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)
10. When two reactants, A and B are mixed to give products C and D, the reaction quotient Q, at the initial stage of the reaction
a) is zero
b) decreases with time
c) is independent of time
d) increases with time
Explanation: Q increases with the formation of products
Q = \[\frac{[Conc. of Products]}{[Conc. of Reactants]}\]