In a reaction, A + B → Product, the rate is doubled when the concentration of B is doubled, and the rate increases by a factor of 8, when the concentrations of both the reactants (A and B) are doubled. The rate law for the reaction can be written as:

1. Rate = k[A][B]²

2. Rate = k[A]²[B]²

3. Rate = k[A][B]

4. Rate = k[A]²[B]

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

When the initial concentration of the reactant is doubled,
the half-life period of a zero-order reaction:

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

If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:

1. t = 2.303/k

2. t = 0.693/k

3. t = 6.909/k

4. t = 4.606/k

The half-life for a zero-order reaction having 0.02 M initial concentration of reactant is 100 s. The rate constant (in mol L^–1 s^–1) for the reaction is:

1. $1.0\times {10}^{-4}$

2. $2.0\times {10}^{-4}$

3. $2.0\times {10}^{-3}$

4. $1.0\times {10}^{-2}$