${{\mathrm{I}}_2}_{\left(\mathrm{s}\right)}$ $+$ $5{{\mathrm{F}}_2}_{\left(\mathrm{g}\right)}$ $$ $\to 2{\mathrm{IF}}_{5\left(g\right)}$ $$

The equilibrium constant K_c expression for the above mentioned reaction is:

$1.$ ${\mathrm{K}}_{\mathrm{C}}=$ $\frac{{\left[{\mathrm{IF}}_5\right]}^2}{{\left[{\mathrm{F}}_2\right]}^5}$
$2.$ ${\mathrm{K}}_{\mathrm{C}}=$ $\frac{{\left[{\mathrm{IF}}_5\right]}^2}{{\left[{\mathrm{F}}_2\right]}^5\left[{\mathrm{I}}_2\right]}$
$3.$ ${\mathrm{K}}_{\mathrm{C}}$ $=\frac{{\left[{\mathrm{F}}_2\right]}^5\left[{\mathrm{I}}_2\right]}{{\left[{\mathrm{IF}}_2\right]}^2}$
$4.$ $$ ${\mathrm{K}}_{\mathrm{C}}$ $=\frac{{\left[{\mathrm{F}}_2\right]}^5}{{\left[{\mathrm{IF}}_5\right]}^2}$

${\mathrm{K}}_{{\mathrm{a}}_1},$ ${\mathrm{K}}_{{\mathrm{a}}_2}$ $\mathrm{and}$ ${\mathrm{K}}_{{\mathrm{a}}_3}$ are the respective ionisation constants for the following reactions.
\(\mathrm{H}_2 \mathrm{~S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}\)
\(\mathrm{HS}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-}\)
\(\mathrm{H}_2 \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}\)
The correct relationship between ${\mathrm{K}}_{{\mathrm{a}}_1},$ ${\mathrm{K}}_{{\mathrm{a}}_2}$ $\mathrm{and}$ ${\mathrm{K}}_{{\mathrm{a}}_3}$ is:

Reaction quotient for the reaction, ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$ is given by , $\mathrm{Q}$ $=$ $\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$ .The reaction will proceed from right to left if K_c value is:

In the reaction A(g) + 2B(g) ⇌ 2C(g) + D(g), the initial concentration of B is twice that of A and, at equilibrium, the concentrations of A and D are equal. The value of the equilibrium constant will be:

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 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 :

(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

The equilibrium reaction that doesn't have equal values for K_c and K_p is: 

1. \(2NO(g) \rightleftharpoons N_2(g) + O_2(g)\)
2. \(SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)\)
3. \(H_2(g) + I_2(g) \rightleftharpoons 2HI (g)\)
4. \(2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)\)

For the reaction ${\mathrm{N}}_2\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_2\left(\mathrm{g}\right)\leftrightharpoons 2\mathrm{NO}\left(\mathrm{g}\right)$ the equilibrium constant is K₁. The equilibrium constant is K₂ for the reaction  $2\mathrm{NO}\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_2\left(\mathrm{g}\right)$ $\leftrightharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right)$ $$
The value of K for the reaction given below will be:
${\mathrm{NO}}_2\left(\mathrm{g}\right)\leftrightharpoons \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)$ $+{\mathrm{O}}_2\left(\mathrm{g}\right)$ $$

1.  $\frac{1}{4}$ $\left(4\right)$ ${\mathrm{K}}_1$ ${\mathrm{K}}_2$

2.  ${\left[\frac{1}{{\mathrm{K}}_1{\mathrm{K}}_2}\right]}^{1/2}$

3.  $\frac{1}{\left({\mathrm{K}}_1{\mathrm{K}}_2\right)}$

4.  $\frac{1}{\left(2{\mathrm{K}}_1{\mathrm{K}}_2\right)}$

For the reaction 2NOCl_(g)⇔2NO_(g)+Cl_2(g), K_C at 427$°$C is \(3\times 10^{-6} \ mol\ L^{-1}\). The value of K_p will be :

1. $1.72\times {10}^{-4}$

2. $7.50\times {10}^5$

3. $2.50\times {10}^{-5}$

4. $2.50\times {10}^{-4}$