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A plot of (1)/(T) Vsk for a reaction giv...

A plot of `(1)/(T)` Vsk for a reaction gives the slope `- 1 xx 10^(4)K`. The energy of activation for the reaction is
(Given `R = 8.314 JK^(-1) " mol"^(-1)`)

A

`8314 "J mol"^(-1)`

B

`1.202 "kJ mol"^(-1)`

C

`12.02 "J mol"^(-1)`

D

`83.14 "kJ mol"^(-1)`

Text Solution

Verified by Experts

Correction in the quenstion: It should be ln k vs 1/T
Plot of ln k vs 1/T gives
Slope `= - (E_(a))/(R)`
`:.-1xx10^(4)= - (E_(a))/(8.314)`
`E_(a)= 8.314xx10^(4)"J mol"^(-1)= 83.14xx10^(3)"J mol"^(-1)`
or `E_(a)= 83.14 "kJ mol"^(-1)`
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a) The rate of a particular reaction doubles when the temperature changes from 300 K to 310 K. 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 changes from 30°C to 50°C. Calculate the energy of activation of the reaction [Given : R = 8.314 "JK"^(-1) "mol"^(-1) ]

(a) Derive an integrated rate equation for the rate constant of a first order reaction. (b) The specific reaction rate of a reaction quadruples when temperature changes from 30^(0) to 50^(0) .Calculate the energy of activation of the reaction. [Given : R = 8.314 JK^(-1)mol^(-1)] .

Show that R=8.314 Jmol^(-1)K^(-1)

The ratio of rate constants of a reaction at 300K and 291K is 2. Calculate the energy of activation. ("Given "R = "8.314JK"^(-1)" mol"^(-1)) .

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