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:

 $I_2 \left(g\right) \leftrightharpoons 2I \left(g\right)$

1. 2.67 $\times$ 10⁴ Pa

2. 1.00 $\times$ 10⁵ Pa

3. 3.63 $\times$ 10⁴ Pa

4. 2.18 $\times$ 10⁵ Pa

Given the reaction 2HI _(g) $\rightleftharpoons$ H_2 (g) + I_2 (g)

A sample of HI_(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI_(g) is 0.04 atm.  The ${\mathrm{K}}_{\mathrm{P}}$ for the given equilibrium would be:

A mixture of 1.57 mol of N₂, 1.92 mol of H_2, and 8.13 mol of NH₃ is introduced into a 20 L vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction N_2 (g) + 3H_2 (g) $\rightleftharpoons$ 2NH_3 (g) is 1.7 × 10².

The direction of the net reaction is:

1. Reaction is at equilibrium.

2. Reaction will proceed in forwarding direction.

3. Reaction will proceed in the backward direction.

4. Data is not sufficient.

Given the reaction:

2BrCl _(g) $\leftrightharpoons$ Br_2 (g) + Cl_2 (g); K_c= 32 at 500 K. If the initial concentration of BrCl is 3.3 × 10^-3 mol L^–1, the molar concentration of BrCl in the mixture at equilibrium would be:

1. 3.0 $\times$ 10^-2 molL^-1

2. 2.0 $\times$ 10^-4 molL^-1

3. 2.5 $\times$ 10^-6 molL^-1

4. 3.0 $\times$ $$10^-4 molL^-1

For the reaction, NO(g) + 1/2O₂ (g) ⇌ NO₂(g)

∆_fG° (NO₂₎ = 52.0 kJ/mol , ∆_fG° (NO) = 87.0 kJ/mol and
∆_fG° (O₂) = 0 kJ/mol. The equilibrium constant for the formation of NO₂ from NO and O₂ at 298K would be:

The equilibrium constant for the following reaction is 1.6 ×10⁵ at 1024K

H_2(g) + Br_2(g)  $\leftrightharpoons$2HBr_(g)

If HBr at pressure 10.0 bar is introduced into a sealed container at 1024 K, the equilibrium pressure of HBr will be:

1. 11.20 bar

2. 5.56 bar

3. 7.30 bar

4. 9.95 bar

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:

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

\(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5}{[\mathrm{NO}]^4\left[\mathrm{H}_2 \mathrm{O}]^6\right.}\)
The balanced chemical equation corresponding to the above-mentioned expression is:

One mole of H₂O and one mole of CO are taken in a 10 L vessel and heated to 725 K. At equilibrium, 40% of water(by mass) reacts with CO according to the equation,

H₂O _(g) + CO _(g)  $\leftrightharpoons$H_2 (g) + CO_2 (g)

The equilibrium constant for the above-mentioned reaction would be: