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Chemical Equilibrium

NEET Chemistry · 93 questions · Page 8 of 10 · Click an option or "Show Solution" to reveal answer

Q71
In aqueous solution the ionization constants for carbonic acid are K1 = 4.2 x 10–7 and K2 = 4.8 x 10–11 Select the correct statement for a saturated 0.034 M solution of the carbonic acid.
A The concentration of CO32−CO_3^{2−} is 0.034 M.
B The concentration of CO32−CO_3^{2−} is greater than that of HCO3−HCO_3^{−}
C The concentration of H+ and HCO3−HCO_3^− are approximately equal.
D The concentration of H+ is double that of CO3−CO_3^−.
Correct Answer
Option C
Solution
H2CO3(aq)0.034−x + H2O(l) ⇌ HCO3−(aq)x + H3O+(aq)x\mathop {{H_2}C{O_3}\left( {aq} \right)}\limits_{0.034 - x} \, + \,{H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {HCO_3^ - \left( {aq} \right)}\limits_x \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_x
K1=[HCO3−][H3O+][H2CO3]{K_1} = {{\left[ {HCO_3^ - } \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {{H_2}C{O_3}} \right]}}
=x×x0.034−x= {{x \times x} \over {0.034 - x}}
⇒4.2×10−7=x20.034\Rightarrow 4.2 \times {10^{ - 7}} = {{{x^2}} \over {0.034}}
⇒x=1.195×10−4\Rightarrow x = 1.195 \times {10^{ - 4}}

As

H2CO3{H_2}C{O_3}

is a weak acid so the concentration of

H2CO3{H_2}C{O_3}

will remain

0.034  0.034\,\,

as

0.034>>x.0.034 > > x.
x=[H+]=[HCO3−]x = \left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right]
=1.195×10−4= 1.195 \times {10^{ - 4}}

Now,

HCO3−x−y(aq)+H2O(l) ⇌ CO32−(aq)y + H3O+(aq)y\mathop {HCO_3^ - }\limits_{x - y} \left( {aq} \right) + {H_2}O\left( l \right)\,\rightleftharpoons\,\mathop {CO_3^{2 - }\left( {aq} \right)}\limits_y \, + \,\mathop {{H_3}{O^ + }\left( {aq} \right)}\limits_y

As

HCO3− HCO_3^ - \,

is again a weak acid (weaker than

H2CO3){H_2}C{O_3})

with

x>>y.x > > y.
K2=[CO32−][H3O+][HCO3−]{K_2} = {{\left[ {CO_3^{{2^ - }}} \right]\left[ {{H_3}{O^ + }} \right]} \over {\left[ {HCO_3^ - } \right]}}
=y×(x+y)(x−y)= {{y \times \left( {x + y} \right)} \over {\left( {x - y} \right)}}

Note :

[H3O+]=H+  \left[ {{H_3}{O^ + }} \right] = {H^ + }\,\,

from first step

(x)(x)

and from second step

(y)=(x+y)\left( y \right) = \left( {x + y} \right)

[As

   x>>y  \,\,\,x > > y\,\,

so

   x+y≃x   \,\,\,x + y \simeq x\,\,\,

and

x−y≃xx - y \simeq x

] So,

   K2≃y×xx=y\,\,\,{K_2} \simeq {{y \times x} \over x} = y
⇒K2=4.8×10−11\Rightarrow {K_2} = 4.8 \times {10^{ - 11}}
=y=[CO32−]= y = \left[ {CO_3^{{2^ - }}} \right]

So the concentration of

   [H+]=[HCO3−]=   \,\,\,\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = \,\,\,

concentrations obtained from the first step.

As the dissociation will be very low in second step so there will be no change in these concentrations.

Thus the final concentrations are

[H+]=[HCO3−]=1.195×10−4\left[ {{H^ + }} \right] = \left[ {HCO_3^ - } \right] = 1.195 \times {10^{ - 4}}
&   [CO32−]=4.8×10−11\& \,\,\,\left[ {CO_3^{2 - }} \right] = 4.8 \times {10^{ - 11}}
Q72
The equilibrium constants KP1 and KP2 for the reactions X ⇋\leftrightharpoons 2Y and Z ⇋\leftrightharpoons P + Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at these equilibria is :
A 1 : 36
B 1 : 1
C 1 : 3
D 1 : 9
Correct Answer
Option A
Solution

Let the initial moles of

XX

be

′a′'a'

and that of

ZZ

be

′b′'b'

the for the given reactions, we have

X         ⇌         2YX\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,2Y
Initial            a  moles       0At  equi.       a(1−α)       2aα(moles)\begin{aligned}& Initial\,\,\,\,\,\,\,\,\,\,\,\,a\,\,moles\,\,\,\,\,\,\,0 \\ & At\,\,equi.\,\,\,\,\,\,\,a\left( {1 - \alpha } \right)\,\,\,\,\,\,\,2a\alpha \\ & (moles)\end{aligned}

Total no. of moles

=a(1−α)+2aα= a\left( {1 - \alpha } \right) + 2a\alpha
=a−aα+2aα= a - a\alpha + 2a\alpha
=a(1+α)= a\left( {1 + \alpha } \right)

Now,

   Kp1=(ny)2nx×(PT1∑n)Δn\,\,\,{K_{{p_1}}} = {{{{\left( {{n_y}} \right)}^2}} \over {{n_x}}} \times {\left( {{{{P_{{T_1}}}} \over {\sum n }}} \right)^{\Delta n}}

or,

   Kp1=(2aα)2.PT1[a(1−α)][a(1+α)]\,\,\,{K_{{p_1}}} = {{{{\left( {2a\alpha } \right)}^2}.{P_{{T_1}}}} \over {\left[ {a\left( {1 - \alpha } \right)} \right]\left[ {a\left( {1 + \alpha } \right)} \right]}}
            \,\,\,\,\,\,\,\,\,\,\,\,
                 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
Z         ⇌         P+QZ\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,P+Q
Initial           b  moles         0           0Ar  equi.      b(1−α)        bα       bα(moles)\begin{aligned}& Initial\,\,\,\,\,\,\,\,\,\,\,b\,\,moles\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0 \\ & Ar\,\,equi.\,\,\,\,\,\,b\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,b\alpha \,\,\,\,\,\,\,b\alpha \\ & (moles)\end{aligned}

Total no. of moles

=b(1−α)+bα+bα= b\left( {1 - \alpha } \right) + b\alpha + b\alpha
=b−bα+bα+bα= b - b\alpha + b\alpha + b\alpha
=b(1+α)= b\left( {1 + \alpha } \right)

Now

   KP2=nQ×nPnz×[PT2∑n ]Δn\,\,\,{K_{{P_2}}} = {{{n_Q} \times {n_P}} \over {{n_z}}} \times {\left[ {{{{P_{{T_2}}}} \over {\sum\nolimits_n \, }}} \right]^{\Delta n}}

or

   KP2=(bα)(bα).PT2[b(1−α)][b(1+α)]\,\,\,{K_{{P_2}}} = {{\left( {b\alpha } \right)\left( {b\alpha } \right).{P_{{T_2}}}} \over {\left[ {b\left( {1 - \alpha } \right)} \right]\left[ {b\left( {1 + \alpha } \right)} \right]}}

or

   \,\,\,
   KP1KP2=4α2.PT1(1−α2)×(1−α)2PT2.α2\,\,\,{{{K_{{P_1}}}} \over {{K_{P2}}}} = {{4{\alpha ^2}.{P_{{T_1}}}} \over {\left( {1 - {\alpha ^2}} \right)}} \times {{{{\left( {1 - \alpha } \right)}^2}} \over {{P_{{T_2}}}.{\alpha ^2}}}
                      \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
=4PT1PT2= {{4{P_{{T_1}}}} \over {{P_{{T_2}}}}}

or

   PT1PT2=19     \,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over 9}\,\,\,\,\,

[ as

   KP1KP1=19  \,\,\,{{{K_{{P_1}}}} \over {{K_{{P_1}}}}} = {1 \over 9}\,\,

given ] or

   PT1PT2=136\,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over {36}}

or

   1:36\,\,\,1:36
Q73
What is the equilibrium expression for the reaction P4 (s) + 5O2 ⇋\leftrightharpoons P4O10 (s)?
A Kc = [P4O10] / 5[P4] [O2]
B Kc = 1/[O2]5
C Kc = [P4O10] / [P4] [O2]5
D Kc = [O2]5
Correct Answer
Option B
Solution

For

P4(s)+5O2(g) ⇌ P4O10(8){P_4}\left( s \right) + 5{O_2}\left( g \right)\,\rightleftharpoons\,{P_4}{O_{10}}\left( 8 \right)
Kc=1(O2)5.{K_c} = {1 \over {{{\left( {{O_2}} \right)}^5}}}.

The solids have concentration unity

Q74
Change in volume of the system does not alter which of the following equilibria?
A N2(g) + O2(g) ⇋\leftrightharpoons 2NO (g)
B PCl5(g) ⇋\leftrightharpoons PCl3 (g) + Cl2 (g)
C N2(g) + 3H2(g) ⇋\leftrightharpoons 2NH3 (g)
D SO2Cl2 (g) ⇋\leftrightharpoons SO2 (g) + Cl2 (g)
Correct Answer
Option A
Solution

In this reaction the ratio of number of moles of reactants to products is same

i.e.  2:2,i.e.\,\,2:2,

hence change in volume will not alter the number of moles.

Q75
For the reaction, 2SO2(g) + O2(g) = 2SO3(g), Δ\Delta H = –57.2 kJ mol–1 and KC = 1.7 × 1016 Which of the following statement is incorrect ?
A The equilibrium will shift in forward direction as the pressure increase.
B The addition of inert gas at constant volume will be not affect the equilibrium constant.
C The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required.
D The equilibrium constant decreases as the temperature increase.
Correct Answer
Option C
Solution

Equilibrium constant has no relation with catalyst.

Catalyst only affects the rate with which a reaction proceeds.

Here we use catalyst V2O5 to speed up the reaction.

Q76
The equilibrium constant for the reaction N2(g) + O2(g) ⇋\leftrightharpoons 2NO(g) at temperature T is 4 ×\times 10-4. The value of Kc for the reaction NO(g) ⇋\leftrightharpoons 121 \over 2N2 (g) + 121 \over 2O2 (g) at the same temperature is :
A 2.5 ×\times 102
B 4 ×\times 10-4
C 50
D 0.02
Correct Answer
Option C
Solution
Kc=[NO]2[N2][O2]=4×10−4{K_c} = {{{{\left[ {NO} \right]}^2}} \over {\left[ {{N_2}} \right]\left[ {{O_2}} \right]}} = 4 \times {10^{ - 4}}
Kc′=[N2]1/2[Q2]1/2[NO]K{'_c} = {{{{\left[ {{N_2}} \right]}^{1/2}}{{\left[ {{Q_2}} \right]}^{1/2}}} \over {\left[ {NO} \right]}}
=1Kc= {1 \over {\sqrt {{K_c}} }}
=14×10−4= {1 \over {\sqrt {4 \times {{10}^{ - 4}}} }}
=50= 50
Q77
Phosphorus pentachloride dissociates as follows, in a closed reaction vessel PCl5 (g) ⇋\leftrightharpoons PCl3 (g) + Cl2 (g) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be
A (xx+1)P\left( {{x \over {x + 1}}} \right)P
B (2x1−x)P\left( {{2x \over {1 - x}}} \right)P
C (xx−1)P\left( {{x \over {x - 1}}} \right)P
D (x1−x)P\left( {{x \over {1 - x}}} \right)P
Correct Answer
Option A
Solution
PCl5(g)1−x ⇌ PCl3(g)x  +  Cl2(g)x\mathop {PC{l_5}\left( g \right)}\limits_{1 - x} \mathop {\,\rightleftharpoons\,PC{l_3}\left( g \right)}\limits_x \,\, + \,\,\mathop {C{l_2}\left( g \right)}\limits_x

Total moles after dissociation

1−x+x+x=1+x1 - x + x + x = 1 + x
PPCl3={P_{PC{l_3}}} =

mole fraction of

PCl3×PCl{}_3 \times

Total pressure

=(x1+x)P= \left( {{x \over {1 + x}}} \right)P
Q78
If the equilibrium constant for A ⇌ B + C is Keq(1)K_{eq}^{(1)} and that of B + C ⇌ P is Keq(2)K_{eq}^{(2)}, the equilibrium constant for A ⇌ P is :
A Keq(1)Keq(2){{K_{eq}^{(1)}} \over {K_{eq}^{(2)}}}
B Keq(1){K_{eq}^{(1)}} + Keq(2){K_{eq}^{(2)}}
C Keq(2){K_{eq}^{(2)}} - Keq(1){K_{eq}^{(1)}}
D Keq(1){K_{eq}^{(1)}} Keq(2){K_{eq}^{(2)}}
Correct Answer
Option D
Solution

We have two reactions: A  ⇌  B+Cwith Keq(1) A \;\rightleftharpoons\; B + C \quad\text{with } K_{eq}^{(1)} B+C  ⇌  Pwith Keq(2) B + C \;\rightleftharpoons\; P \quad\text{with } K_{eq}^{(2)} We want the equilibrium constant for the overall reaction A  ⇌  P. A \;\rightleftharpoons\; P. How to combine the equilibrium constants When two reactions are added to yield a net (overall) reaction, the equilibrium constant of the net reaction is the product of the equilibrium constants of the individual reactions.

Formally: (Reaction 1)+(Reaction 2)  ⇒  (Net Reaction), \text{(Reaction 1)} + \text{(Reaction 2)} \;\Rightarrow\; \text{(Net Reaction)}, and Knet  =  K(Reaction 1)×K(Reaction 2). K_{\text{net}} \;=\; K_{\text{(Reaction 1)}} \times K_{\text{(Reaction 2)}}. Applying it here Reaction 1: A→B+CA \to B + C has Keq(1).K_{eq}^{(1)}. Reaction 2: B+C→PB + C \to P has Keq(2).K_{eq}^{(2)}. When we add them, A⏟Reactant 1  ⟶  B+C⏟Products of Reaction 1=Reactants of Reaction 2  ⟶  P⏟Product of Reaction 2. \underbrace{A}_{\text{Reactant 1}} \;\longrightarrow\; \underbrace{B + C}_{\text{Products of Reaction 1}=\text{Reactants of Reaction 2}} \;\longrightarrow\; \underbrace{P}_{\text{Product of Reaction 2}}. So the net reaction is A→PA \to P, and its equilibrium constant is Keq(net)  =  Keq(1)×Keq(2). K_{eq}^{\text{(net)}} \;=\; K_{eq}^{(1)} \times K_{eq}^{(2)}. Thus, the correct answer is: Keq(1) Keq(2). \boxed{K_{eq}^{(1)} \, K_{eq}^{(2)}}. Hence, the answer (Option D) is correct.

Q79
The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal : Fe2O3(s) + 3 CO(g) ⇌\rightleftharpoons 2 Fe(1) + 3 CO2(g) Using the Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium ?
A Removal of CO
B Removal of CO2
C Addition of CO2
D Addition of Fe2O3
Correct Answer
Option D
Solution

Addition of a solid component to a system at constant pressure has no effect on the equilibrium.

Therefore, addition of Fe2O3 will not disturb the equilibrium.

Q80
The equilibrium constant (KC) for the reaction N2(g) + O2(g) →\to 2NO(g) at temperature T is 4 ×\times 10–4. The value of KC for the reaction, NO(g) →\to 1/2N2(g) + 1/2O2(g) at the same temperature is :
A 0.02
B 2.5 ×\times 102
C 4 ×\times 10-4
D 50.0
Correct Answer
Option D
Solution

For the reaction

N2+O2→2NO{N_2} + {O_2} \to 2NO
 K=4×10−4\,K = 4 \times {10^{ - 4}}

Hence for the reaction

NO→12N2+12O2NO \to {1 \over 2}{N_2} + {1 \over 2}{O_2}
K′=1KK' = {1 \over {\sqrt K }}
=14×10−4= {1 \over {\sqrt {4 \times {{10}^{ - 4}}} }}
=50= 50
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