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

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 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

In a zero order reaction for every 10° rise of temperature, the rate is doubled. If the

temperature is increased from 10°C to 100°C, the rate of the reaction will become

1. 256 times

2. 512 times

3. 64 times

4. 128 times

Half-life period of a first order reaction is 1386s. The specific rate constant of the reaction

is

1. 5.0 x 10^-3s^-1

2. 0.5 x 10^-2s^-1

3. 0.5 x 10^-3s^-1

4. 5.0 x 10^-2s^-1

In the reaction, BrO^-₃(aq) + 5Br^-(aq) + 6H⁺ $\to$3Br₂(l) + 2H₂O(l)

The rate of appearance of bromine (Br₂) is related to rate of disappearance of bromide

ions as following 

1. d[Br₂]/dt = -(3/5)d[Br^-]/dt

2. d[Br₂]/dt = (5/3)d[Br^-]/dt

3. d[Br₂]/dt = -(5/3)d[Br^-]/dt

4. d[Br₂]/dt = (3/5)d[Br^-]/dt

Consider the reaction

N₂(g) +3H₂(g) $\to$2NH₃(g)

The equality relationship between $\frac{\mathrm{d}\left[{\mathrm{NH}}_3\right]}{\mathrm{dt}}$ and $-\frac{\mathrm{d}\left[{\mathrm{H}}_2\right]}{\mathrm{dt}}$ is:

$$\begin{gathered} 1. +\frac{\mathrm{d}\left[{\mathrm{NH}}_3\right]}{\mathrm{dt}} = -\frac{1 \mathrm{d}[\mathrm{H}{}_2]}{3 \mathrm{dt}} \\ 2. +\frac{\mathrm{d}\left[{\mathrm{NH}}_3\right]}{\mathrm{dt}} = -\frac{2 \mathrm{d}\left[{\mathrm{H}}_2\right]}{3 \mathrm{dt}} \\ 3. +\frac{\mathrm{d}[\mathrm{NH}{}_3]}{\mathrm{dt}} = -\frac{3 \mathrm{d}\left[{\mathrm{H}}_2\right]}{2 \mathrm{dt}} \\ 4. +\frac{\mathrm{d}\left[{\mathrm{NH}}_3\right]}{\mathrm{dt}} = -\frac{\mathrm{d}\left[{\mathrm{H}}_2\right]}{\mathrm{dt}} \end{gathered}$$

The rate constant for a reaction of zero-order in A is 0.0030 mol L^-1 s^-1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M?

1. 8.3 sec

2. 0.83 sec

3. 83 sec

4. 10.3 sec

What is the activation energy for reverse reaction on the basis of given data ?

N₂O₄(g) $\to$ 2NO₂(g)  ΔH = +54kJ

E_a(forward) = +57.2 kJ

1. -54 kJ

2.  +3.2 kJ

3. 60.2 kJ

4. 111.2 kJ