$\mathrm{If\; A}$ $+$ $B$ $\leftrightharpoons$ $C$ $+$ $D$ $Cons\tan t$ $=$ $K_1$
$E$ $+$ $F$ $\rightleftharpoons$ $G$ $+$ $H$ $Cons\tan t$ $=$ $K_2$

then C + D + E + F ⇒ product. The constant of reaction will be:

1.  $\frac{K_1}{K_2}$

2.  $\frac{K_2}{K1}$

3. $K_1{K}_2$

4. None of these 

The equilibrium constant Kp for the following reaction is:

${{\mathrm{MgCO}}_3}_{\left(\mathrm{s}\right)}\rightleftharpoons {\mathrm{MgO}}_{\left(\mathrm{s}\right)}+{{\mathrm{CO}}_2}_{\left(\mathrm{g}\right)}$

1. $\mathrm{Kp }= {\mathrm{P}}_{{\mathrm{CO}}_2}$

2. $\mathrm{Kp}={\mathrm{P}}_{{\mathrm{CO}}_2}\times \frac{{\mathrm{P}}_{{\mathrm{CO}}_2}\times {\mathrm{P}}_{\mathrm{MgO}}}{{\mathrm{P}}_{{\mathrm{MgCO}}_3}}$

3. $\mathrm{Kp}=\frac{{\mathrm{P}}_{{\mathrm{CO}}_2}+{\mathrm{P}}_{\mathrm{MgO}}}{{\mathrm{P}}_{{\mathrm{MgCO}}_3}}$

4. $\mathrm{Kp}=\frac{{\mathrm{P}}_{{\mathrm{MgCO}}_3}}{{\mathrm{P}}_{{\mathrm{CO}}_2}\times {\mathrm{P}}_{\mathrm{MgO}}}$

The correct relation between dissociation constants of a di-basic acid is:

1. $K{a}_1=K{a}_2$

2. $K{a}_1>K{a}_2$

3. $K{a}_1<K{a}_2$

4. $K{a}_1=\frac{1}{K{a}_2}$

For any reversible reaction, if we increase the concentration of the reactants, the effect on equilibrium constant will:
1. Depend on the amount of concentration
2. Remain unchanged
3. Decrease
4. Increase

The ionization constant of ${\mathrm{CH}}_3\mathrm{COOH}$ is 1.7 × ${10}^{-5}$ and the concentration of ${\mathrm{H}}^{+ }$ions is 3.4 × ${10}^{-4}$. The initial concentration of ${\mathrm{CH}}_3\mathrm{COOH}$ will be:
1.  3.4 × ${10}^{-4}$
2.  3.4× ${10}^{-3}$
3.  6.8 × ${10}^{-4}$
4.  6.8 × ${10}^{-3}$

For a reaction , ${\mathrm{BaO}}_2\left(\mathrm{s}\right)\rightleftharpoons \mathrm{BaO}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$; ∆H = + ve. At equilibrium condition, the pressure of O₂ depends on the:

1. Increased mass of BaO₂

2. Increased mass of BaO

3. Increased temperature on equilibrium.

4. Increased mass of BaO₂ and BaO both.

The reaction quotient (Q) for the reaction:

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

is given by $\mathrm{Q}=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$. The reaction will proceed from right to left if:
1. Q = K_C
2. Q < K_C
3. Q > K_C
4. Q = 0

(where K_C is the equilibrium constant)