Find the equilibrium constant Kc for the reaction:

N₂O₄ ⇌ 2NO₂

At 280°C, the equilibrium mixture in 2 L of chloroform contains 0.2 mol N₂O₄ and 2 × 10⁻³ mol NO₂.

1. 1$\times$10${}^{-2}$

2. 2$\times$10${}^{-3}$

3. 1$\times$10${}^{-5}$

4. 2$\times$10${}^{-5}$

For the follwing gaseous equilibria X,Y and Z at 300K 

X : 2SO${}_2$$+O_2\rightleftharpoons 2S{O}_3$

Y : $PC{I}_5\rightleftharpoons PC{I}_3+C{I}_2$

Z : 2HI $\rightleftharpoons$ H${}_2$ + I${}_2$

Ratio of K${}_p$ and K${}_c$ in the increasing order is: 

1. X=Y=Z

2. X<Y<Z

3. X<Z<Y

4. Z<Y<X

$N_2+<mspace\; width="1em"></mspace>3H_2\rightleftharpoons 2N{H}_3$
1 mole \(N_2\) and \(3H_2\) are at 4 atm.
Equilibrium pressure is found to be 3 atm. find the \(K_p\)

1. \(\frac{1}{0.5(1.5)^3}\) 

2. \(\frac{1}{0.5(0.5)^3}\) 

3. \(\frac{3\times3}{0.5(0.5)^3}\) 

4. None of these 

4 moles of A are mixed with 4 moles of B,

when 2 moles of C are formed at equilibrium

according to the reaction A +B $\rightleftharpoons$C+ D 

The value of the equilibrium constant is: 

1. 4

2. 1

3. 1/2

4. 1/4

Find the species that can act as both a Brønsted–Lowry acid and a Brønsted–Lowry base.

In the reaction HCN + H${}_2$O $\rightleftharpoons <mspace\; width="1em"></mspace>H_3{O}^{+}+<mspace\; width="1em"></mspace>C{N}^{-}$ the conjugate acid- base pair is 

1. HCN,H${}_3$O${}^{+}$

2. HCN,CN${}^{-}$

3. $H_{2}O,<mspace\; width="1em"></mspace>C{N}^{-}$

4. $C{N}^{-},H_3{O}^{+}$

At a certain temperature the following equilibrium is established, CO(g) + N0₂(g) $\stackrel{}{\rightleftharpoons }$ 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) 1.2

(2) 2.25

(3) 2.1

(4) 3.6

Find the correct increasing order of pH for 0.1 M solutions of HCl, NH₄Cl, NaCl, and NaCN.

1. NaCl < NH₄Cl < NaCN < HCl
2. HCl < NH₄Cl < NaCl < NaCN
3. NaCN < NH₄Cl < NaCl < HCl
4. HCl < NaCl < NaCN < NH₄Cl

The solubility product of CuS, Ag${}_2$S. HgS are 10${}^{-37},{10}^{-44}$ and 10${}^{-54}$ respectively. The solubilities of these sulphides are in the order of: 

1. Ag${}_2$S > CuS > HgS

2. Ag${}_2$S > HgS > CuS

3. HgS > Ag${}_2$S > CuS

4. CuS > Ag${}_2$S > HgS 

In which of the following solvents will AgBr have the highest solubility 

1. 10${}^{-3}$ M NaBr

2. 10${}^{-3}$ M NH4OH 

3. Pure water 

4. 10${}^{-3}$ M HBr