In a first-order reaction A $\to$ products, the concentration of the reactant decreases to 6.25 % of its initial value in 80 minutes. The value of the rate constant, if the initial concentration is 0.2 mole/litre, will be:

1. $2.17$ $\times$ ${10}^{-2}$ $mi{n}^{-1}$

2. $3.46$ $\times$ ${10}^{-2}$ $mi{n}^{-1}$

3. $3.46$ $\times$ ${10}^{-3}mi{n}^{-1}$

4. $2.16$ $\times$ ${10}^{-3}$ $mi{n}^{-1}$

If a reaction A + B $\to$ C is exothermic to the extent of 30 kJ/mol and the forward reaction has an activation energy of 70 kJ/mol, the activation energy for the reverse reaction will be:

1. 30 kJ/mol                                       

2. 40kJ/mol

3. 70 kJ/mol                                       

4. 100 kJ/mol

The decomposition of hydrocarbons follows the equation: k = (4.5 × 10¹¹s^–1) e^–28000K/T

The activation energy (E_a) for the reaction would be:
1. 232.79 kJ mol^–1
2. 245.86 kJ mol^–1
3. 126.12 kJ mol^–1
4. 242.51 kJ mol^–1

An increase in the concentration of the reactants of a reaction leads to a change in:

The plot of ln k vs \({1 \over T}\) for the following reaction 
\(2N_2O_5(g) \rightarrow 4NO_2 (g) + O_2(g) \) gives a straight line with the slope of the line equal to \(-1.0 \times 10^4 K \).
The activation energy for the reaction in J mol^–1 is:
(Given R = 8.3 J K^–1 mol^–1)

What does Z_AB represent in the collision theory of chemical reactions?

If a reaction A + B → C is exothermic to the extent of 30 kJ mol⁻¹ and the forward reaction has an activation energy of 249 kJ mol⁻¹, the activation energy for the reverse reaction in kJ mol^-1 will be: