${{\mathrm{I}}_2}_{\left(\mathrm{s}\right)}$ $+$ $5{{\mathrm{F}}_2}_{\left(\mathrm{g}\right)}$ $$ $\to 2{\mathrm{IF}}_{5\left(g\right)}$ $$

The equilibrium constant K_c expression for the above mentioned reaction is:

1.	\(\mathrm{K_{C} = \dfrac{\left[IF_{5}\right]^{2}}{\left[F_{2}\right]^{5}}}\)	2.	\(\mathrm{K_{C} = \dfrac{\left[IF_{5}\right]^{2}}{\left[F_{2}\right]^{5} \left[I_{2}\right]}}\)
3.	\(\mathrm{K_{C} = \dfrac{\left[F_{2}\right]^{5} \left[I_{2}\right]}{\left[IF_{2}\right]^{2}}}\)	4.	\(\mathrm{K_{C} = \dfrac{\left[F_{2}\right]^{5}}{\left[IF_{5}\right]^{2}}}\)

The solution that has the lowest pH is: 

(assuming 100% dissociation)

The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. The pK_a of bromacetic acid will be: 

1. 1.98

2. 2.75

3. 4.56

4. 7.15

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.

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

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: