## Chemical Equilibrium Questions and Answers Part-4

1. The equilibrium constant for the reversible reaction $N_{2}+3H_{2}\rightleftharpoons2NH_{3}$    is K and for the reaction
$1\diagup2 N_{2}+3\diagup2 H_{2}\rightleftharpoons NH_{3}$    , the equilibrium constant is K'. K and K' will be related as
a) K = K'
b) $K' =\sqrt{K}$
c) $K =\sqrt{K'}$
d) K × K' = 1

Explanation: $K' =\sqrt{K}$

2. If $K_{1} and K_{2}$  are respective equilibrium constants for the two reactions $XeF_{6}\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons XeOF_{4}\left(g\right)+2HF\left(g\right)$
$XeO_{4}\left(g\right)+XeF_{6}\left(g\right)\rightleftharpoons XeOF_{4}\left(g\right)+XeO_{3}F_{2}\left(g\right)$
the equilibrium constant for the reaction
$XeO_{4}\left(g\right)+2HF\left(g\right)\rightleftharpoons XeO_{3}F_{2}\left(g\right)+H_{2}O\left(g\right)$
will be
a) $\frac{K_{1}}{K_2^2}$
b) $K_{1}.K_{2}$
c) $\frac{K_{1}}{K_{2}}$
d) $\frac{K_{2}}{K_{1}}$

Explanation: Reaction (II) and reverse of reaction (I) gives the desired reaction hence K = $K_{2} \times \frac{1}{K_{1}} = \frac{K_{2}}{K_{1}}$

3. A cylinder fitted with a movable piston contains liquid water in equilibrium with water vapour at 25°C. Which operation result in a decrease in the equilibrium vapour pressure
a) Moving the piston downward a short distance
b) Removing a small amount of vapour
c) Removing a small amount of the liquid water
d) Dissolving salt in the water

Explanation: Dissolution of salt lowers the V.P. It is also effected by temperature

4. The volume of the reaction vessel containing an equilibrium mixture in the reaction $SO_{2}CI_{2}\left(g\right)\rightleftharpoons SO_{2}\left(g\right) + CI_{2}\left(g\right)$
is increased when the equilibrium is re-established
a) The amount of $SO_{2}\left(g\right)$  will decrease
b) The amount of $SO_{2}CI_{2}\left(g\right)$  will increase
c) The amount of $CI_{2}\left(g\right)$  will increase
d) The amount of $CI_{2}\left(g\right)$  will remain unchanged

Explanation: It will decrease the concentration. The equilibrium will shift in the direction where more moles are formed to keep Kc constant.

5. In gaseous equilibrium the correct relation between $K_{C} and K_{P}$   is
a) $K_{c}=K_{p}\left(RT\right)^{\triangle n}$
b) $K_{p}=K_{c}\left(RT\right)^{\triangle n}$
c) $\frac{K_{c}}{RT}=\left(K_{p}\right)^{\triangle n}$
d) $\frac{K_{p}}{RT}=\left(K_{c}\right)^{\triangle n}$

Explanation: Relation is $K_{p}=K_{c}\left(RT\right)^{\triangle n}$

6. In which of the following reaction $K_{P}>K_{C}$
a) $N_{2}+3H_{2}\rightleftharpoons 2NH_{3}$
b) $H_{2}+1_{2}\rightleftharpoons 2HI$
c) $PCl_{3}+Cl_{2}\rightleftharpoons PCl_{5}$
d) $2SO_{3}\rightleftharpoons O_{2}+2SO_{2}$

Explanation: $\triangle$n = 3 – 2 = 1
$K_{p}=K_{c}\left(RT\right)$    hence $K_{P}>K_{C}$

7. For reaction $PCl_{3}\left(g\right)+CI_{2}\left(g\right)\rightleftharpoons PCl_{5}\left(g\right)$      , the value of $K_{C}$ at 250°C is 26 mol–1 litre1. The value of $K_{P}$ at this temperature will be
a) $0.61 atm^{-1}$
b) $0.57 atm^{-1}$
c) $0.83 atm^{-1}$
d) $0.46 atm^{-1}$

Explanation: $\triangle$n = –1
Kp = 26 × (0.0821 × 523)–1 = 0.61 atm–1

8. The equilibrium constant for the reaction, $N_{2}\left(g\right)+O_{2}\left(g\right)\rightleftharpoons 2NO\left(g\right) is 4 ×10^{-4}$       at 2000 K. In presence of a catalyst, equilibrium is attained ten times faster. Therefore, the equilibrium constant, in presence of the catalyst, at 2000 K is
a) $40 × 10^{-4}$
b) $4 × 10^{-4}$
c) $4 × 10^{-3}$
d) difficult to compute without more data

Explanation: Kc is not influenced by presence of a catalyst

9. For a chemical reaction $2A + B\rightleftharpoons C$  , the thermodynamic equilibrium constant $K_{P}$ is
a) in $atm^{-2}$
b) in $atm^{-3}$
c) in $atm^{-1}$
d) dimensionless

Explanation: Unit of Kp = $\left(Atm\right)^{\triangle n}$ = (Atm)–2 ($\triangle n$ = moles of products – moles of reactants)
Q = $\frac{[Conc. of Products]}{[Conc. of Reactants]}$