In a reaction with second-order kinetics where molecules X convert to Y, increasing the concentration
of X to three times its original value will affect the rate of Y formation by which factor?

1. 6 times

2. 9 times

3. 12 times

4. 3 times

The rate of a chemical reaction doubles when the temperature increases by 10 K (from 298 K). What is the activation energy of the reaction?

1. 65.6 Jmol^-1

2. 52.9 kJmol^-1

3. 45.9 kJmol^-1

4. 35.7 Jmol^-1

The activation energy for the reaction 2HI(g) → H₂(g)+ I₂(g) is 209.5 kJ mol⁻¹ at 581K. What is the fraction of molecules of reactants having energy equal to or greater than activation energy?

$1. 2.31 \times {10}^{-16}$
$2. 3.52 \times {10}^{-18}$
$3. 1.47 \times {10}^{-19}$
$4. 2.13 \times {10}^{-20}$

The rate constant for a first order reaction is 60 s^–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16^th value?

$1. 2.3 \times {10}^{-2} \mathrm{s}$
$2. 4.6 \times {10}^{-2 }\mathrm{s}$
$3. 3.1 \times {10}^{-3} \mathrm{s}$
$4. 1.4 \times {10}^{-2} \mathrm{s}$

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:
1. 232.79 kJ mol^–1
2. 245.86 kJ mol^–1
3. 126.12 kJ mol^–1
4. 242.51 kJ mol^–1

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