For the reversible reaction:
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) + heat 

The equilibrium shifts in a forward direction:

1. by increasing the concentration of $N{H}_3\left(g\right)$
2. by decreasing the pressure.
3. by decreasing the concentration of $N_2\left(g\right)$ $and$ $H_2\left(g\right)$
4. by increasing pressure and decreasing temperature.

For a given exothermic reaction, K_p and K_p^’ are the equilibrium constants at temperatures T₁ and T₂ respectively. Assuming that the heat of reaction is constant in temperatures range between T₁ and T₂, it is a readily observation that:
1. $K_p>K_p^{\text{'}}$
2. $K_p<K_p^{\text{'}}$
3. $K_p=K_p^{\text{'}}$
4. $K_p=\frac{1}{K_p^{\text{'}}}$

The strongest acid among the following compounds is:

pH of a saturated solution of Ba(OH)₂ is 12. The value of solubility product K_sp of Ba (OH)₂ is:
1. $3.3\times {10}^{-7}$
2. $5.0\times {10}^{-7}$
3. $4.0\times {10}^{-6}$
4. $5.0\times {10}^{-6}$

Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value?

1. \(\mathrm{B a C l_{2}}\)
2. \(\mathrm{A l C l_{3}}\)
3. \(\mathrm{L i C l}\)
4. \(\mathrm{B e C l_{2}}\)${\mathrm{}}_{}$

Buffer solutions have constant acidity and alkalinity because:
 

A buffer solution is prepared in which the concentration of NH₃ is 0.30 M and the concentration of  $N{H}_4^{+ }$ is 0.20 M. If the equilibrium constant, K_b for NH₃ equals 1.8×10^–5, then what is the pH of this solution? 
(log 1.8 = 0.25; log 0.67 = –0.176)

1.  9.43
2.  11.72
3.  8.73
4.  9.08

The value of $∆\mathrm{H}$ for the reaction is less than zero. Formation of $X_2\left(g\right)$ $+$ $4$ $Y_2\left(g\right)$ $\leftrightharpoons 2X{Y}_4\left(g\right)$ will be favoured at:

For the reaction ${\mathrm{N}}_2\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_2\left(\mathrm{g}\right)\leftrightharpoons 2\mathrm{NO}\left(\mathrm{g}\right)$ the equilibrium constant is K₁. The equilibrium constant is K₂ for the reaction  $2\mathrm{NO}\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_2\left(\mathrm{g}\right)$ $\leftrightharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right)$ $$
The value of K for the reaction given below will be:
${\mathrm{NO}}_2\left(\mathrm{g}\right)\leftrightharpoons \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)$ $+{\mathrm{O}}_2\left(\mathrm{g}\right)$ $$

1.  $\frac{1}{4}$ $\left(4\right)$ ${\mathrm{K}}_1$ ${\mathrm{K}}_2$

2.  ${\left[\frac{1}{{\mathrm{K}}_1{\mathrm{K}}_2}\right]}^{1/2}$

3.  $\frac{1}{\left({\mathrm{K}}_1{\mathrm{K}}_2\right)}$

4.  $\frac{1}{\left(2{\mathrm{K}}_1{\mathrm{K}}_2\right)}$