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 ${\mathrm{mol}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{K}}^{-1}$  and 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 

Consider the equilibrium reaction: A(g) ⇌ B(g) with an enthalpy change (ΔH) of -42 kJ/mol.

Determine the activation energies for the forward and backward reactions, given that the ratio of the activation energy of the forward reaction to the activation energy of the backward reaction is 2 : 3.