Chemical Equilibrium — Page 6 | NEET Chemistry

Q51

The equilibrium constant for the reversible reaction 2A(g)

\rightleftharpoons

2B(g) + C(g) is K1

{3 \over 2}

A(g)

\rightleftharpoons

{3 \over 2}

B(g) +

{3 \over 4}

C(g) is K2. K1 and K2 are related as :

A

{K_1} = \sqrt {{K_2}}

B

{K_2} = \sqrt {{K_1}}

C

{K_2} = K_1^{3/4}

D

{K_1} = K_2^{3/4}

Correct Answer

Option C

Solution

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

\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 $\dfrac{3}{4}$ times of eq. (i), hence, $K_{2}=\left(K_{1}\right)^{\dfrac{3}{4}}$

Q52

For the reaction SO2 (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

S{O_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right)\,\rightleftharpoons\,S{O_3}\left( g \right)

{K_p} = {K_C}{\left( {RT} \right)^x}

where

x = \Delta {n_g} =

number of gaseous moles in product

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

$-$ number of gaseous moles in reactant

= 1 - \left( {1 + {1 \over 2}} \right)

= 1 - {3 \over 2} = - {1 \over 2}

Q53

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

{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 \times {10^{ - 3}}\,mol/L

Q54

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,

{P_T}V = {n_T}RT

\Rightarrow 6 \times 100 = (9 + x) \times 0.082 \times 610

\Rightarrow x = 3

Now,

{K_P} = {{{P_{PC{l_3}}} \times {P_{C{l_2}}}} \over {{P_{PC{l_5}}}}}

= {{\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]}}

= {{27} \over {12}}

= {9 \over 4}

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

Q55

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

{{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 4}

B

{{K_P^{{1 \over 2}}{P^2}} \over 4}

C

{{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 16}

D

{{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 =

{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 =

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

Formula of Keq =

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

=

{1 \over {{K_p}}}

$\Rightarrow$

{1 \over {{K_p}}}

=

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

$\Rightarrow$

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

= Kp $\times$ 27 $\times$

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

$\Rightarrow$ PNH3 =

\sqrt {{K_p}} \times {\left( {27} \right)^{{1 \over 2}}} \times {\left( {{{{P_{Total}}} \over 4}} \right)^{{4 \over 2}}}

=

{{{3^{{3 \over 2}}}K_p^{{1 \over 2}}P_{Total}^2} \over {16}}

Q56

Given below are two statements : Statement I : A catalyst cannot alter the equilibrium constant

\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 (

K_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.

Q57

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) ⇌

{1 \over 2}

N2(g) +

{3 \over 2}

H2(g)

A 8

B

{1 \over 8}

C

{1 \over 4}

D

{1 \over {64}}

Correct Answer

Option B

Solution

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

{1 \over {{K_C}}}

Multiplying by

{1 \over 2}

, reaction becomes NH3(g) ⇌

{1 \over 2}

N2(g) +

{3 \over 2}

H2(g) ; $\therefore$ New KC =

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

=

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

=

{1 \over 8}

Q58

The equilibrium constant at 298 K for a reaction A + B

\leftrightharpoons

C + D is 100. If the initial concentration of all the four species were 1M each, then equilibrium concentration of D (in mol L–1) will be:

A 0.818

B 1.818

C 1.182

D 0.182

Correct Answer

Option B

Solution

Given, $\therefore$

\,\,\,{K_c} = {\left( {{{1 + a} \over {1 - a}}} \right)^2} = 100

$\therefore$

\,\,\,{{1 + a} \over {1 - a}} = 10

On solving

a=0.81

{\left[ D \right]_{At\,eq}} = 1 + a = 1 + 0.81 = 1.81

Q59

For the reaction 2NO2 (g)

\leftrightharpoons

2NO (g) + O2 (g), (Kc = 1.8

\times

10-6 at 184oC) (R = 0.0831 kJ/(mol. K)) When Kp and Kc are compared at 184oC , it is found that :

A Kp is greater than Kc

B Kp is less than Kc

C Kp = Kc

D Whether Kp is greater than, less than or equal to Kc depends upon the total gas pressure

Correct Answer

Option A

Solution

For the reaction : -

2N{O_2}\left( g \right)\rightleftharpoons2NO\left( g \right) + {O_2}\left( g \right)

Given

{K_c} = 1.8 \times {10^{ - 6}}\,\,

at

\,\,{184^ \circ }C

R=0.0831

\,\,kJ/mol.k

{K_p} = 1.8 \times {10^{ - 6}} \times 0.0831 \times 457

= 6.836 \times {10^{ - 6}}

\left[ {} \right.

as

\,\,\,\,\,{184^ \circ }C = \left( {273 + 184} \right) = 457\,k,\,\,

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. {\Delta n = \left( {2 + 1, - 1} \right) = 1} \right]

Hence it is clear that

{K_p} > {K_c}

Q60

A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T. The equilibrium pressure is 10 bar in this vessel. Kp for this reaction is :

A 5

B 10

C 25

D 100

Correct Answer

Option C

Solution

To determine the equilibrium constant, $K_p$ , for the decomposition reaction of the solid compound XY into its gaseous components X and Y, we need to consider the following reaction:

\text{XY (s)} \rightleftharpoons \text{X (g)} + \text{Y (g)}

For a reaction where a solid decomposes into gases, the equilibrium constant $K_p$ is defined in terms of the partial pressures of the gaseous products.

Since XY is a solid, its activity is considered to be 1 and does not appear in the equilibrium expression.

Hence the expression for $K_p$ is:

K_p = P_X \cdot P_Y

Where $P_X$ and $P_Y$ are the partial pressures of the gases X and Y, respectively.

Given that the total pressure at equilibrium is 10 bar, and assuming that X and Y are present in equal amounts due to the stoichiometry of the decomposition reaction (i.e., 1:1), we can write:

P_X = P_Y = \frac{10}{2} = 5 \, \text{bar}

Substituting these partial pressures into the expression for $K_p$ gives:

K_p = P_X \cdot P_Y = 5 \, \text{bar} \cdot 5 \, \text{bar} = 25 \, \text{bar}^2

Therefore, the equilibrium constant $K_p$ for this reaction is 25. Thus, the correct answer is: Option C - 25