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NEET NEET Questions — Free MCQ Practice

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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.

A.

$\frac{31}{27}$

B.

$\frac{60}{40}$

C.

$\frac{31}{9}$

D.

$\frac{62}{27}$

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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

A.

PCl₅ will remain unchanged

B.

Cl₂ will be greater

C.

PCl₅ will become less

D.

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

A.

yield is maximum at this temperature

B.

catalyst is active only at this temperature

C.

energy needed for the reaction is easily obtained at this temperature

D.

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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For the equilibrium of the reaction, HgO(s) $\overset{}{⇌}$ Hg(g) + $\frac{1}{2}$ O₂(g), k_P for the reaction at total pressure of P is:

A.

$K_{P} = \frac{2}{3^{3 /2}} p^{3 /2}$

B.

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

C.

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

D.

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

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Explanation

HgO(s) $\overset{}{⇌}$ Hg (g) + $\frac{1}{2}$ O₂ (g)

1 0 0

t =t , (1-x) X X/2

K_p = P_Hg(g) × P(O₂)^1/2

Total moles at equilibrium = $\frac{3 x}{2}$

P_Hg = $\frac{x}{3 x / 2} = P = \frac{2}{3} P$

PO₂ = $\frac{x /2}{3 x /2} P = \frac{1}{3} P$

K_p = $\frac{2}{3} P {(\frac{1}{3} P)}^{1 / 2} = \frac{2}{3^{3 /2}} P^{3 /2}$

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The value of k_p for the reaction at 27°C Br₂(l) + Cl₂(g) $\overset{}{⇌}$ 2BrCl(g) is '1 atom'. At equilibrium in a closed container partial pressure of BrCI gas is 0.1 atm and at this temperature the vapour pressure of Br₂(l) is also 0.1 atm. Then what will be minimum moles of Br₂(l) to be added to 1 mole of Cl₂ , initially, to get above equilibrium situation :

A.

$\frac{10}{6}$ moles

B.

$\frac{5}{6}$ moles

C.

$\frac{15}{6}$ moles

D.

2 moles

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Explanation

Br₂(l) + Cl₂(g) $\overset{}{⇌}$ 2BrCl(g)

t = 0 1 0

(1 – x) 2x

K_p = $\frac{(P_{BrCl})^{2}}{P_{{Cl}_{2}}}$ = 1 so, $P_{{Cl}_{2}}$ = ${{(P}_{BrCl})}^{2}$ = 0.01 atm

them at equilibrium, $\frac{n_{BrCl}}{n_{{Cl}_{2}}} = \frac{0 .1}{0 .01} = 10 = \frac{2 x}{1 - x}$

so,10 – 10x = 2xorx = $\frac{10}{12} = \frac{5}{6}$ moles

Moles of Br(l) required for maintaining vapour pressure of 0.1 atm

= 2 × $\frac{5}{6}$ moles = $\frac{10}{6}$ moles = moles of BrCl(g).

Moles required for taking part in reaction = moles of Cl₂ used up = $\frac{5}{6}$ moles.

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5 mol PCI₅(g) and one mole N₂ gas is placed in a closed vessel. At equilibrium PCI₅(g) decomposes 20% and total pressure in to the container is found to be 1 atm. The k_P for equilibrium

PCl5(g) $\overset{}{⇌}$ PCl₃(g) + Cl₂(g)

A.

$\frac{1}{24}$ atm

B.

$\frac{1}{4}$ atm

C.

$\frac{1}{16}$ atm

D.

$\frac{1}{28}$ atm

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The degree of dissociation of water in a 0.1 M aqueous solution of HCl at a certain temperature t°C is 3.6 x 10^–15. The temperature t must be :

A.

< 25°C

B.

= 25°C

C.

> 25°C

D.

insufficient data to predict

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Explanation

K_w = 55.5 × 3.6 × 10^–15× 0.1 = 2 × 10^–14

Hence temperature must be > 25°C

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Which one is the correct expression below for the solution containing 'n' number of weak acids?

A.

${[H}^{+}] = \sqrt{\sum_{i =1}^{n} \frac{K_{i}}{C_{i}}}$

B.

${[H}^{+}] = \sqrt{\sum_{i =1}^{n} K_{i} C_{i}}$

C.

$[H^{+}] = \sum_{i = 1}^{n} K_{i} C_{i}$

D.

none of these

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Explanation

HA $\overset{}{⇌}$ H⁺ + A^– ; K₁ = $\frac{[H^{+}] [A^{-}]}{[HA]}$

HB $\overset{}{⇌}$ H⁺ + B^– ; K₂ = $\frac{[H^{+}] [B^{-}]}{[HB]}$

By mass balance,

[HA]_initial = [HA]_eq + [A^–] = C₁

[HB]_initial = [HB]_eq + [B^–] = C₂

By charge balance,

[H⁺] = $\frac{K_{1} [HA]}{[H^{+}]} + \frac{K_{2} [HB]}{[H^{+}]}$

[H⁺]² = K₁ [HA] + K₂ [HB] = K₁ {C₁ – [A^–]} + K₂ {C₂ – [B^–[}

If K₁, K₂ are very less then

[H⁺] = $\sqrt{K_{1} C_{1} + K_{2} C_{2} + . ... K_{n} C_{n}}$ = $\sqrt{\sum_{i = 1}^{n} K_{i} C_{i}}$ for ‘n’ number of weak acids

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The pH of glycine at the first half equivalence point is 2.34 and that at second half equivalence point is 9.60.At the equivalence point (The first inflection point) The pH is :

A.

3.63

B.

2.34

C.

5.97

D.

11.94

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Explanation

pH = pKa₁ = 2.34.

pH = pKa₂ = 9.6.

pH = $\frac{{pKa}_{1} + {pKa}_{2}}{2}$ = 5.97.

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A 1.458 g of Mg reacts with 80.0 ml of a HCI solution whose pH is –0.477. The change in pH after all Mg has reacted. (Assume constant volume. Mg = 24.3 g/mol.)(log 3 = 0.477)

A.

–0.176

B.

+0.477

C.

–0.2385

D.

0.3

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Explanation

Mg(aq) + 2HCl(aq) → MgCl₂(aq) + H₂

$\frac{1 .458}{24 .3}$

Millimoles of HCl = 3 × 80 = 240 mM

Moles of HCl after reaction = 240 – 60 × 2 = 120

New Molarity = $\frac{120}{80}$ = 1.5 M

pH = – log[H⁺] = – log 1.5 = – 0.176

Change is pH = – 0.176 – (– 0.477) = 0.3

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