For the reaction, $\mathrm{aA}+\mathrm{bB}\to \mathrm{cC}+\mathrm{dD}$ the plot of log k vs $\frac{1}{T}$ is given below :

Find the temperature(K) at which the rate constant of the reaction is 10^–4s^–1 ?

(Rounded-off to the nearest integer) 

[Given: The rate constant of the reaction is ${10}^{-5}{\mathrm{s}}^{-1}\text{\hspace{0.17em}at\hspace{0.17em}}500\mathrm{K}\text{.]\hspace{0.17em}}$

The rate constant of a reaction increases by five times on increase in temperature from 27°C to 52°C. The value of activation energy in kJ mol^–1 is-

(Rounded-off to the nearest integer) [R = 8.314 J K^–1 mol^–1]

1. 50

2. 56

3. 52

4. 60

The rate constant (k) of a reaction is measured at different temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in kJ mol^–1 is:
(R is gas constant)

1.  2R
2.  R
3.  1/R
4.  2/R

The rate of a reaction is decreased by 3.555 times when the temperature was changed from 40°C to 30°C. The activation energy (in kJ ${\mathrm{mol}}^{-1}$) of the reaction is:
(Take R=8.314 J ${\mathrm{mol}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{K}}^{-1}$ In 3.555=1.268)

1. 100 kJ/mol
2. 120 kJ/mol
3. 95 kJ/mol
4. 108 kJ/mol

Two reactions, R₁ and R₂ have identical pre-exponential factors. Activation energy of R₁ exceeds that of R₂ by 10 kJ mol^–1. If k₁ and k₂ are rate constants for reactions R₁ and R₂ respectively at 300 K, then ln(k₂/k₁) is equal to:

(R = 8.314 J mol^–1 K^–1).

1. 6

2. 4

3. 8

4. 12

Which of the following statements is correct regarding the equation k = Ae^-Ea/RT ? 

1. k is the equilibrium constant 

2. A is adsorption factor 

3. E_a is the energy of activation 

4. R is the Rydberg constant