5 mole of NH₄HS (s) start to decompose at a particular temperature in a closed vessel. If pressure of NH_3(g) in the vessel is 2 atm, then K_p for the reaction,

NH_4(s)HS ⇌ NH_3(g) + H₂S_(g), will be –

Solution

P_NH₃ = P_H₂S = 2 atm

∴ K_p = P_NH₃ × P_H₂S = 4

Pure ammonia is placed in a vessel at a temperature where its dissociation constant (a) is appreciable. At equilibrium :

K_p does not change significantly with pressure.
does not change with pressure.
concentration of NH₃ does not change with pressure.
concentration of hydrogen is less than that of nitrogen.

Solution

Statement (a) is correct and the rest statements are wrong.

K_p depends only on temperature hence at constant temp.

K_p will not change.

Some chemists at ISRO wished to prepare a saturated solution of a silver compound and they wanted it to have the highest concentration of silver ion possible. Which of the following compounds, would they use ?

K_sp(AgCl) = 1.8 × 10^-10

K_sp(AgBr) = 5.0 × 10^-13

K_sp(Ag₂CrO₄) = 2.4 × 10^-12

AgCl
AgBr
Ag₂CrO₄
Any of these

Solution

To find which of the given compound among AgCl, AgBr and AgCrO₄ would they use, first find which has highest conc. of Ag⁺ ions.

For this,we have to calculate the solubility of Ag⁺ ions (let s be the solubility of Ag⁺ ions)

For AgCl,

AgCl ⇌ Ag⁺ + Cl^-

⇒ K_sp = (s)(s) = s²

⇒ \(s = \sqrt{K_{sp}} = \sqrt[6]{\left ( 1.8 \times 10^{-10} \right )^{3}}\)

For AgBr,

⇒ \(s = \sqrt{K_{sp}} = \sqrt[6]{\left ( 5.0 \times 10^{-13} \right )^{3}}\)

For Ag₂CrO₄

\(s = \sqrt[3]{K_{sp}} = \sqrt[6]{\left ( 2.4 \times 10^{-12} \right )^{2}}\)

So, solution of silver compound can be made from Ag₂CrO₄ which has highest solubility among the given silver halides.

K₁ and K₂ are equilibrium constant for reactions (i) and (ii)

N_2(g) + O_2(g) ⇌ 2NO_(g) ……..(i)

NO_(g) ⇌ ¹⁄₂N_2(g) + ¹⁄₂O_2(g) ……..(ii)

Then,

\(K_{1} = \left (\frac{1}{K_{2}} \right )^{2}\)
K₁ = K₂²
\(K_{1} = \left (\frac{1}{K_{2}} \right )\)
K₁ = K₂⁰

Solution

For reaction (i)

\(K_{1} = \frac{\left [ NO \right ]^{2}}{\left [ N_{2} \right ]\left [ O_{2} \right ]}\)

and for reaction (ii)

\(K_{2} = \frac{\left [ N_{2} \right ]^{1/2}\left [ O_{2} \right ]^{1/2}}{\left [ NO \right ]}; K_{1} = \left (\frac{1}{K_{2}^{2}} \right )\)

For which one of the following reactions K_p = K_c?

PCl₅ ⇌ PCl₃ + Cl₂
N₂ + O₂ ⇌ 2NO
N₂ + 3H₂ ⇌ 2NH₃
2SO₃ ⇌ 2SO₂ + O₂

Solution

K_p = K_c(RT)^Δn_g

∴ K_p = K_conly when Δn_g = 0

For reaction , N₂ + O₂ ⇌ 2NO

Δn_g = 2 - 2 = 0

∴ K_p = K_c(RT)⁰ or K_p = K_c

The equilibrium constant for a reaction A + 2B ⇌ 2C is 40. The equilibrium constant for reaction C ⇌ B + ¹⁄₂A is

40
(¹⁄₄₀)²
¹⁄₄₀
¹⁄_{(40)^¹⁄₂}

Solution

Given A + 2B ⇌ 2C

\(K = \frac{\left [ C \right ]^{2}}{\left [ A \right ]\left [ B \right ]^{2}} .................(i)\)

C ⇌ B + ¹⁄₂A

\(K' = \frac{\left [ B \right ]\left [ A \right ]^{1/2}}{\left [ C \right ]} .................(ii)\)

Dividing (ii) by (i)

\(K' = \left [\frac{1}{K} \right ]^{1/2} = \left [\frac{1}{40} \right ]^{1/2}\)

Which of the following is correct for the reaction?

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

K_p = K_c
K_p < K_c
K_p > K_c
Pressure is required to predict the correlation

Solution

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

Δn = n_product - n_reactants = 2 - 4 = -2

∴ K_p = K_c(RT)^-2 or K_p = K_c/(RT)²

Thus, K_p < K_c

The solubility of Ca₃(PO₄)₂ is ‘s’ moles per litre. Its solubility product is

72 s⁵
108 s⁵
9 s²
8 s³

Solution

The solubility equilibrium of Ca₃(PO₄)₂ is represented as,

Ca₃(PO₄)₂ ⇌ 3 Ca²⁺ + 2 PO₄^3-

3 Ca²⁺ → 3s
2 PO₄^3- → 2s

∴ K_sp = [Ca²⁺]³[PO₄^3-]²

= [3s]³[2s]² = 108 s⁵

For the following reaction in gaseous phase

CO_(g) + ¹⁄₂ O_2(g) → CO_2(g), K_P/K_C is

(RT)^1/2
(RT)^-1/2
RT
(RT)^-1

Solution

For a gaseous phase reaction K_p and K_c are related as

K_p = K_c(RT)^Δn_g

For the given reaction,

CO_(g) + ¹⁄₂ O_2(g) → CO_2(g)

Δn_g = 1 – (1 + 0.5) = – 0.5 or -¹⁄₂

K_p = K_c(RT)^-¹⁄₂

or K_P/K_C = (RT)^-1/2

The solubility of PbCl₂ is given by

\(\left [ \frac{K_{sp}}{4} \right ]^{1/3}\)
[8 K_sp]^1/2
[K_sp]^1/2
\(\sqrt{K_{sp}}\)

Solution

Let s be the solubility of PbCl₂ and K_sp be its solubility product

The solubility equilibrium of PbCl₂ is represented as

PbCl₂ ⇌ Pb⁺² + 2Cl^-

∴ K_sp = [Pb⁺²][Cl^-]² = (s)(2s)² = 4s³

⇒ s = \(\left [ \frac{K_{sp}}{4} \right ]^{1/3}\)

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