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Chemical Equilibrium

JEE Chemistry · 56 questions · Page 2 of 6 · Click an option or "Show Solution" to reveal answer

Q11
For the reaction, CO(g) + Cl2(g) ⇋\leftrightharpoons COCl2(g) the KpKc{{{K_p}} \over {{K_c}}} is equal to :
A RT\sqrt {RT}
B RT
C 1/RT
D 1.0
Correct Answer
Option C
Solution
Kp=Kc(RT)Δn;{K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};

Here

Δn=1−2=−1\Delta n = 1 - 2 = - 1

∴\therefore

kpKc=1RT{{{k_p}} \over {{K_c}}} = {1 \over {RT}}
Q12
At −20∘C-20^{\circ} \mathrm{C} and 1 atm1 \mathrm{~atm} pressure, a cylinder is filled with equal number of H2,I2\mathrm{H}_2, \mathrm{I}_2 and HI\mathrm{HI} molecules for the reaction H2( g)+I2( g)⇌2HI(g)\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}), the Kp\mathrm{K}_{\mathrm{p}} for the process is x×10−1x \times 10^{-1}. x=\mathrm{x}= __________. [Given : R=0.082 L atm K−1 mol−1\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}]
A 2
B 1
C 10
D 0.01
Correct Answer
Option C
Solution

At

−20∘C-20^{\circ} \mathrm{C}

and a pressure of

1 atm1 \mathrm{~atm}

, a cylinder contains equal quantities of

H2,I2,\mathrm{H}_2, \mathrm{I}_2,

and

HI\mathrm{HI}

molecules. The equilibrium constant for the reaction

H2( g)+I2( g)⇌2HI(g)\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})

denoted as

Kp\mathrm{K}_{\mathrm{p}}

, is given by the expression

x×10−1x \times 10^{-1}

. To solve for

xx

, use the provided equilibrium expression:

KP=KC=[HI]2[H2][I2]=1=10×10−1\mathrm{K_{P} = \mathrm{K_C}} =\frac{\left[\mathrm{HI}\right]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=1 = 10 \times 10^{-1}

Therefore,

x=10x = 10

.

Q13
The equilibrium constant for the reaction SO3 (g) ⇋\leftrightharpoons SO2 (g) + 121 \over 2 O2 (g) is Kc = 4.9 ×\times 10–2. The value of Kc for the reaction 2SO2 (g) + O2 (g) ⇋\leftrightharpoons 2SO3 (g) will be :
A 416
B 9.8 ×\times 10-2
C 4.9 ×\times 10-2
D 2.40 ×\times 10-3
Correct Answer
Option A
Solution
SO3(g) ⇌ SO2(g)  +  12O2(g)S{O_3}\left( g \right)\,\rightleftharpoons\,S{O_2}\left( g \right)\,\, + \,\,{1 \over 2}{O_2}\left( g \right)
Kc=[SO2][O2]1/2[SO3]{K_c} = {{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{{{1}/{2}}}}} \over {\left[ {S{O_3}} \right]}}
=4.9×10−2;= 4.9 \times {10^{ - 2}};

On taking the square of the above reaction

[SO2]2[O2][SO3]2{{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]} \over {{{\left[ {S{O_3}} \right]}^2}}}
=24.01×10−4= 24.01 \times {10^{ - 4}}

now

KC′K{'_C}

for

  2SO2(g)  +  O2(g) ⇌ 2SO3\,\,2S{O_2}\left( g \right)\,\, + \,\,{O_2}\left( g \right)\,\rightleftharpoons\,2S{O_3}
=[SO3]2[SO2]2[O2]= {{{{\left[ {S{O_3}} \right]}^2}} \over {{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}
=124.01×10−4= {1 \over {24.01 \times {{10}^{ - 4}}}}
=416= 416
Q14
The equilibrium constant for the reversible reaction 2A(g) ⇌\rightleftharpoons 2B(g) + C(g) is K1 32{3 \over 2}A(g) ⇌\rightleftharpoons 32{3 \over 2}B(g) + 34{3 \over 4}C(g) is K2. K1 and K2 are related as :
A K1=K2{K_1} = \sqrt {{K_2}}
B K2=K1{K_2} = \sqrt {{K_1}}
C K2=K13/4{K_2} = K_1^{3/4}
D K1=K23/4{K_1} = K_2^{3/4}
Correct Answer
Option C
Solution

2A(g)=2B( g)+C(g)K1...(i)2 \mathrm{A}(\mathrm{g})=2 \mathrm{B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \quad K_{1} \quad\quad ...(i)

32A(g)=32B(g)+34C(g)K2...(ii)\frac{3}{2} \mathrm{A}(\mathrm{g})=\frac{3}{2} \mathrm{B}(\mathrm{g})+\frac{3}{4} \mathrm{C}(\mathrm{g}) \quad K_{2}\quad\quad...(ii)

eq. (ii) is 34\dfrac{3}{4} times of eq. (i), hence, K2=(K1)34K_{2}=\left(K_{1}\right)^{\dfrac{3}{4}}

Q15
For the reaction SO2 (g) + 12O2(g)⇋{1 \over 2} O_2(g) \leftrightharpoons SO3(g). if KP = KC(RT)x where the symbols have usual meaning then the value of x is: (assuming ideality)
A -1
B -1/2
C 1/2
D 1
Correct Answer
Option B
Solution
SO2(g)+12O2(g) ⇌ SO3(g)S{O_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right)\,\rightleftharpoons\,S{O_3}\left( g \right)
Kp=KC(RT)x{K_p} = {K_C}{\left( {RT} \right)^x}

where

x=Δng=x = \Delta {n_g} =

number of gaseous moles in product

                            \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

−- number of gaseous moles in reactant

=1−(1+12)= 1 - \left( {1 + {1 \over 2}} \right)
=1−32=−12= 1 - {3 \over 2} = - {1 \over 2}
Q16
For the reaction equilibrium N2O4 (g) ⇋\leftrightharpoons 2NO2 (g) the concentrations of N2O4 and NO2 at equilibrium are 4.8 ×\times 10-2 and 1.2 ×\times 10-2 mol L-1 respectively. The value of Kc for the reaction is
A 3 ×\times 10-1 mol L-1
B 3 ×\times 10-3 mol L-1
C 3 ×\times 103 mol L-1
D 3.3 ×\times 102 mol L-1
Correct Answer
Option B
Solution
Kc=[NO2]2[N2O4]=[1.2×10−2]2[4.8×10−2]{K_c} = {{{{\left[ {N{O_2}} \right]}^2}} \over {\left[ {{N_2}{O_4}} \right]}} = {{\left[ {1.2 \times {{10}^{ - 2}}} \right]^2} \over {\left[ {4.8 \times {{10}^{ - 2}}} \right]}}
=3×10−3 mol/L= 3 \times {10^{ - 3}}\,mol/L
Q17
4.0 moles of argon and 5.0 moles of PCl5 are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The Kp for the reaction is : [Given : R = 0.082 L atm K−-1 mol−-1]
A 2.25
B 6.24
C 12.13
D 15.24
Correct Answer
Option A
Solution

Here 4 moles of inert gas argon also present.

∴\therefore Total moles of mixture present at equilibrium, nT = 5 + x + 4 = 9 + x At equilibrium, total pressure (pT) = 6 atm Volume (v) = 100 L Temperature = 610 K ∴\therefore Using ideal gas equation,

PTV=nTRT{P_T}V = {n_T}RT
⇒6×100=(9+x)×0.082×610\Rightarrow 6 \times 100 = (9 + x) \times 0.082 \times 610
⇒x=3\Rightarrow x = 3

Now,

KP=PPCl3×PCl2PPCl5{K_P} = {{{P_{PC{l_3}}} \times {P_{C{l_2}}}} \over {{P_{PC{l_5}}}}}
=[39+3×6]×[39+3×6][5−39+3×6]= {{\left[ {{3 \over {9 + 3}} \times 6} \right] \times \left[ {{3 \over {9 + 3}} \times 6} \right]} \over {\left[ {{{5 - 3} \over {9 + 3}} \times 6} \right]}}
=2712= {{27} \over {12}}
=94= {9 \over 4}

= 2.25 atm Note : Inert gas always contribute to total mole and pressure calculation.

Q18
Consider the reaction N2(g) + 3H2(g) ⇌\rightleftharpoons 2NH3(g) The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that PNH3 << Ptotal at equilibrium)
A 332KP12P24{{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 4}
B KP12P24{{K_P^{{1 \over 2}}{P^2}} \over 4}
C 332KP12P216{{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 16}
D KP12P216{{K_P^{{1 \over 2}}{P^2}} \over 16}
Correct Answer
Option C
Solution

N2(g) + 3H2(g) ⇌\rightleftharpoons 2NH3(g) ; Keq = Kp Write this equation reverse way, 2NH3(g) ⇌\rightleftharpoons N2(g) + 3H2(g) ; Keq =

1Kp{1 \over {{K_p}}}

.tg .tg 2NH3(g) ⇌ N2(g) + 3H2(g) At t = 0 Po 0 0 At t = teq PNH3 p 3p At equillibrium PTotal = PNH3 + PN2 + PH2 = PNH3 + p + 3p (As PNH3 << Ptotal so we can ignore PNH3) ∴\therefore PTotal = 4p ⇒\Rightarrow p =

Ptotal4{{{{P_{total}}} \over 4}}

Formula of Keq =

pN2×(pH2)3(pNH3)2{{{p_{{N_2}}} \times {{\left( {{p_{{H_2}}}} \right)}^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}

=

1Kp{1 \over {{K_p}}}

⇒\Rightarrow

1Kp{1 \over {{K_p}}}

=

p×27p3(pNH3)2{{p \times 27{p^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}

⇒\Rightarrow

(pNH3)2{{{\left( {{p_{N{H_3}}}} \right)}^2}}

= Kp ×\times 27 ×\times

(Ptotal4)4{{{\left( {{{{P_{total}}} \over 4}} \right)}^4}}

⇒\Rightarrow PNH3 =

Kp×(27)12×(PTotal4)42\sqrt {{K_p}} \times {\left( {27} \right)^{{1 \over 2}}} \times {\left( {{{{P_{Total}}} \over 4}} \right)^{{4 \over 2}}}

=

332Kp12PTotal216{{{3^{{3 \over 2}}}K_p^{{1 \over 2}}P_{Total}^2} \over {16}}
Q19
Given below are two statements : Statement I : A catalyst cannot alter the equilibrium constant (Kc)\left(\mathrm{K}_{\mathrm{c}}\right) of the reaction, temperature remaining constant. Statement II : A homogenous catalyst can change the equilibrium composition of a system, temperature remaining constant. In the light of the above statements, choose the correct answer from the options given below
A Statement I is true but Statement II is false
B Statement I is false but Statement II is true
C Both Statement I and Statement II are true
D Both Statement I and Statement II are false
Correct Answer
Option A
Solution

Let's analyze both statements: Statement I: "A catalyst cannot alter the equilibrium constant (

KcK_c

) of the reaction, temperature remaining constant."

A catalyst speeds up both the forward and reverse reactions equally, allowing the system to reach equilibrium faster.

However, it does not change the energy difference between reactants and products (i.e., the Gibbs free energy change), which directly determines the equilibrium constant.

Therefore, this statement is true.

Statement II: "A homogenous catalyst can change the equilibrium composition of a system, temperature remaining constant."

A homogeneous catalyst acts in the same phase as the reactants and, like any catalyst, it only helps the reaction reach equilibrium more quickly.

It does not alter the equilibrium position, meaning the relative concentrations of reactants and products at equilibrium remain unchanged if the temperature is constant.

Thus, this statement is false.

In summary: Statement I is true.

Statement II is false.

The correct answer is: Option A.

Q20
The value of KC is 64 at 800 K for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) The value of KC for the following reaction is : NH3(g) ⇌ 12{1 \over 2}N2(g) + 32{3 \over 2}H2(g)
A 8
B 18{1 \over 8}
C 14{1 \over 4}
D 164{1 \over {64}}
Correct Answer
Option B
Solution

N2(g) + 3H2(g) ⇌ 2NH3(g) ; KC 2NH3(g) ⇌ N2(g) + 3H2(g) ;

1KC{1 \over {{K_C}}}

Multiplying by

12{1 \over 2}

, reaction becomes NH3(g) ⇌

12{1 \over 2}

N2(g) +

32{3 \over 2}

H2(g) ; ∴\therefore New KC =

(1KC)12{\left( {{1 \over {{K_C}}}} \right)^{{1 \over 2}}}

=

(164)12{\left( {{1 \over {64}}} \right)^{{1 \over 2}}}

=

18{1 \over 8}
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