Consider the following reaction 
\(\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) ; \Delta H^0=+58 \mathrm{~kJ}\)
${}_{}$Also, consider the following stimuli on the above equilibrium.

(I)	Temperature is decreased.
(II)	Pressure is increased by adding N2 at constant temperature.

For each of the above cases (I, II), the direction in which the equilibrium shifts is:
1. (I) Towards reactant, (II) No change.
2. (I) Towards product, (II) Towards reactant.
3. (I) Towards product, (II) No change.
4. (I) Towards reactant, (II) Towards product.

For a reaction; $X<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>Y<mspace\; width="1em"></mspace>\leftrightharpoons <mspace\; width="1em"></mspace>2Z$, 1.0 mol of X, 1.5 mol of Y and 0.5 mol of Z were taken in a 1 L vessel and allowed to react. At equilibrium, the concentration of Z was 1.0 mol L^–1 . The equilibrium constant of the reaction is $\frac{x}{15}$. The value of x is:

1. 24

2. 13

3. 16

4. 19

The solubility of Ca(OH)₂ in water is: 

[Given: K_sp Ca(OH)₂ in water = 5.5 × 10^–6]

1. 1.77 × 10^–6

2. 1.11 × 10^–6

3. 1.11 × 10^–2

4. 2.77 × 10^–2

At 1990 K and 1 atm pressure, there are an equal number of Cl₂ molecules and Cl atoms in the reaction mixture.
The value of K_P for the reaction ${\mathrm{Cl}}_{2\left(\mathrm{g}\right)}\rightleftharpoons 2{\mathrm{Cl}}_{\left(\mathrm{g}\right)}$ under the above conditions is x × 10^–1. The value of x is:

1. 4

2. 8

3. 5

4. 10

For the reaction A_(g)$\to$(B)_(g), the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of $∆$_rG for the reaction at 300 K and 1 atm in J mol^–1 is – xR, where x is:
(R = 8.31 J mol^–1 K^–1 and ln 10 = 2.3)

1. 1400

2. 1380

3. 1360

4. 1340

An aqueous solution contains 0.10 M H₂S and 0.20 M HCl. If the equilibrium constant for the formation of HS^– from H₂S is 1.0 × 10^–7 and that of S^2– from HS^– ions is 1.2 × 10^–13 then the concentration of S^2– ions in aqueous solution will be:

1. 5 × 10^–8 M

2. 3 × 10^–20 M

3. 6 × 10^–21 M

4. 5 × 10^–19 M

An aqueous solution contains an unknown concentration of Ba²⁺. When 50 mL of a 1 M solution of Na₂SO₄ is added, BaSO₄ just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO₄ is 1 × 10^–10. What is the original concentration of Ba²⁺ ?

1. 5 × 10^–9 M

2. 2 × 10^–9 M

3. 1 × 10^–9 M

4. 1.0 × 10^–10 M

The solubility product of Pbl₂ is 8.0 × 10^–9 . The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^–6 mol/L. The value of x is:

(Rounded off to the nearest integer)$\text{\hspace{0.17em}[Given:\hspace{0.17em}}\sqrt{2}=1.41\text{\hspace{0.17em}]\hspace{0.17em}}$

Arrange the following solution in the decreasing order of pOH :

(A). 0.01 M HCl

(B). 0.01 M NaOH

(C). 0.01 M ${\mathrm{CH}}_3\mathrm{COONa}$

(D). 0.01 M NaCl

1. (B) > (C) > (D) > (A)

2. (A) > (C) > (D) > (B)

3. (B) > (D) > (C) > (A)

4. (A) > (D) > (C) > (B)

The correct expression for the following reaction is:

Fe₂N_(s) + \(\frac{3}{2}\)H_2(g) \(\leftrightharpoons \) 2Fe_(s) + NH_3(g)