : `R to P` The rate is given by : `(-d[R])/(dt) prop [R]` `(-d[R])/(dt) = k[R]` where k is rate constant `(-d[R])/([R]) =kdt` multiply is given by -ve sign `(d[R])/([R]) = - kdt` on integration `int(d[R])/([R]) = -k int dt` `In [R] = -kt + I` Whent `t = 0, R = [R]_(0)`, where `[R]_0` = INITIAL concentration of the reactant `:. [R]_(0) = -k(0) + I "" .......(1)` `I = ln [R]_(0)` Substituting the value of I in the above eqn, `ln [R] = -kt + In [R]` `kt = In[R]_0 - In [R]` `kt-ln [R]_(0) - ln [R]` `kt = ln ([R]_0)/([R])` `k = 1/t ln ([R]_0)/([R])` Changing into common log `k = (2.303)/(t) "log" ([R]_0)/([R])"".........(2)` (b) `"log" (k_2)/(k_1) = (E_a)/(2.303 R) [(T_2 - T_1)/(T_1T_2)]` Given : `(k_2)/(k_1) = 4, T_1 = 30^@C = 30 + 273 = 303 K, T_2 = 50^@C = 50 + 273 = 273 = 323 K , R = 8.314 K^(-1) mol^(-1)` `"log" 4 = (E_a)/(2.303 xx 8.314) [(323 - 303)/(303 xx 323)] ` `E_a = ("log " 4 xx 2.303 xx 8.314 xx 303 xx 323)/(20 ) = 56,414 J//mol = 56.414 KJ//mol` .