Rate constant of reaction at 300 K and 400 K are 0.0345 S^(-1) and 0.1365 S^(-1) respectively. Calculate the activation energy for the reaction. [Given : R = 8.314 J K^(-1) mol^(-1)]

Rate constant of reaction at 300 K and 400 K are 0.0345 S^(-1) and 0.1

1.	Rate constant of reaction at 300 K and 400 K are 0.0345 S^(-1) and 0.1365 S^(-1) respectively. Calculate the activation energy for the reaction. [Given : R = 8.314 J K^(-1) mol^(-1)]
Answer» Solution :(A) `logK_(2)/K_(1) = E_(a)/2.303R[T_(2)-T_(1)/T_(1)T_(2)]` `log0.136/0.034 = E_(a)/2.303 xx 8.314[400-300/300xx400]` `E_(a)= 13.8 kJ//mol` (b) Consider a zero order reaction `R rarr P` Zero order reaction is one in which rate is proportional to zeroth power of reactant concentration. According to rate law for zero order reaction, `Rate = k[R]^(0)`Rate = `k xx 1` where k is rate CONSTANT or velocity constant. But rate is defined by, Rate `= -d[R]/dt:.-d[R]/dt = k` Rearranging the equation, we get`d[R] = -kdt` On integration ` int d[R] = -kint dt[R] = -kt + I` where I is called integration constant To get, I, when t = 0,`[R] = [R]_(0)` the initial concentration of the reactant `[R]_(0) = -k xx 0 + II = [R]_(0)` Substitute I VALUE in equation(1), we get `[R] = -kt + [R]_(0)[R]_(0)-[R] = kt` `k = [R]_(0) - [R]/t` On rearranging, `[R] = -kt + [R]_(0)` This equation is in y = mx + c FORM and suggests that A plot of [R] versus .t. gives a STRAIGHT LINE with a slope = -k and y-intercept =` [R]_(0)`