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Equilibrium NEET Questions — Free MCQ Practice

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Which is not an acid salt?

A.

NaH₂PO₂

B.

NaH₂PO₃

C.

NaH₂PO₄

D.

NaHSO₃

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Explanation

(a) H₃PO₂ is monobasic acid and thus, it forms only one normal salt.

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The following equilibria are given

N₂ + 3H₂ $⇌$ 2NH₃, K₁

N₂ + O₂ $⇌$ 2NO, K₂

H₂ + $\frac{1}{2}$ O₂ $⇌$ H₂O, K₃

The equilibrium constant of the reaction, 2NH₃ + $\frac{5}{2}$ O₂ $⇌$ 2NO + 3H₂O in terms of K₁, K₂and K₃is

A.

$K_{1} K_{3}^{3} / K_{2}$

B.

$K_{2} K_{3}^{3} / K_{1}$

C.

K₁K₂K₃

D.

K₁K₂/K₃

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Explanation

(b) For equilibrium,

(i) N₂ (g) + 3H₂(g) $⇌$ 2NH₃ (g),

K₁ = [NH₃]²/[N₂][H₂]³ ...(i)

(ii) N₂(g) + O₂(g) $⇌$ 2NO (g),

K₂ = [NO]²/[N₂][O₂] .....(ii)

(iii) H₂(g) + $\frac{1}{2}$ O₂(g) $⇌$ H₂O (g),

K₃ = [H₂O]/[H₂][O₂]^1/2 ......(iii)

For the reaction,

2NH₃ (g) + 5/2 O₂ (g) $⇌$ 2NO (g) + 3H₂O (g)

K = [NO]²[H₂O]³/[NH₃]²[O₂]^5/2 ....(iv)

Equation (iii) multiplied by 3

3H₂ + 3/2 O₂ (g) $⇌$ 3H₂O

then, $K_{3}^{3}$ = [H₂O]³/[H₂]³[O₂]^3/2

From eqs. (i), (ii) and (v)

K = (K₂ x $K_{3}^{3}$ )/K₁

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The pK_a for acid A is greater than pK_a for acid B. the strong acid is:

A.

acid A

B.

acid B

C.

are equally strong

D.

none of these

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Explanation

(b) Higher pK_a (-logK_a) means lower K_a for acid.

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A solution of FeCl₃ in water acts as acidic due to:

A.

acidic impurities

B.

ionization

C.

hydrolysis of Fe³⁺

D.

dissociation

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Explanation

(c) Fe³⁺ + 3H₂O $\to$ Fe(OH)₃ + 3H⁺

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Which equilibrium can be described as Lewis acid-base reaction but not Bronsted acid-base reaction?

A.

H₂O + CH₃COOH $⇌$ H₃O⁺ + CH₃COO^-

B.

2NH₃ + H₂SO₄ $⇌$ $2 {NH}_{4}^{+}$ + ${SO}_{4}^{2 -}$

C.

NH₃ + CH₃COOH $⇌$ ${NH}_{4}^{+}$ + CH₃COO^-

D.

[Cu(H₂O)₄]²⁺ + 4NH₃ $⇌$ [Cu(NH₃)₄]²⁺ + 4H₂O

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Explanation

(d) It involves gain and loss of electron pair (Lewis concept).

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In the reaction, PCl₅ $⇌$ PCl₃ + Cl₂, the amounts of PCl₅, PCl₃ and Cl₂ at equilibrium are 2 mole each and the total pressure is 3 atm. The equilibrium constant K_p is:

A.

1.0 atm

B.

2.0 atm

C.

3.0 atm

D.

6.0 atm

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Explanation

(a) K_p = (ⁿCl₂ x ⁿPCl₃)/ⁿPCl₅ x [p/ $Σ n$ ]¹ = 2x2/2 x[3/6]¹ = 1 atm

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An aqueous solution of hydrogen sulphide shows the equilibrium,

H₂S $⇌$ H⁺ + HS^-

If dilute hydrochloic acid is added to an aqueous solution of hydrogen sulphide without any change in temperature, then:

A.

the equilibrium constant will change

B.

the concentration of HS^- will increase

C.

the concentration of undissociated hydrogen sulphide will decrease

D.

the concentration of HS^- will decrease

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Explanation

(d) K_a for H₂S = [H⁺][HS^-]/H₂S ;

An increase in [H⁺] will show a decrease in [HS^-] to maintain constant K_a value.

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When HCl gas is passed through a saturated solution of common salt, pure NaCl is precipitated because:

A.

the impurities dissolve in HCl

B.

HCl is highly soluble in H₂O

C.

the product of [Na⁺] and [Cl^-] exceeds the solubility product of NaCl

D.

the solubility product of NaCl is lowered by the chloride ion from aqueous HCl

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Explanation

(c) NaCl (s) $⇌$ Na⁺ (aq) + Cl^- (aq); HCl $\to$ H⁺ + Cl^-

The increase in [Cl^-] brings in an increase in [ Na⁺][Cl^-] which will lead for backward reaction because,

K_spNaCl = [Na⁺][Cl^-]

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The aqueous solution of a salt is alkaline. This shows that salt is made from:

A.

a strong acid and strong base

B.

a strong acid and weak base

C.

a weak acid and weak base

D.

a weak acid and strong base

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Explanation

(d) e.g., CH₃COONa;

CH₃COO^- + H₂O $⇌$ CH₃COOH + OH^-

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40% of a racemic mixture of 0.2 mole of N₂ and 0.6 mole of H₂ react to give NH₃ according to the equation, N₂ (g) + H₂(g) $⇌$ 2NH₃ (g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:

A.

4:5

B.

5:4

C.

7:10

D.

8:5

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Explanation

(a) N₂ + 3H₂ $⇌$ 2NH₃

Initially at eq. 0.2 0.6 0

(0.2-a) (0.6-3a) 2a

Total mixture is 0.8; 40% of it reacts, i.e, (0.8x40)/100 reacts to give (0.8x40)/100 x 1/2 mole of NH₃

or NH3 formed is 0.16 mole

2a=0.16

.^.. a=0.08

Initial mole = 0.8

Final mole = (0.2-0.08) + (0.6-0.24) +0.16 = 0.12 + 0.36 +0.16 =0.64

.^.. Ratio of final mole to initial mole = 0.64/0.8 = =0.8 = 4/5

If you didn't understand why 1/2 was multiplied to number of moles of mixture utilized to get moles of Ammonia produced, then read below:

See the number of moles used in the first equation.

'a' mole of Nitrogen and '3a' moles of Hydrogen produce '2a' moles of Ammonia.

So, total '4a' moles of mixture result in '2a' moles of Ammonia.

Therefore, the ratio of moles of Ammonia produced from moles of mixture utilized is 2a/4a = 1/2

Equilibrium MCQs for NEET — Chemistry Questions with Answers

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