₉₂U²³⁵, nucleus absorb a neutron and disintegrate in ₅₄Xe¹³⁹, ₃₈Sr⁹⁴ and x . The product x is:

1. 3 - neutrons

2. 2 - neutrons

3. α - particle

4. β - particle

$\mathrm{In\; the\; reaction,\; 3}\mathrm{A}\to 2\mathrm{B\; the}$ rate of reaction $\frac{+\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}$ $$ is equal to:-

1. $-\frac{3}{2}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

2. $-\frac{2}{3}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

3. $-\frac{1}{3}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

4. $+2\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$

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

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 the forward reaction A → B is Ea.
Which of the following statements about the activation energy for the reverse reaction is correct?

1. It is the negative of Ea

2. It is always less than Ea

3. It can be less than or more than Ea

4. It is always double of Ea

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: