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)

1.	1.08 × 10–10 mol2 L–2	2.	1.08 × 10–12 mol2 L–2
3.	1.08 × 10–14 mol2 L–2	4.	1.08 × 10–8 mol2 L–2

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

Among the following, the correct order of acidity is:

1. HClO < HClO₂ < HClO₃ < HClO₄
2. HClO₂ < HClO < HClO₃ < HClO₄
3. HClO₄ < HClO₂ < HClO  < HClO₃
4. HClO₃ < HClO₄ < HClO₂ < HClO

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:

Consider the nitration of benzene using mixed conc. H₂SO₄ and HNO₃.
If a large amount of KHSO₄ is added to the mixture, the rate of nitration will be:

If the equilibrium constant for N₂(g) + O₂ (g) ⇄ 2NO(g) is K, the equilibrium constant for \(\frac{1}{2}\)N₂(g) + \(\frac{1}{2}\)O₂(g) ⇄ NO(g) will be?

1. $K^{\frac{1}{2}}$

2. $\frac{1}{2}K$

3. K

4. $K^2$

Aqueous solution of which of the following compounds is the best conductor of electric current?
1. Acetic acid, \(\mathrm{C_{2} H_{4} O_{2}}\)${\mathrm{}}_{}$
2. Hydrochloric acid, \(\mathrm{HCl}\)
3. Ammonia, \(\mathrm{N H_{3}}\)${\mathrm{}}_{}$
4. Fructose, \(\mathrm{C_{6} H_{12} O_{6}}\)${\mathrm{}}_{}$
 

The value of the equilibrium constant for a particular reaction is 1.6 × 10¹². When the system is in equilibrium, it will include:
1. All reactants
2. Mostly reactants
3. Mostly products
4. Similar amounts of reactants and products

For the reversible reaction:
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) + heat 

The equilibrium shifts in a forward direction:

1. by increasing the concentration of $N{H}_3\left(g\right)$
2. by decreasing the pressure.
3. by decreasing the concentration of $N_2\left(g\right)$ $and$ $H_2\left(g\right)$
4. by increasing pressure and decreasing temperature.

For a given exothermic reaction, K_p and K_p^’ are the equilibrium constants at temperatures T₁ and T₂ respectively. Assuming that the heat of reaction is constant in temperatures range between T₁ and T₂, it is a readily observation that:
1. $K_p>K_p^{\text{'}}$
2. $K_p<K_p^{\text{'}}$
3. $K_p=K_p^{\text{'}}$
4. $K_p=\frac{1}{K_p^{\text{'}}}$