1.
Reaction$$Ba{O_2}\left( s \right) \rightleftharpoons BaO\left( s \right) + {O_2}\left( g \right),$$ $$\Delta H = + ve$$
In equilibrium condition, pressure of $${O_2}$$ depends on
According to law of mass action, the rate of forward reaction $$ = {r_1}$$
$$\eqalign{
& {r_1} \propto \left[ {Ba{O_2}} \right] \cr
& {\text{or}}\,\,\,{r_1} = {k_1}\left[ {Ba{O_2}} \right] \cr} $$
$$Ba{O_2}$$ is solid substance in pure state concentration $$ = 1\,m$$
then, $${r_1} = {k_1}$$
Similarly the rate of backward reaction $$ = {r_2}$$
$$\eqalign{
& {r_2} \propto \left[ {BaO} \right]\left[ {{O_2}} \right] \cr
& {\text{or}}\,\,\,{r_2} = {k_2}\left[ {BaO} \right]\left[ {{O_2}} \right] \cr} $$
$$\because $$ Concentration of solid $$\left[ {BaO} \right] = 1\,\,\,\,\left[ {{O_2}\left( g \right)} \right]$$
$$\therefore \,\,\,{r_2} = {k_2}\left[ {{O_2}} \right]$$
$$\eqalign{
& {\text{At equilibrium,}} \cr
& {r_1} = {r_2} \cr
& {K_1} = {K_2}\left[ {{O_2}} \right] \cr
& {\text{or}}\,\,\,{K_1} = {K_2} \cdot \,{p_{{O_2}}} \cr
& {\text{where,}}\,\,{p_{{O_2}}} = {\text{partial pressure of}}\,{O_2} \cr
& {\text{or}}\,\,\,\,\,\frac{{{K_1}}}{{{K_2}}} = {p_{{O_2}}}\,{\text{(equilibrium constant)}} \cr
& \because \,\,\frac{{{K_1}}}{{{K_2}}} = K\,\,\,{\text{or}}\,\,K = {p_{{O_2}}} \cr} $$
So, from the above it is clear that pressure of $${O_2}$$ does not depend upon the concentration of reactants. The given equation is an endothermic reaction. If the temperature of such reaction is increased, then dissociation of $$Ba{O_2}$$ would increase and more $${O_2}$$ is produced.
2.
The equilibrium constant at $$298 K$$ for a reaction $$A + B \rightleftharpoons C + D$$ is 100. If the initial concentration of all the four species were $$1 M$$ each, then equilibrium concentration of $$D\left( {{\text{in}}\,{\text{mol}}\,{L^{ - 1}}} \right)$$ will be :
3.
The value of equilibrium constant of the reaction, $$HI\left( g \right) \rightleftharpoons \frac{1}{2}{H_2}\left( g \right) + \frac{1}{2}{I_2}\left( g \right)$$ is 8.0.
The equilibrium constant of the reaction, $${H_2}\left( g \right) + {I_2}\left( g \right) \rightleftharpoons 2HI\left( g \right)$$ will be
$$\eqalign{
& HI\left( g \right) \rightleftharpoons \frac{1}{2}{H_2}\left( g \right) + \frac{1}{2}{I_2}\left( g \right) \cr
& K = \frac{{{{\left[ {{I_2}} \right]}^{\frac{1}{2}}}{{\left[ {{H_2}} \right]}^{\frac{1}{2}}}}}{{\left[ {HI} \right]}}\,\,\,...{\text{(i)}} \cr
& {H_2}\left( g \right) + {I_2}\left( g \right) \rightleftharpoons 2HI\left( g \right) \cr
& K' = \frac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}\,\,\,...{\text{(ii)}} \cr
& {\text{From Eqs}}{\text{. (i) and (ii)}} \cr
& K \times \sqrt {K'} = 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K' = \frac{1}{{{K^2}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{{{\left( 8 \right)}^2}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{64}} \cr} $$
4.
The expression for equilibrium constant, $${K_c}$$ for the following reaction is $$2Cu{\left( {N{O_3}} \right)_{2\left( s \right)}} \rightleftharpoons $$ $$2Cu{O_{\left( s \right)}} + 4N{O_{2\left( g \right)}} + {O_{2\left( g \right)}}$$
A.
$${K_c} = \frac{{{{\left[ {Cu{O_{\left( s \right)}}} \right]}^2}{{\left[ {N{O_{2\left( g \right)}}} \right]}^4}\left[ {{O_{2\left( g \right)}}} \right]}}{{{{\left[ {Cu{{\left( {N{O_3}} \right)}_{2\left( s \right)}}} \right]}^2}}}$$
B.
$${K_c} = \frac{{{{\left[ {N{O_{2\left( g \right)}}} \right]}^4}\left[ {{O_{2\left( g \right)}}} \right]}}{{{{\left[ {Cu{{\left( {N{O_3}} \right)}_{2\left( s \right)}}} \right]}^2}}}$$
C.
$${K_c} = {\left[ {N{O_{2\left( g \right)}}} \right]^4}\left[ {{O_{2\left( g \right)}}} \right]$$
D.
$${K_c} = \frac{{{{\left[ {Cu{O_{\left( s \right)}}} \right]}^2}}}{{{{\left[ {Cu{{\left( {N{O_3}} \right)}_{2\left( s \right)}}} \right]}^2}}}$$
Answer :
$${K_c} = {\left[ {N{O_{2\left( g \right)}}} \right]^4}\left[ {{O_{2\left( g \right)}}} \right]$$
$$2Cu{\left( {N{O_3}} \right)_{2\left( s \right)}} \rightleftharpoons $$ $$2Cu{O_{\left( s \right)}} + 4N{O_{2\left( g \right)}} + {O_{2\left( g \right)}}$$
Since conc. of solids is taken as 1, the expression for $${K_c}$$ becomes $${K_c} = {\left[ {N{O_{2\left( g \right)}}} \right]^4}\left[ {{O_{2\left( g \right)}}} \right]$$
5.
$$5\,moles$$ of $$PC{l_5}$$ are heated in a closed vessel of 5 litre capacity. At equilibrium $$40\% $$ of $$PC{l_5}$$ is found to be dissociated. What is the value of $${K_c}?$$
6.
For the reaction : $$2Ba{O_2}\left( s \right) \rightleftharpoons 2BaO\left( s \right) + {O_2}\left( g \right);$$ $$\Delta H = + ve.$$ In equilibrium condition, pressure of $${O_2}$$ is dependent on
For the reaction
$$2Ba{O_2}\left( s \right) \rightleftharpoons 2BaO\left( s \right) + {O_2}\left( g \right);\Delta H = + ve.$$
At equilibrium $${K_p} = {P_{{O_2}}}$$
Hence, the value of equilibrium constant depends only upon partial pressure of $${O_2}.$$ Further on increasing temperature formation of $${O_2}$$ increases as this is an endothermic reaction. Hence, pressure of $${O_2}$$ is dependent on temperature.
7.
The $$\% $$ yield of ammonia as a function of time in the reaction $${N_{2\left( g \right)}} + 3{H_{2\left( g \right)}} \rightleftharpoons 2N{H_{3\left( g \right)}},$$ $$\Delta H < 0$$ at $$\left( {P,{T_1}} \right)$$ is given below.
If this reaction is conducted at $$\left( {P,{T_2}} \right),$$ with $${T_2} > {T_1},$$ the $$\% $$ yield of ammonia as a function of time is represented by
$$\Delta H < 0$$
Initially, with increase in temperature $$\left( {{T_2} > {T_1}} \right)\% $$ yield increases. Afterwards, equilibrium is reached and if the temperature is increased, i.e., heat is supplied to the system, then according to Le Chateliers principle, the equilibrium will shift in the backward direction, where the heat is absorbed. Hence, the % yield decreases.
8.
Which one of the following information can be obtained on the basis of Le-Chatelier’s
principle?
A.
Dissociation constant of a weak acid
B.
Entropy change in a reaction
C.
Equilibrium constant of a chemical reaction
D.
Shift in equilibrium position on changing value of a constant
Answer :
Shift in equilibrium position on changing value of a constant
Le-Chatelier's and Braun French chemists made certain generalisations to explain the effect of changes in concentrations, temperature or pressure on the state of system in equilibrium. When a system is subjected to a change in one of these factors, the equilibrium gets disturbed and the system re-adjusts itself until it return to equilibrium.
9.
For the reaction equilibrium $${N_2}{O_4}\left( g \right) \rightleftharpoons 2\,N{O_2}\left( g \right)$$ the concentrations of $${N_2}{O_4}$$ and $$N{O_2}$$ at equilibrium are $$4.8 \times {10^{ - 2}}$$ and $$1.2 \times {10^{ - 2}}mol\,{L^{ - 1}}$$ respectively. The value of $${K_c}$$ for the reaction is
10.
For the reaction $$:2N{O_{2\left( g \right)}} \rightleftharpoons 2N{O_{\left( g \right)}} + {O_{2\left( g \right)}},$$
$$\left( {{K_c} = 1.8 \times {{10}^{ - 6}}\,{\text{at}}\,{{184}^ \circ }C} \right)\left( {R = 0.0831\,kJ/\left( {mol.\,K} \right)} \right)$$
When $${K_p}$$ and $${K_c}$$ are compared at $${{{184}^ \circ }C}$$ it is found that
A.
Whether $${K_p}$$ is greater than, less than or equal to $${K_c}$$ depends upon the total gas pressure