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Chemical Equilibrium

NEET Chemistry · 93 questions · Page 1 of 10 · Click an option or "Show Solution" to reveal answer

Q1
For the reaction A(g)⇌2 B( g)\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g}), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 , at 1000 K. [Given : R=0.0831 L atm mol−1 K−1\mathrm{R}=0.0831 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ] KpK_p for the reaction at 1000K1000 K is
A 0.033
B 0.021
C 83.1
D 2.077×1052.077 \times 10^5
Correct Answer
Option A
Solution

Calculate KC K_C : The equilibrium constant in terms of concentrations (KC K_C ) is given by the ratio of the forward reaction rate constant (kf k_f ) to the backward reaction rate constant (kb k_b ): KC=kfkb=12500 K_C = \dfrac{k_f}{k_b} = \dfrac{1}{2500} Convert KC K_C to KP K_P : To convert the concentration equilibrium constant (KC K_C ) to the pressure equilibrium constant (KP K_P ), use the following relation: KP=KC(RT)Δng K_P = K_C (RT)^{\Delta n_g} where: R=0.0831 L atm mol−1 K−1 R = 0.0831 \, \mathrm{L \, atm \, mol^{-1} \, K^{-1}} T=1000 K T = 1000 \, \mathrm{K} Δng=2−1=1\Delta n_g = 2 - 1 = 1 (change in the number of moles of gas, Δng \Delta n_g , from reactants to products) Substituting these values in, we get: KP=12500×0.0831×1000 K_P = \dfrac{1}{2500} \times 0.0831 \times 1000 Calculate KP K_P : Solving the above expression gives: KP=0.033 K_P = 0.033 Therefore, the equilibrium constant KP K_P for the reaction at 1000 K is 0.033.

Q2
For the reaction in equilibrium N2( g)+3H2( g)⇌2NH3( g),ΔH=−Q\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}), \Delta \mathrm{H}=-\mathrm{Q} Reaction is favoured in forward direction by:
A use of catalyst
B decreasing concentration of N2\mathrm{N}_2
C low pressure, high temperature and high concentration of ammonia
D high pressure, low temperature and higher concentration of H2\mathrm{H}_2
Correct Answer
Option D
Solution

The given chemical reaction is an example of the synthesis of ammonia, commonly known as the Haber process:

N2( g)+3H2( g)⇌2NH3( g),ΔH=−Q\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}), \Delta \mathrm{H}=-\mathrm{Q}

This reaction is exothermic, as indicated by the negative enthalpy change (

ΔH=−Q\Delta \mathrm{H} = -\mathrm{Q}

).

In order to determine which conditions favor the forward reaction, we need to consider Le Chatelier's principle, which states that the system will adjust to counteract any changes imposed upon it.

Let's analyze each option: Option A: Use of catalyst A catalyst does not favor the forward or reverse direction of a reaction.

It only speeds up the rate at which equilibrium is achieved.

Option B: Decreasing concentration of

N2\mathrm{N}_2

Decreasing the concentration of

N2\mathrm{N}_2

would shift the equilibrium to the left, favoring the reverse reaction to produce more

N2\mathrm{N}_2

and

H2\mathrm{H}_2

.

Option C: Low pressure, high temperature, and high concentration of ammonia Low pressure and high temperature would favor the reverse reaction.

Moreover, a high concentration of ammonia would also shift the equilibrium to the left.

Thus, this set of conditions does not favor the forward reaction.

Option D: High pressure, low temperature and higher concentration of

H2\mathrm{H}_2

High pressure favors the formation of ammonia because there are fewer moles of gas on the product side (2 moles) as compared to the reactant side (4 moles, i.e., 1 mole of

N2\mathrm{N}_2

and 3 moles of

H2\mathrm{H}_2

). Low temperature favors the exothermic forward reaction. Additionally, a higher concentration of

H2\mathrm{H}_2

will shift the equilibrium to the right, forming more ammonia.

Therefore, the correct answer is: Option D: High pressure, low temperature and higher concentration of

H2\mathrm{H}_2
Q3
In which of the following equilibria, Kp\mathrm{K}_p and Kc\mathrm{K}_{\mathrm{c}} are NOT equal?
A PCl5( g)⇌PCl3( g)+Cl2( g)\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}
B H2( g)+I2( g)⇌2HI(g)\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}
C CO(g)+H2O(g)⇌CO2( g)+H2( g)\mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})}
D 2BrCl(g)⇌Br2( g)+Cl2( g)2 \mathrm{BrCl}_{(\mathrm{g})} \rightleftharpoons \mathrm{Br}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}
Correct Answer
Option A
Solution

To determine in which of the given equilibria

Kp\mathrm{K}_p

and

Kc\mathrm{K}_{\mathrm{c}}

are not equal, it's important to understand the relationship between these two equilibrium constants.

This relationship is expressed by the equation:

Kp=Kc(RT)Δn\mathrm{K}_p = \mathrm{K}_c (RT)^{\Delta n}

where

RR

is the gas constant,

TT

is the temperature in Kelvin, and

Δn\Delta n

is the change in the number of moles of gas (number of moles of gaseous products minus number of moles of gaseous reactants).

If

Δn=0\Delta n = 0

, then

Kp\mathrm{K}_p

and

Kc\mathrm{K}_{\mathrm{c}}

are equal because

(RT)0=1(RT)^0 = 1

. However, if

Δn≠0\Delta n \neq 0

, the constants will not be the same, and the degree to which they differ will depend on the temperature and the value of

Δn\Delta n

. Now, let's analyze each option: Option A:

PCl5( g)⇌PCl3( g)+Cl2( g)\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}

Reactant side moles = 1, Product side moles = 2;

Δn=2−1=1\Delta n = 2 - 1 = 1

. Option B:

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

Reactant side moles = 2, Product side moles = 2;

Δn=2−2=0\Delta n = 2 - 2 = 0

. Option C:

CO(g)+H2O(g)⇌CO2( g)+H2( g)\mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})}

Reactant side moles = 2, Product side moles = 2;

Δn=2−2=0\Delta n = 2 - 2 = 0

. Option D:

2BrCl(g)⇌Br2( g)+Cl2( g)2 \mathrm{BrCl}_{(\mathrm{g})} \rightleftharpoons \mathrm{Br}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}

Reactant side moles = 2, Product side moles = 2;

Δn=2−2=0\Delta n = 2 - 2 = 0

. From this analysis, it is evident that

Kp\mathrm{K}_p

and

Kc\mathrm{K}_{\mathrm{c}}

are not equal in Option A where

Δn=1\Delta n = 1

. In all other options, since

Δn=0\Delta n = 0

,

Kp\mathrm{K}_p

is equal to

Kc\mathrm{K}_{\mathrm{c}}

. Thus, the correct answer is Option A.

Q4
For the reaction 2 A⇌B+C,Kc=4×10−32 \mathrm{~A} \rightleftharpoons \mathrm{B}+\mathrm{C}, \mathrm{K}_{\mathrm{c}}=4 \times 10^{-3}. At a given time, the composition of reaction mixture is: [A]=[B]=[C]=2×10−3M. [A]=[B]=[C]=2 \times 10^{-3} \mathrm{M} \text{. } Then, which of the following is correct?
A Reaction is at equilibrium.
B Reaction has a tendency to go in forward direction.
C Reaction has a tendency to go in backward direction.
D Reaction has gone to completion in forward direction.
Correct Answer
Option C
Solution

To determine which option is correct regarding the reaction state and its direction, we need to calculate the reaction quotient QcQ_c and compare it to the equilibrium constant KcK_c.

The reaction given is: 2A⇌B+C 2 \mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C} The equilibrium constant expression KcK_c for this reaction is: Kc=[B][C][A]2 K_c = \dfrac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}]^2} Given that Kc=4×10−3K_c = 4 \times 10^{-3} and the concentrations of A, B, and C at this time are each 2×10−32 \times 10^{-3} M, we can substitute these values into the expression for KcK_c to calculate the reaction quotient QcQ_c: Qc=(2×10−3M)(2×10−3M)(2×10−3M)2 Q_c = \dfrac{(2 \times 10^{-3} \mathrm{M})(2 \times 10^{-3} \mathrm{M})}{(2 \times 10^{-3} \mathrm{M})^2} Simplifying, we find: Qc=4×10−6M24×10−6M2=1 Q_c = \dfrac{4 \times 10^{-6} \mathrm{M}^2}{4 \times 10^{-6} \mathrm{M}^2} = 1 Comparing QcQ_c with KcK_c: Qc=1 Q_c = 1 Kc=4×10−3 K_c = 4 \times 10^{-3} Since Qc>KcQ_c > K_c (1 > 0.004), the reaction quotient is greater than the equilibrium constant.

This indicates that the concentration of products (B and C) is too high relative to the concentration of reactants (A) for the system to be at equilibrium under these conditions.

This means that the reaction has a tendency to move in the backward direction to reach equilibrium, reducing the concentration of the products (B and C) and increasing the concentration of the reactant (A).

Therefore, the correct answer to the given question is: Option C : Reaction has a tendency to go in backward direction.

Q5
Consider the following reaction in a sealed vessel at equilibrium with concentrations of N2=3.0×10−3M,O2=4.2×10−3M\mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M} and NO=2.8×10−3M\mathrm{NO}=2.8 \times 10^{-3} \mathrm{M}. 2NO(g)⇌N2( g)+O2( g)2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} If 0.1 mol L L−10.1 \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1} of NO(g)\mathrm{NO}_{(\mathrm{g})} is taken in a closed vessel, what will be degree of dissociation (α\alpha) of NO(g)\mathrm{NO}_{(\mathrm{g})} at equilibrium?
A 0.00889
B 0.0889
C 0.8889
D 0.717
Correct Answer
Option D
Solution
2NO(g)⇌N2( g)+O2( g)Kc=[N2][O2][NO]2=3×10−3×4.2×10−32.8×10−3×2.8×10−3=1.607\begin{aligned} &2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\\ &\begin{aligned} \mathrm{K}_c & =\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2} \\ & =\frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}} \\ & =1.607 \end{aligned} \end{aligned}
Kc=0.05α×0.05α(0.1−0.1α)2\mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{(0.1-0.1 \alpha)^2}
Kc=0.05α×0.05α0.01(1−α)21.607=(0.05)2α20.01(1−α)2α2(1−α)2=1.607×(0.1)2(0.05)2α1−α=1.27×0.10.05α1−α=2.54α=2.54−2.54α3.54α=2.54α=2.543.54=0.717\begin{gathered} \mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{0.01(1-\alpha)^2} \\ 1.607=\frac{(0.05)^2 \alpha^2}{0.01(1-\alpha)^2} \\ \frac{\alpha^2}{(1-\alpha)^2}=\frac{1.607 \times(0.1)^2}{(0.05)^2} \\ \frac{\alpha}{1-\alpha}=\frac{1.27 \times 0.1}{0.05} \\ \frac{\alpha}{1-\alpha}=2.54 \\ \alpha=2.54-2.54 \alpha \\ 3.54 \alpha=2.54 \\ \alpha=\frac{2.54}{3.54}=0.717 \end{gathered}
Q6
For a weak acid HA, the percentage of dissociation is nearly 1% at equilibrium. If the concentration of acid is 0.1 mol L−1^{-1}, then the correct option for its Ka_a at the same temperature is :
A 1×10−41\times10^{-4}
B 1×10−61\times10^{-6}
C 1×10−51\times10^{-5}
D 1×10−31\times10^{-3}
Correct Answer
Option C
Solution
Ka=Cα2\mathrm{K_a=C\alpha^2}
Ka=(0.1)×(0.01)2\mathrm{K_a=(0.1)\times(0.01)^2}
Ka=1×10−5\mathrm{K_a=1\times10^{-5}}
Q7
K p for the following reaction is 3.0 at 1000 K. CO 2 (g) + C(s) ⇌\rightleftharpoons 2CO(g) What will be the value of K c for the reaction at the same temperature? (Given : R = 0.083 L bar K −-1 mol −-1 )
A 3.6
B 0.36
C 3.6 ×\times 10 −-2
D 3.6 ×\times 10 −-3
Correct Answer
Option C
Solution

CO 2 (g) + C(s) ⇌\rightleftharpoons 2CO(g)

Δ\Delta

n g = 2 −- 1 = 1 K p = K c (RT)

Δ\Delta

n g K p = K c (RT) [∵\because K p = 3]

Kc=KpRT=30.083×1000{\mathrm{K_c}} = {{{\mathrm{K_p}}} \over {\mathrm{RT}}} = {3 \over {0.083 \times 1000}}

= 0.036 = 3.6 ×\times 10 −-2

Q8
3O 2 (g) ⇌\rightleftharpoons 2O 3 (g) for the above reaction at 298 K, K c is found to be 3.0 ×\times 10 −-59 . If the concentration of O 2 at equilibrium is 0.040 M then concentration of O 3 in M is
A 4.38 ×\times 10 −-32
B 1.9 ×\times 10 −-63
C 2.4 ×\times 10 31
D 1.2 ×\times 10 21
Correct Answer
Option A
Solution

3O 2 (g) ⇌\rightleftharpoons 2O 3 (g)

Kc=[O3]2[O2]3{K_c} = {{{{[{O_3}]}^2}} \over {{{[{O_2}]}^3}}}
[O3]2=Kc[O2]3=3×10−59×(0.04)3{[{O_3}]^2} = {K_c}{[{O_2}]^3} = 3 \times {10^{ - 59}} \times {(0.04)^3}
[O3]2=1.9×10−63=19×10−64{[{O_3}]^2} = 1.9 \times {10^{ - 63}} = 19 \times {10^{ - 64}}
[O3]=4.38×10−32[{O_3}] = 4.38 \times {10^{ - 32}}

Concentration of O 3 at equilibrium

=4.38×10−32= 4.38 \times {10^{ - 32}}

M

Q9
Which one of the following conditions will favour maximum formation of the product in the reaction A 2(g) + B 2(g) ⇌ X 2(g) , Δ\Delta r H = –X kJ ?
A Low temperature and high pressure
B Low temperature and low pressure
C High temperature and high pressure
D High temperature and low pressure
Correct Answer
Option A
Solution

On increasing the pressure and decreasing the temperature, equilibrium will shift in forward direction.

Q10
The equilibrium constants of the following are N 2 + 3H 2 ⇌\rightleftharpoons 2NH 3 ; K 1 N 2 + O 2 ⇌\rightleftharpoons 2NO ; K 2 H 2 + 12{1 \over 2}O 2 ⇌\rightleftharpoons H 2 O; K 3 The equilibrium constant (K) of the reaction : 2NH 3 + 52{5 \over 2} O 2 ⇌\rightleftharpoons 2NO + 3H 2 O will be
A K 2 K 3 3 /K 1
B K 2 K 3 /K 1
C K 2 3 K 3 /K 1
D K 1 K 3 3 /K 2
Correct Answer
Option A
Solution

2NH 3 ⇌\rightleftharpoons N 2 + 3H 2 ;

1K1{1 \over {{K_1}}}

N 2 + O 2 ⇌\rightleftharpoons 2NO ; K 2 3H 2 +

32{3 \over 2}

O 2 ⇌\rightleftharpoons 3H 2 O; (K 3 ) 3 By adding all equations, we get 2NH 3 +

52{5 \over 2}

O 2 ⇌\rightleftharpoons 2NO + 3H 2 O ∴\therefore K =

K2×(K3)3K1{{{K_2} \times {{\left( {{K_3}} \right)}^3}} \over {{K_1}}}
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