$2\mathrm{A}$ $\to$ $\mathrm{B}$ $+$ $\mathrm{C}$; It would be a zero-order reaction when:

1.	The rate of reaction is proportional to the square of the concentration of A
2.	The rate of reaction remains the same at any concentration of A
3.	The rate remains unchanged at any concentration of B and C
4.	The rate of reaction doubles if the concentration of B is doubled

If at a given instant, for the reaction 2N₂O₅ → 4NO₂ + O₂ rate and rate constant are 1.02 × 10^-4  and 3.4 × 10^-5 sec ^-1 respectively, then the concentration of ${\mathrm{N}}_2{\mathrm{O}}_5$ at that time will be: 

1. 1.732 mol L⁻¹

2. 3.0 mol L⁻¹

3. $1.02$ $\times$ ${10}^{-4}$ mol L⁻¹

4. $3.4$ $\times$ ${10}^5$ mol L⁻¹

When a biochemical reaction is carried out in a laboratory outside the human body in the absence of an enzyme, then the rate of reaction obtained is ${10}^{-6}$ times. The activation energy of a reaction in the presence of an enzyme is:

1. $\frac{6}{RT}$
2. P is required.
3. Different from $E_a$ obtained in the laboratory.
4. Data is insufficient.

The activation energy for a simple chemical reaction A → B is E_a in a forward direction. The activation energy for the reverse reaction:

1. Is negative of E_a

2. Is always less than E_a

3. Can be less than or more than E_a

4. Is always double of E_a

The reaction A → B follows first-order kinetics. The time taken for 0.8 mol of A to produce 0.6 mol of B is 1 hour. The time taken for the conversion of 0.9 mol of A to produce 0.675 mol of B will be: 

If the rate of the reaction is equal to the rate constant, the order of the reaction is:

The temperature dependence of the rate constant (k) of a chemical reaction is written in terms of the Arrhenius equation,
k = A.e^–E*/RT. The activation energy (E*) of the reaction can be calculated by plotting: 

1. $\mathrm{k}$ $\mathrm{vs}$ $\mathrm{T}$

2. $\mathrm{k}$ $\mathrm{vs}$ $\frac{1}{\log }$ $\mathrm{T}$

3. $\log$ $\mathrm{k}$ $\mathrm{vs}$ $\frac{1}{\mathrm{T}}$

4. $\log$ $\mathrm{k}$ $\mathrm{vs}$ $\frac{1}{\mathrm{log\; T}}$

The radioisotope, tritium $\left({}_1{}^3\mathrm{H}\right)$ has a half-life of 12.3 years. If the initial amount of tritium is 32 mg, how many milligrams of it would remain after 49.2 years:

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