(a) PCl_{5 (g)} $\leftrightharpoons$ PCl_{3 (g)} + Cl_{2 (g)}

(b) CaO _(s) + CO_{2 (g)} $\leftrightharpoons$  CaCO_{3 (s)}

(c) 3Fe _(s) + 4H₂O _(g) $\leftrightharpoons$ Fe₃O_{4 (s)} + 4H_{2 (g)}

The effect of an increase in the volume on the number of moles of products in the above-mentioned reactions would be, respectively:

1. a) Increase,  b) decrease, c) same

2. a) Decrease,  b) same, c) increase

3. a) Increase,  b) increase, c) same

4. a) Increase,  b) decrease, c) increase

For the given reaction: 

PCl_{5 (g)}  $\leftrightharpoons$PCl_{3 (g)} + Cl_{2 (g)}, ∆_rH° = 124.0 kJ mol^–1 and K_c = 8.3 $\times$ 10^-3 mol L^-1 at 473 K

The effect on K_c if (i) pressure is increased  and (ii) the temperature is increased will be, respectively-

1. (i) Will increase; (ii) will decrease

2. (i) Will decrease; (ii) will remain the same 

3. (i) Will remain the same; (ii) will increase

4. (i) will remain the same; (ii) Will decrease

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

2.  HSO₄^-, CO₃^-

3. SO₄^-2, CO₃^2-

4. HS₂O₄^-, CO₃^2-

OH^– , F^– , H⁺ and BCl₃

The first ionization constant of H₂S is 9.1 × 10^–8. The concentration of HS^– ion in its 0.1 M solution will be:

The ionization constant of acetic acid is 1.74 × 10^–5. The pH of acetic acid in its 0.05 M solution will be:
1. 7.81

The ionization constant of propanoic acid is 1.32 × 10^–5. The degree of ionization of 0.05M acid solution will be:

1. $\mathrm{\alpha }$ = 0.63 × 10^–2

2. $\mathrm{\alpha }$ = 1.63 × 10^–4

3. $\mathrm{\alpha }$ = 1.63 × 10^–2

4. $\mathrm{\alpha }$ = 0.05 × 10^–2

The salt that gives a neutral solution in water is: 

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

The solubility product for a salt of type AB is $4\times {10}^{-8}$.  The molarity of its standard solution will be:
1. $2\times {10}^{-4}$ $\mathrm{mol}/\mathrm{L}$

2. $16\times {10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

3. $2\times {10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

4. $4\times {10}^{-4}$ $\mathrm{mol}/\mathrm{L}$