On increasing the pressure, the direction in which the gas phase reaction proceeds to re-establish equilibrium is predicted by applying Le-Chatelier's principle. Consider the reaction,

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

Which of the following is correct, if the total pressure at which the equilibrium is established is increased without changing the temperature?

1.	K will remain the same.
2.	K will decrease.
3.	K will increase.
4.	K will increase initially and then decrease, when pressure is very high.

 At 450 K, K_p= 2.0 × 10¹⁰ bar^-1 for the given reaction at equilibrium

$2{{\mathrm{SO}}_2}_{(\mathrm{g})\hspace{0.17em}}+$ ${{\mathrm{O}}_2}_{\left(\mathrm{g}\right)}$ $\rightleftharpoons$ $2{{\mathrm{SO}}_3}_{\left(\mathrm{g}\right)}$ $$

The value of  K_c at this temperature would be :

$1.$ $7.48$ $\times {10}^{12}$ ${\mathrm{M}}^{-1}$
$2.$ $6.56$ $\times$ ${10}^{11}$ ${\mathrm{M}}^{-1}$
$3.$ $7.48$ $\times$ ${10}^{11}$ ${\mathrm{M}}^{-1}$
$4.$ $1.23$ $\times$ ${10}^{10}$ ${\mathrm{M}}^{-1}$

For the reaction: ${\mathrm{FeO}}_{\left(\mathrm{s}\right)}$ $+$ ${\mathrm{CO}}_{\left(\mathrm{g}\right)}$ $\rightleftharpoons$ ${\mathrm{Fe}}_{\left(\mathrm{s}\right)}$ $\hspace{0.17em}+\hspace{0.17em}{{\mathrm{CO}}_2}_{\left(\mathrm{g}\right)},$ ${\mathrm{K}}_{\mathrm{p}}$ $=$ $0.265$  at 1050 K. If the initial partial pressures are p_CO= 1.4 atm and ${\mathrm{p}}_{{\mathrm{CO}}_2}$= 0.80 atm, the partial pressure of CO₂ at equilibrium at 1050 K would be:

For the reaction, 

${\mathrm{N}}_2$ $\left(\mathrm{g}\right)$ $+$ $3{\mathrm{H}}_2\left(\mathrm{g}\right)$ $\rightleftharpoons$ $2{\mathrm{NH}}_3$ $\left(\mathrm{g}\right);$ $$
${\mathrm{K}}_{\mathrm{c}}=$ $0.061$ ${\mathrm{mol}}_{}^{-2}{\mathrm{L}}^2$  at 500K. At a particular instant of time, [N₂] = 3.0 mol L^–1, [H₂] = 2.0 mol L^–1  and
[NH₃] = 0.5 mol L^–1 .

True statement among the following is: 

1. Reaction is at equilibrium.

2. Reaction will proceed in the forward direction.

3. Reaction will proceed in the backward direction.

4. Can't predict the direction of the reaction.
 

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass. 

${\mathrm{C}}_{\left(\mathrm{s}\right)}$ $+$ ${{\mathrm{CO}}_2}_{\left(\mathrm{g}\right)}$ $\rightleftharpoons 2{\mathrm{CO}}_{\left(\mathrm{g}\right)}$

At the specified temperature, K_c for this reaction would be:

$1.$ $0.25$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$
$2.$ $0.34$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$
$3.$ $0.15$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$
$4.$ $1.25$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}$

For the reaction, $3{\mathrm{O}}_2$ $\left(\mathrm{g}\right)$ $\rightleftharpoons$ $2{\mathrm{O}}_3$ $\left(\mathrm{g}\right),$ ${\mathrm{K}}_{\mathrm{c}}$ $=$ $2.0$ $\times$ ${10}^{-50}$ L/mol at 25^°C. If the equilibrium concentration of O₂ in air at 25^°C is $1.6$ $\times$ ${10}^{-2}$, the concentration of O₃ would be: 

$1.$ $2.86$ $\times {10}^{28}$ $\mathrm{M}$
$2.$ $28.66$ $\times$ ${10}^{-28}$ $\mathrm{M}$
$3.$ $1.43$ $\times$ ${10}^{-28}$ $\mathrm{M}$
$4.$ $2.86$ $\times$ ${10}^{-28}$ $\mathrm{M}$

Consider the following reaction:
\(CO_{\left(\right. g \left.\right)}\) \(+\) \(3 {H_{2}}_{\left(g\right)}\) \(\rightleftharpoons\) \({CH_{4}}_{\left(\right. g \left.\right)}\) \(\)\(+\) \(H_{2} O_{\left(\right. g \left.\right)}\) \(\)

If the reaction mixture contains 0.30 mol of CO, 0.10 mol of H_2, and 0.02 mol of H₂O, and an unknown amount of CH₄ at equilibrium at 1300 K in a 1L flask, the concentration of CH₄ in the mixture will be -

(K_c = 3.90 at 1300 K) $\mathrm{}$

1 . \(5 . 85\) \(\times\) \(10^2\) \(M\)
2 . \(5 . 85\) \(\times\) \(10^{- 1}\) \(M\)
3 . \(5 . 85\) \(\times\) \(10^3\) \(M\)
4 . \(5 . 85\) \(\times\) \(10^{- 2}\) \(M\)

The concentration of hydrogen ion in a sample of soft drink is $3.8$ $\times {10}^{-3}$ $\mathrm{M}$. The  pH of the soft drink will be: