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:
1. Size and shape of a pure substance are always fixed.
2. Volume of solids and liquids is relatively fixed.
3. Charges and masses of pure substances are always fixed.
4. All of the above

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

For the reaction: H_{2 (g)} + I_{2 (g)}  $\rightleftharpoons$2HI _(g) ; K_c = 54.8 at 700K .

If 0.5 mol L^–1 of HI_(g) is present at equilibrium at 700 K, the concentration of H_2(g) and I_2(g)  would be:

1. [H₂]= [I₂]=0.05 mol L^-1

2. [H₂]= 0.5 mol L^-1, [I₂] = 0.05 mol L^-1

3. [H₂] = 0.068 mol L^-1, [I₂]= 0.55 mol L^-1

4. [H₂] = [I₂] =0.068 mol L^-1

2ICl_(g) $\leftrightharpoons$I_2(g) + Cl_2(g); K_c = 0.14

Given the reaction:
CH₃COOH_(l) + C₂H₅OH_(l) ⇌ CH₃COOC₂H_5(l) + H₂O_(l)

At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. The equilibrium constant of the reaction will be:

(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