For the following reaction, 
 H$$₂(g)+I₂(g)⇌2HI(g) $at$ $250°C,$

The effect on the state of equilibrium on doubling the volume of the system will be:

Given an endothermic reaction:

CH_{4 (g)} + H₂O _(g)  $\leftrightharpoons$CO _(g) + 3H_{2 (g)}

If the temperature is increased, then 

1. The equilibrium will not be disturbed.

2. The equilibrium will shift in the backward direction.

3. The equilibrium will shift in the forward direction.

4. None of the above

Given a hypothetical reaction :
${\mathrm{AB}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{B}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{AB}}_3\left(\mathrm{g}\right);$ $∆\mathrm{H}=-\mathrm{x}$ $\mathrm{kJ}$
More AB₃​ could be produced at equilibrium by :

Mark the conditions that favour the maximum product formation in the given reaction.
 $A_2$ $\left(g\right)$ $$ $+$ $B_2$ $\left(g\right)$ $$ $\leftrightharpoons$ $X_2$ $(g),$ ${∆}_{r}H$ $=$ $-X$ $k$ $J$ $$ 

1. Low temperature and High pressure
2. Low temperature and Low pressure
3. High temperature and High pressure
4. High temperature and Low pressure

The addition of a small amount of argon at a constant volume will not affect the equilibrium of the following reaction:

Consider the following reactions.

(i) COCl_{2 (g)}  $\leftrightharpoons$CO _(g) + Cl_{2 (g)}

(ii) CO_{2 (g)} + C _(s) $\leftrightharpoons$ 2CO _(g)

(iii) 2H_{2 (g)} + CO _(g) $\leftrightharpoons$ CH₃OH _(g)

Which reaction(s) will shift in a backward manner when the pressure is increased?

1. Only (iii)

2. Only (ii)

3. Both (i) and (ii) 

4. None of the above

(a) PCl_{5 (g)} $\leftrightharpoons$ PCl_{3 (g)} + Cl_{2 (g)}

(b) CaO _(s) + CO_{2 (g)} $\leftrightharpoons$  CaCO_{3 (s)}

(c) 3Fe _(s) + 4H₂O _(g) $\leftrightharpoons$ Fe₃O_{4 (s)} + 4H_{2 (g)}

The effect of an increase in the volume on the number of moles of products in the above-mentioned reactions would be, respectively:

1. a) Increase,  b) decrease, c) same

2. a) Decrease,  b) same, c) increase

3. a) Increase,  b) increase, c) same

4. a) Increase,  b) decrease, c) increase

For the reaction,

  PCl₅(g) $\rightleftharpoons$PCl₃(g) + Cl₂ (g)

The forward reaction at constant temperature is favoured by:

1. Introducing an inert gas at a constant volume

2. Introducing chlorine gas at a constant volume

3. Introducing an inert gas at constant pressure

4. None of the above

Inert gas is added to the equilibrium ${\mathrm{PCl}}_5(\mathrm{g}) \rightleftharpoons {\mathrm{PCl}}_3(\mathrm{g})$ $+$ ${\mathrm{Cl}}_2\left(\mathrm{g}\right)$ at constant pressure. The degree of dissociation will :

1. Remain unchanged

2. Decrease

3. Increase

4. Decrease or increase but cannot be predicted with certainty

According to Le-Chatelier's principle, adding heat to a solid $\rightleftharpoons$liquid equilibrium will cause the :

1. Temperature to increase

2. Temperature to decrease

3. Amount of liquid to decrease

4. Amount of solid to decrease