Mechanism of a hypothetical reaction

${\mathrm{X}}_2 + {\mathrm{Y}}_2 \stackrel{ }{\to } 2\mathrm{XY}$ is given below

(i) ${\mathrm{X}}_2 \stackrel{ }{\rightleftharpoons } \mathrm{X} + \mathrm{X} \left(\mathrm{fast}\right)$

(ii) $\mathrm{X} + {\mathrm{Y}}_2 \stackrel{ }{\to } \mathrm{XY} + \mathrm{Y} \left(\mathrm{slowest}\right)$

(iii) $\mathrm{X} + \mathrm{Y} \stackrel{ }{\to }\mathrm{XY} \left(\mathrm{fast}\right)$

The overall order of the reaction will be

1. 1

2.  2

3. 0

4. 1.5

A first order reaction has a specific reaction rate of 10^-2s^-1. How much time will it take for 20 g of the reactant to reduce to 5 g?

(a) 238.6 s      (b) 138.6 s        (c) 346.5  s        (d) 693.0 s

The decomposition of phosphine (PH₃) on tungsten at low pressure is a first-order reaction. It is because the

(1) rate is proportional to the surface coverage

(2) rate is inversely proportional to the surface coverage

(3) rate is independent of the surface coverage

(4) rate of decomposition is very slow

The rate of a first-order reaction is 0.04 mol L^-1 s^-1 at 10 sec and 0.03 mol L^-1 s^-1 at 20 sec after initiation of the reaction. The half-life period of the reaction is

1. 34.1 s

2. 44.1 s

3. 54.1 s

4. 24.7 s

The activation energy of a reaction can be determined from the slope of the graph between : 

1. ln K vs T

2. ln $\frac{\mathrm{K}}{\mathrm{T}}$ vs T

3. ln K vs $\frac{1}{\mathrm{T}}$

4. $\frac{\mathrm{T}}{\ln \mathrm{K}} vs \frac{1}{T}$

If the half-life is independent of its initial concentration, then the order of the reaction is:

1. 0 

2. 1

3. 3

4. 2

The rate constant of the reaction A $\to$ B is 0.6 x 10^-3 molar per second. If the

concentration of A is 5 M then concentration of B after 20 min is

1. 1.08 M

2. 3.60 M

3. 0.36 M

4. 0.72 M

If the rate of reaction doubles when the temperature is raised from 20 °C to 35 °C, then the activation energy for the reaction will be :
(R = 8.314 J mol^-1 K^-1)

1. 342 kJ mol^-1

2. 269 kJ mol^-1

3. 34.7 kJ mol^-1

4. 15.1 kJ mol^-1

A reaction having equal energies of activation for forward and reverse reactions has

1. $∆$S = 0

2. $∆$G = 0

3. $∆$H = 0

4. $∆$H = $∆$G = $∆$S = 0

For the reaction,

N₂O₅(g) → 2NO₂(g) + \(\frac{1}{2}\)O₂(g)

the value of the rate of disappearance of ${\mathrm{N}}_2{\mathrm{O}}_5$ is given as $6.\mathrm{25 }\times {10}^{-3}$ $mol$ $L^{-1}{s}^{-1}$. The rate of formation of $N{O}_2$ $and$ $O_2$ is given respectively as:

1. 6.25 x 10^-3 mol L^-1s^-1 and 6.25 x 10^-3 mol L^-1s^-1

2. 1.25 x 10^-2 mol L^-1s^-1 and 3.125 x 10^-3 mol L^-1s^-1$\mathrm{}{\mathrm{}}^{}$

3. 6.25 x 10^-3 mol L^-1s^-1 and 3.125 x 10^-3 mol L^-1s^-1

4. 1.25 x 10^-2 mol L^-1s^-1 and 6.25 x 10^-3 mol L^-1s^-1