The half-life for radioactive decay of ¹⁴C is 5730 years. A wood sample contains only 80% of the ¹⁴C. The age of the wood sample would be-

1. 1898 years

2. 1765 years

3. 1931 years

4. 1860 years

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

The decomposition of A into the product has a value of k as 4.5 × 10³ s^–1 at 10°C and energy of activation of 60 kJ mol^–1. The temperature at which the rate constant becomes 1.5 × 10⁴ s^–1   would be -

1. \(24 \text { K} \)
2. \(24 ^\circ \text { C} \)
3. \(31 ^\circ \text { C} \)
4. \(38 ^\circ \text { C} \)

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. The energy of activation of the reaction would be -

1. 65.93 kJ mol^–1

2. 52.85 kJ mol^–1

3. 55.46 kJ mol^–1

4. 60.93 kJ mol^–1

The decomposition of NH₃ on a platinum surface is a zero-order reaction. The rates of production of N₂ and H₂ will be respectively:
(given ; k = 2.5 × 10^–4 mol^–1 L s^–1 ) 

The rate equation of a reaction is expressed as, Rate = \(k(P_{CH_{3}OCH_{3}})^{\frac{3}{2}}\)

(Unit of rate = bar min^–1)

The units of the rate constant will be:
1. bar^1/2 min    
2. bar² min^–1   
3. bar^–1 min^–2  
4. bar^–1/2 min^–1