For a general reaction A $\to$ B, the plot of the concentration of A vs. time is given in the figure.
 

The slope of the curve will be:

1.	-k	2.	-k/2
3.	-k2	4.	-k/3

A first-order reaction takes 40 min for 30 % decomposition. The half life of the reaction will be: 

The rate constant for the decomposition of hydrocarbons is 2.418 × 10^–5 s^–1 at 546 K. If the energy of activation is 179.9 kJ/mol, the value of the pre-exponential factor will be:

$1.$ $4.0$ $\times$ ${10}^{12}$ ${\mathrm{s}}^{-1}$
$2.$ $7.8$ $\times$ ${10}^{-13}$ ${\mathrm{s}}^{-1}$
$3.$ $3.8$ $\times$ ${10}^{-12}$ ${\mathrm{s}}^{-1}$
$4.$ $4.7$ $\times$ ${10}^{12}$ ${\mathrm{s}}^{-1}$

For a reaction A → Product, with k = 2.0 × 10^–2 s^–1, if the initial concentration of A is 1.0 mol L^-1, the concentration of A after 100 seconds would be :

The decomposition of sucrose follows the first-order rate law. For this decomposition, t_1/2 is 3.00 hours. The fraction of a sample of sucrose that remains after 8 hours would be:

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:

The rate constant for the first-order decomposition of H₂O₂ is given by the equation: \(log \ k \ = \ 14.34 \ - \ 1.25 \ \times \ 10^{4}\frac{K}{T}\). The value of E_a for the reaction would be:

1. 249.34 kJ mol^-1

2. 242.64 J mol^-1

3. -275.68 kJ mol^-1

4. 239.34 kJ mol^-1

${}_{}$

1. Pseudo-first-order reaction

2. First-order reaction

3. Second order reaction

4. Third-order reaction

The rate constant for a first-order reaction is $4.606\times {10}^{-3}$ $s^{-1}$. The time required to reduce 2.0 g of the reactant to 0.2 g will be:

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