Chemical Equilibrium MCQ Questions & Answers in Physical Chemistry | Chemistry
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141.
For the reaction $$S{O_{2\left( g \right)}} + \frac{1}{2}{O_{2\left( g \right)}} \rightleftharpoons S{O_{3\left( g \right)}},$$ if $${K_p} = {K_c}{\left( {RT} \right)^x}$$ where the symbols have usual meaning then the value of $$x$$ is ( assuming ideality ) :
A
$$- 1$$
B
$$ - \frac{1}{2}$$
C
$$\frac{1}{2}$$
D
$$1$$
Answer :
$$ - \frac{1}{2}$$
$$\eqalign{
& S{O_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons S{O_3}\left( g \right) \cr
& {K_p} = {K_C}{\left( {RT} \right)^x} \cr} $$
where $$x = \Delta {n_g} = $$ number of gaseous moles in product - number of gaseous moles in reactant
$$\eqalign{
& = 1 - \left( {1 + \frac{1}{2}} \right) \cr
& = 1 - \frac{3}{2} \cr
& = - \frac{1}{2} \cr} $$
142.
Let the solubility of an aqueous solution of $$Mg{\left( {OH} \right)_2}$$ be $$x$$ then its $${k_{sp}}$$ is
143.
Study the given figure and label $$X, Y$$ and $$Z.$$
$$X$$
$$Y$$
$$Z$$
(a)
Backward reaction
Forward reaction
Products
(b)
Forward reaction
Backward reaction
Equilibrium
(c)
Reversible reaction
Irreversible reaction
Equilibrium
(d)
Forward reaction
Forward reaction
Backward reaction
A
(a)
B
(b)
C
(c)
D
(d)
Answer :
(b)
The concentration of reactants decreases and that of products increases with time. Rate of reaction increases with time.
At equilibrium, $${{\text{R}}_f} = {R_b}$$
144.
$$18.4\,g$$ of $${N_2}{O_4}$$ is taken in a $$1\,L$$ closed vessel and heated till the equilibrium is reached.
$${N_2}{O_{4\left( g \right)}} \rightleftharpoons 2N{O_{2\left( g \right)}}$$
At equilibrium it is found that $$50\% $$ of $${N_2}{O_4}$$ is dissociated. What will be the value of equilibrium constant?
A
0.2
B
2
C
0.4
D
0.8
Answer :
0.4
$${K_c} = \frac{{0.2 \times 0.2}}{{0.1}} = 0.4$$
145.
A liquid is in equilibrium with its vapour at its boiling point. On the average, the molecules in the two phases have equal :
A
inter-molecular forces
B
potential energy
C
total energy
D
kinetic energy
Answer :
total energy
Vapours and liquid are at the same temperature.
146.
$${K_1},{K_2}$$ and $${K_3}$$ are the equilibrium constants of the following reactions (I), (II) and (III) respectively :
$$\eqalign{
& \left( {\text{I}} \right)\,{N_2} + 2{O_2} \rightleftharpoons 2N{O_2} \cr
& \left( {{\text{II}}} \right)\,2N{O_2} \rightleftharpoons {N_2} + 2{O_2} \cr
& \left( {{\text{III}}} \right)\,\,N{O_2} \rightleftharpoons \frac{1}{2}{N_2} + {O_2} \cr} $$
The correct relation from the following is
A
$${K_1} = \frac{1}{{{K_2}}} = \frac{1}{{{K_3}}}$$
B
$${K_1} = \frac{1}{{{K_2}}} = \frac{1}{{{{\left( {{K_3}} \right)}^2}}}$$
147.
A vessel at $$1000 K$$ contains $$C{O_2}$$ with a pressure of 0.5 atm. Some of the $$C{O_2}$$ is converted into $$CO$$ on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of $$K$$ is :
148.
$$0.6\,mole$$ of $$PC{l_5},0.3\,mole$$ of $$PC{l_3}$$ and $$0.5\,mole$$ of $$C{l_2}$$ are taken in a $$1\,L$$ flask to obtain the following equilibrium : $$PC{l_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}}$$
If the equilibrium constant $${K_c}$$ for the reaction is 0.2, predict the direction of the reaction.
A
Forward direction
B
Backward direction
C
Direction of the reaction cannot be predicted
D
Reaction does not move in any direction.
Answer :
Backward direction
$$\eqalign{
& PC{l_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}} \cr
& {Q_c} = \frac{{0.5 \times 0.3}}{{0.6}} = 0.25 \cr} $$
$${K_c} = 0.2,$$ Since, $${Q_c} > {K_c}$$ reaction will proceed in backward direction.
149.
Using the Gibbs energy change $$\Delta {G^ \circ } = + 63.3\,kJ$$ for the following reaction,
$$A{g_2}C{O_3}\left( s \right) \rightleftharpoons $$ $$2A{g^ + }\left( {aq} \right) + CO_3^{2 - }\left( {aq} \right)$$ the $${K_{sp}}$$ of $$A{g_2}C{O_3}\left( s \right)$$ in water at $${25^ \circ }C$$ is $$\left( {R = 8.314\,J{K^{ - 1}}mo{l^{ - 1}}} \right)$$