Class 11 Chemistry MCQ – Law of Chemical Equilibrium and Equilibrium Constant

This set of Class 11 Chemistry Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Law of Chemical Equilibrium and Equilibrium Constant”.

1. A reaction is given by aA + bB → cC + dD. How do you represent an equilibrium constant?
a) [A]a[B]b/[C]c[D]d
b) [C]c[D]d/[A]a[B]b
c) [A][B]/[C][D]
d) [C][D]/[A][B]
View Answer

Answer: b
Explanation: An equilibrium constant in a chemical reaction As given by the ratio of Forward reaction rate by backward reaction rate its units are (mol/L)Δng, where Δng is moles of products – moles of reactants which are in the gaseous state only.

2. The value of equilibrium constant for different chemical reactions at a particular temperature is ____________
a) constant
b) unique
c) the same
d) cannot say
View Answer

Answer: b
Explanation: The value of the equilibrium constant changes with temperature i.e. it’s different for different temperatures and is unique for every reaction at a constant temperature or given temperature.

3. What does it indicate having a higher equilibrium constant?
a) reaction occurs faster
b) rate of backward reaction is faster
c) both the backward and forward reactions are equal
d) reaction may be slower than usual
View Answer

Answer: a
Explanation: As we know that the equilibrium constant is the ratio of the forward reaction rate to the backward reaction rate. Having higher equilibrium constant depicts that the reaction that is forward occurring faster then the backward one and the extension of completion of the reaction is more.
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4. The equilibrium constant of a reaction is 20 units and the equilibrium constant of other reaction is 30 units when both the reactions are added up together then the equilibrium constant of the resultant reaction is given by _____________
a) 20 units
b) 600 units
c) 50 units
d) 10 units
View Answer

Answer: b
Explanation: When the reaction can be expressed as the sum of two other reactions, the equilibrium constant of the overall reaction is equal to the product of equilibrium constants of individual reactions, so 20 units x 30 units = 600 units.

5. If the initial concentrations of reactants in a reaction increase then the equilibrium constant ____________
a) also increases
b) decreases
c) remains constant
d) may increase or decrease
View Answer

Answer: c
Explanation: The equilibrium constant of a chemical reaction is independent of the initial concentrations of the reactants. So when the initial concentrations of the reactants increase in a reaction the equilibrium constant remains the same.
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6. If we use activities in place of molar concentration in an equilibrium constant expression then what is the dimension of k?
a) it becomes dimensionless
b) the same units of molar concentration
c) units are related to the activity
d) cannot predict it
View Answer

Answer: a
Explanation: In equilibrium constant expression of a chemical reaction, (equilibrium constant is represented by the letter K) if the activities are used in place of the molar concentration K becomes dimensionless, that means it has no dimensions.

7. What is the equilibrium constant of the following reaction: 4NH3 + 5O2 → 4NO + 6H2O?
a) [NO][H2O]/[NH3][O2]
b) [C]c[D]d/[A]a[B]b
c) [NO4[H2]6/NH3]4[O2]5
d) [NO]4[H2O]6/[NH3]4[O2]5
View Answer

Answer: d
Explanation: the above given chemical reaction 4NH3 + 5O2 → 4NO + 6H2O is in the form of aA + bB → cC + dD. We know that the equilibrium constant of this reaction is [C]c[D]d/[A]a[B]b, show the required answer is [NO]4[H2O]6/[NH3]4[O2]5.
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8. for the reaction H2(g) + Br2(g) ↔ 2HBr(g) the equilibrium constant is given as 0.04 then what is the equilibrium constant for the reaction 2HBr(g) ↔ H2(g) + Br2(g)?
a) 0.04
b) 4
c) 25
d) 100
View Answer

Answer: c
Explanation: In a chemical reaction if the equilibrium constant for a backward reaction is given by k then the equilibrium constant for the forward reaction is given by 1/k, so your the resultant answer becomes 1/0.04 = 25.

9. Given that at 800k the concentrations are as follows: N2=3.0 × 10–3M, O2 = 4.2 × 10–3M and NO = 2.8 × 10–3M, what is the equilibrium constant for the reaction N2(g) + O2(g) \(\rightleftharpoons \) 2NO(g)?
a) 0.622
b) 0.6
c) 0.63
d) 0.94
View Answer

Answer: a
Explanation: For the reaction, N2(g) + O2(g) \(\rightleftharpoons \) 2NO(g), the equilibrium constant is given by [NO]2/[N2][O2]. So the equilibrium constant K is given by (2.8 × 10–3 M)(2.8 × 10–3 M)/(3.0 × 10–3 M)(4.2 × 10–3 M) = 0.622, there are no units as Δng = 0.
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10. If the chemical reaction is: aA + bB → cC + dD and K is the equilibrium constant. Then what is the equilibrium constant of the reaction naA + nbB → ncC + ndD?
a) K
b) nK
c) Kn
d) K/n
View Answer

Answer: c
Explanation: If a chemical reaction is multiplied by a constant then the equilibrium constant of the particular chemical reaction is raised to the power of the constant by which the chemical reaction is multiplied, so here and is a constant and equilibrium constant of the reaction is kn.

Sanfoundry Global Education & Learning Series – Chemistry – Class 11.

To practice all chapters and topics of class 11 Chemistry, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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