Consider the following graph:

The point that represents reaction in equilibrium is-

1. C

2. A

3. B

4. D

The concentration vs time graph for the Haber process is given below:

The curve representing hydrogen gas is

1. Y

2. Z

3. X

4. In the Haber process, hydrogen gas is not used

The pK_b of dimethylamine and pk_a of acetic acid are 3.27 and 4.77 respectively at T (K).
The correct option for the pH of dimethylammonium acetate solution is:

The following concentrations were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium at 500K.
[N₂] = 1.5 × 10^–2M. [H₂] = 3.0 ×10^–2 M and
[NH₃] = 1.2 ×10^–2M.
The equilibrium constant value is:

The value of K_c = 4.24 at 800K for the reaction,

CO(g) + H₂O(g) $\leftrightharpoons$CO₂ (g) + H₂ (g)

If only CO and H₂O are present initially at concentrations of 0.10M each.

The equilibrium concentrations of CO₂, and CO are-

1. 0.067 M; 0.033 M
2. 0.76 M; 0.029 M
3. 0.65 M; 0.027 M
4. 0.061 M; 0.30 M

The value of K_c for the reaction 2A $\leftrightharpoons$B + C is 2 × 10^–3. At a given time, the composition of the reaction mixture is [A] = [B] = [C] = 3 × 10^–4 M.
The reaction will proceed:

1. In the backward direction
2. In the forward direction
3. In both direction
4. Cannot be predicted

The incorrect match among the following  is:

a. HO^– = Lewis base
b. F^– = Lewis base
c. H⁺ = Lewis acid
d. BCl₃ = Lewis base

1. Both a and b
2. Only b
3. Only d
4. Both a and d

The solubility product of mercurous iodide is $4.5\times {10}^{-29}$. The solubility of mercurous iodide will be:

1. $6.5\times {10}^{-7}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

2. $4.09\times {10}^{-8}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

3. $4.09\times {10}^{-7}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

4. $6.5\times {10}^{-8}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

The solubility product of silver chromate is $1.1\times {10}^{-12}$. The solubility of silver chromate will be:

1. $6.5\times {10}^{-5}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

2. $6.5\times {10}^{-6}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

3. $5.5\times {10}^{-5}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

4. $5.5\times {10}^{-6}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

The species that can act as Bronsted acids as well as bases is/are:

1.  ${\mathrm{H}}_2\mathrm{O}$

2.  ${\mathrm{HCO}}_3^{-},$ ${\mathrm{HSO}}_4^{-}$ $$

3.  ${\mathrm{NH}}_3$

4.  All of the above.