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 :

1.	Using a catalyst	2.	Removing some of B2 gas
3.	Increasing the temperature	4.	Increasing the pressure

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

At room temperature, MY and NY₃, two nearly insoluble salts, have the same K_sp values of 6.2 × 10^-13. The true statement regarding MY and NY₃ is:

The solubility of BaSO₄ in water is $2.42$ $\times$ ${10}^{-3}$ g/ litre at 298 K. The value of the solubility product will be: (Molar mass of BaSO₄ = 233 gmol^–1)

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. $2.50\times {10}^{-4}$

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:

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:

The number of ${\mathrm{H}}^{\oplus }$ ions present in 1 ml of a solution whose pH= 4 , is given as:

($N_A=$ $Avogadro\text{'}s$ $number$)

1. ${10}^{-7}{N}_A$

2. ${10}^{-8}{N}_A$

3. ${10}^{-16}{N}_A$

4. ${10}^{-14}{N}_A$

If the pH of an acidic buffer is 5.7 and ${\mathrm{pK}}_{\mathrm{a}}$ is 5 then the ratio of $\frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$ will be: