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Equilibrium MCQs for NEET — Chemistry Questions with Answers

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The hydrogen ion concentration of a 10^-8 M HCl aqueous solution at 298 K(K_w 10^-14)is:

1.0×10^-6 M

1.0525 x 10^-7 M

9.525 x 10^-8 M

1.0×10^-8 M

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Explanation

(b) In aqueous solution of 10^-8 M HCl, [H⁺] is based upon the concentration of H⁺ ion of water 10^-8 M HCl and concentration of H⁺ ion of water K_w of H₂O=10^-14=[H⁺][OH^-] or[H⁺]=10^-7 M (due to its neutral behaviour)

So, in aqueous solution of 10^-8 M HCI,

$[H^{+}] = [H^{+}] of HCl + [H^{+}] of water = 10^{- 8} + 10^{- 7} = 11 \times 10^{- 8} M = 1.10 \times 10^{- 7} M$

Hence, answer is nearer to (b).

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Two flasks A and B of equal volume containing 1 mole and 2 mole of O₃ respectively, are heated to the sametemperature. When the reaction 2O₃ $⇋_{}^{}$ 3O₂ practically stops, then both the flasks shall have

the same ratio [O₂] /[O₃]

the same ratio : [O₂]³/²/[O₃]

only O₂

the same time to reach equilibrium

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Explanation

For reaction

$\frac{[O_{2}]^{3}}{[O_{3}]^{2}} =$ constantso $\sqrt{K_{C}}$ = $\frac{[O_{2}]^{3 /2}}{[O_{3}]}$ = constant

= will be same for both the containers

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A 10L container at 300K contains CO₂ gas at pressure of 0.2 atm and an excess solid CaO (neglect thevolume of solid CaO). The volume of container is now decreased by moving the movable piston fitted in the container. What will be the maximum volume of container when pressure of CO₂ attains its maximum value given that CaCO₃ (s) $⇋_{}^{}$ CaO(s) + CO₂(g)K_ph = 0.800 atm

5 L

2.5 L

1 L

The information is insufficient.

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Explanation

K_p = 0.800 atm = $P_{{CO}_{2}}$ = maximum pressure of CO₂ in the container to calculate maximum volumeof container the $P_{{CO}_{2}}$ = 0.8 atm and none of $P_{{CO}_{2}}$ = 0.8 should get converted into CaCO₃(s).

so V(0.800 atm) = (10 L) (0.2 atm)

so v = 2.5 L

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At a certain temperature the following equilibrium is established, CO(g) + N0₂(g) $\overset{}{⇌}$ CO₂(g)+ NO(g)One mole of each of the four gas is mixed in one litre container and the reaction is allowed to reach equilibrium state. When excess of baryta water (Ba(OH)₂) is added to the equilibrium mixture, the weight of white ppt (BaCO₃) obtained is 236.4 gm. The equilibrium constant K_c of the reaction is (Ba = 137)

1.2

2.25

2.1

3.6

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Explanation

	CO(g)	+	NO₂(g)	$\overset{}{⇌}$	CO₂(g)	+	NO(g)
t =	0.1 mole		1 mole		1 mole		1 mole
Al eq.	1 – x		1 – x		1 + x		1 + x

CO₂ + Ba(OH)₂ $\overset{}{\to}$ BaCO₃

mole of BaCO₃ = $\frac{236 .4}{197}$ = 1.2

So mole of CO₂ at eq. = 1.2

or 1 + × = 1.2

x = 0.2

K_C = ${(\frac{1 + x}{1 - x})}^{2} = {(\frac{1 .2}{0 .8})}^{2} =$ 2.25

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At temperature T^, the compound AB₂ (g) dissociates according to the reaction, 2AB₂ (g) $\overset{}{⇌}$ 2 AB(g) + B₂(g). With a degree of dissociation x^, which is small compared with unity. Deduce the expression for x in terms of the equilibrium constant, K_p and the total pressure, P.

$x = {[\frac{2 K_{p}}{P}]}^{- 1 /3}$

$x = {[\frac{K_{p}}{P}]}^{1 /3}$

$x = {[\frac{K_{p}}{2P}]}^{1 /3}$

$x = {[\frac{6 K_{p}}{3 P}]}^{1 /3}$

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Explanation

2AB₂ (g) $\to_{}^{}$ 2AB (g) + B₂ (g)

Mole before dissociation 1 0 0

Mole after dissociation (1 – x) x $\frac{x}{2}$

Total mole at equilibrium (Σn) = 1 - x + x + $\frac{x}{2}$ = 1 + $\frac{x}{2}$

Now, K_p = $\frac{n_{B_{2}} \times {(n_{AB})}^{2}}{{(n_{{AB}_{2}})}^{2}} \times {[\frac{P}{\sum n}]}^{Δn}$

K_p = $\frac{\frac{x}{2} . (x)^{2}}{{(1 - x)}^{2}} \times {[\frac{p}{1 + \frac{x}{2}}]}^{1}$

K_p = $\frac{x^{3} p}{2}$ $[∵ x is small, ∴ 1 - x and 1 + \frac{x}{2} = 1]$

x = $^{3} \sqrt{\frac{2 K_{p}}{p}}$

$x = {[\frac{K_{p}}{2P}]}^{1 /3}$

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Equilibrium constant for the given reaction is kc = 10²⁰ at temperature 300 K,A(s) + 2B (aq.) $\overset{}{⇌}$ 2C (s) + D (aq.)K = 10²⁰ The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mole/litre of B at 300 K is

$~ 4 \times 10^{- 11}$

$~ 2 \times 10^{- 10}$

$~ 2 \times 10^{- 11}$

$~ 10^{- 10}$

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Explanation

A(s) + 2B(aq) $\to_{}^{}$ 2C(s) + D(aq)

Initial 1 $\frac{1}{3}$ 0 0

At eq. 1 – x $\frac{1}{3} - 2 x$ 2x x

$\approx$ a

x $\approx$ 1/3

10²⁰ = $\frac{\frac{1}{3}}{{[B]}^{2}}$

10²⁰ = $\frac{\frac{1}{3}}{a^{2}}$

a² = $\frac{1}{3 \times 10^{20}}$ = $\frac{10^{- 20}}{3}$

a = $\frac{10^{- 10}}{\sqrt{3}} \approx$ 4 × 10^–11 M

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10l box contain O₃ and O₂ at equilibrium at 2000 K. The ΔG° = –534.52 kJ at 8 atm equilibrium pressure.The following equilibrium is present in the container. 2O₃(g) $\overset{}{⇌}$ 3O₂(g). The partial pressure of O₃ will be (In 10 = 2.3, R = 8.3 Jmole^–1K^–1):

8 ×10⁴

22.62 × 10^-7

9.71 × 10^-6

9.71 × 10^-2

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Explanation

ΔG° = – RT In K = – 2.3 × 2000 × 8.3 log K

$\frac{534 .52 × 10^{3}}{2 .3 × 8 .3 × 2000} = \log K$

K = 10¹⁴

2O₃(g) $\overset{}{⇌}$ 3O₂(g) K = 10¹⁴

So $P_{O_{2}}$ <<< $P_{O_{2}}$

So $P_{O_{2}}$ + $P_{O_{2}}$ = 8

$P_{O_{2}}$ = 8 atm

K = 10¹⁴ = $\frac{{(P_{O_{2}})}^{3}}{{(P_{O_{3}})}^{2}} = \frac{{(8)}^{3}}{P_{O_{3}}}$

$P_{O_{3}}$ = 22.62 × 10^–7 atm.

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Solid ammonium carbamate dissociates to give ammonia and carbon dioxide as follows:

NH₂COONH₄(s) $\overset{}{⇌}$ 2 NH3(g) + CO₂(g) At equilibrium, ammonia is added such that partial pressures of NH3 now equals the original total pressure. Calculate the ratio of the total pressures now to the original total pressure.

$\frac{31}{27}$

$\frac{60}{40}$

$\frac{31}{9}$

$\frac{62}{27}$

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Explanation

NH₂COONH₄(s) $\overset{}{⇌}$ 2NH₃(g) + CO₂(g)

Initial 2P P’

K_p = ${(P_{{NH}_{3}})}^{2} (P_{{CO}_{2}})$

K_p = (2P)² (P) …… (i)

P_T(initial) = 3P

NH₂COONH₄(s) $\overset{}{⇌}$ 2NH₃(g) + CO₂(g)

Final 3P P’

K_p = (3p)² (p’) …… (ii)

From eq. (i) and (ii)

(2P)²= (3P)² (P')

P’ = $\frac{4P}{9}$

$\frac{p_{T} (New)}{p_{T} (Old)} = \frac{3 P + P'}{3 P} = \frac{3 P + \frac{4 P}{9}}{3 P} = \frac{31}{27}$

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The reactions, PCl₅ (g) $\overset{}{⇌}$ PCl₃(g) + Cl₂(g) and Cl₂(g) $\overset{}{⇌}$ CO₂(g) + Cl₂(g) are simultaneously in equilibrium in an equilibrium box at constant volume. A few moles of CO(g) are later introduced into the vessel. After some time, the new equilibrium concentration of

PCl₅ will remain unchanged

Cl₂ will be greater

PCl₅ will become less

PCl₅ will become greater

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Explanation

If CO is added 2nd equilibrium will proceed in the backward direction and the concentration of Cl₂ will decrease. This Cl₂ will be further formed by the decomposition of PCl₅.

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In the Haber process for the industrial manufacture of ammonia involving the reaction, N₂ + 3H₂ $\overset{}{⇌}$ 2NH₃ at 200 atm pressure in the presence of a catalyst, temperature of about 500°C. This is considered as optimum temperature for the process because

yield is maximum at this temperature

catalyst is active only at this temperature

energy needed for the reaction is easily obtained at this temperature

rate of the catalytic reaction is fast enough while the yield is also appreciable for this exothermic reaction at this temperature.

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Explanation

Formation of ammonia is an exothermic process, therefore, it is favorable at a lower temperature. But at lower temperature rate of the reaction becomes slow.

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Yes. You can attempt every Equilibrium question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.