For a reaction , ${\mathrm{BaO}}_2\left(\mathrm{s}\right)\rightleftharpoons \mathrm{BaO}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$; ∆H = + ve. At equilibrium condition, the pressure of O₂ depends on the:

1. Increased mass of BaO₂

2. Increased mass of BaO

3. Increased temperature on equilibrium.

4. Increased mass of BaO₂ and BaO both.

The compound with the highest pH among the following is:

1. CH₃COOK

2. Na₂CO₃

3. NH₄Cl

4. NaNO₃

If a solution of 0.1 N NH₄OH and 0.1 N NH₄Cl has pH 9.25, then pK_b of NH₄OH is:

1. 9.25

2. 4.75

3. 3.75

4. 8.25

A compound, among the following, that cannot be classified as a protonic acid is:

If the solubility product of CuS is 6 × 10^–16, the maximum molarity of CuS in an aqueous solution will be:
1. 1.45 × 10⁻⁸ mol L⁻¹
2. 3.45 × 10⁻⁸ mol L⁻¹
3. 2.45 × 10⁻⁸ mol L⁻¹
4. 4.25 × 10⁻⁸ mol L⁻¹

The volume of the container containing a liquid and its vapours at a constant temperature is suddenly increased. What would be the effect of the change on vapour pressure?
1. It would decrease initially.
2. It would increase initially.
3. It would remain the same.
4. None of the above

For the following reaction, 

2SO₂(g) + O₂(g) $\leftrightharpoons$2SO₃(g)

The value of K_c at equilibrium with a concentration of [SO₂]= 0.60M,[O₂] = 0.82M and [SO₃] = 1.90M

would be:
1. 8.5
2. 9.4
3. 12.2
4. 16.3

Pure liquids and solids are ignored while writing the expression for the equilibrium constant because:

Given the reaction,
\(2 \mathrm{~N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \leftrightharpoons 2 \mathrm{~N}_2 \mathrm{O}_{(\mathrm{g})}\)
If a mixture of 0.482 mol N₂ and 0.933 mol of O₂ is placed in a 10 L vessel and allowed to form N₂O (K_c= 2.0 × 10^–37 L mol^–1), the concentration of N₂O at equilibrium will be:
1. \(6.6 \times 10^{-21} \mathrm{M} \)
2. \(0.6 \times 10^{-21} \mathrm{M} \)
3. \(4.6 \times 10^{-11} \mathrm{M} \)
4. \(3.6 \times 10^{-31} \mathrm{M}\)

For a reaction, 2NO _(g) + Br_{2 (g)}  $\leftrightharpoons$2NOBr _(g)

When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. The concentration of NO and Br₂ at equilibrium will be:

1. NO = 0.0352 mol; ${\mathrm{Br}}_2$= 0.0178 mol

2. NO = 0.352 mol; ${\mathrm{Br}}_2$= 0.178 mol

3. NO = 0.0634 mol; ${\mathrm{Br}}_2$= 0.0596 mol

4. NO = 0.634 mol; ${\mathrm{Br}}_2$= 0.596 mol