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 \hspace{0.17em}$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

At a certain temperature and pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms. The K_p for the equilibrium of the reaction would be:

 $I_2<mspace\; width="1em"></mspace>\left(g\right)<mspace\; width="1em"></mspace>\leftrightharpoons <mspace\; width="1em"></mspace>2I<mspace\; width="1em"></mspace>\left(g\right)$

1. 2.67 $\times$ 10⁴ Pa

2. 1.00 $\times$ 10⁵ Pa

3. 3.63 $\times$ 10⁴ Pa

4. 2.18 $\times$ 10⁵ Pa

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

For the reaction, 2NOCl (g) $\rightleftharpoons$ 2NO (g) + Cl₂ (g); K_p= 1.8 × 10^–2 atm at 500 K.

The value of K_c for above mentioned reaction would be:

$1.$ $4.33$ $\times$ ${10}^{-4}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$
$2.$ $4.33$ $\times$ ${10}^4$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$
$3.$ $1.65$ $\times$ ${10}^{-5}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$
$4.$ $2.39$ $\times$ ${10}^{-3}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$

For the following equilibrium, K_c= 6.3 × 10¹⁴ at 1000 K

$\mathrm{NO}\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_3$ $\left(\mathrm{g}\right)$ $\to$ ${\mathrm{NO}}_2\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_2$ $\left(\mathrm{g}\right)$ $$

The value of K_c for the reverse reaction is:

$1.$ $2.33$ $\times$ ${10}^{-16}$
$2.$ $1.59$ $\times$ ${10}^{-15}$
$3.$ $2.67$ $\times$ ${10}^{-13}$
$4.$ $4.47$ $\times$ ${10}^{14}$

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

For the equilibrium reaction:

2N₂(g) + O₂(g) ⇌ 2N₂O(g)

A mixture containing 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L vessel and allowed to reach equilibrium. If the equilibrium constant is:

Kc = 2.0 × 10⁻³⁷ L mol⁻¹

Calculate the equilibrium concentration of N₂O.

Options:

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 of volume 1 litre 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