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Chemical Equilibrium Practice Questions & Answers

Module 10: Chemical Equilibrium

Understanding reversible reactions and dynamic equilibrium states.

Topics:

  • Equilibrium Constants: Kc (concentration) and Kp (pressure).
  • Reaction Quotient (Q): Predicting the direction of a reaction.
  • Le Châtelier’s Principle: Effect of concentration, pressure, and temperature changes.
  • ICE Tables: Calculating equilibrium concentrations.

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Which of the following statements best defines dynamic equilibrium in a chemical reaction?

  • The reaction has stopped completely and no products or reactants are being formed.

  • The concentrations of reactants and products are equal.

  • The rate of the forward reaction is equal to the rate of the reverse reaction.

  • The rate constants of the forward and reverse reactions are equal.

View Answer & Explanation
Correct Answer: Option C -

The rate of the forward reaction is equal to the rate of the reverse reaction.

Explanation:

At dynamic equilibrium, the reaction continues to occur in both directions, but the rates of the forward and reverse reactions are equal (Rfwd=RrevR_{fwd} = R_{rev}Rfwd=Rrev), resulting in constant concentrations.

Consider the reversible reaction: N2O4(g)2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)N2O4(g)2NO2(g) At equilibrium, which quantity remains constant with time?

  • Only the concentration of N2O4\text{N}_2\text{O}_4N2O4

  • Only the concentration of NO2\text{NO}_2NO2

  • The concentrations of both N2O4\text{N}_2\text{O}_4N2O4 and NO2\text{NO}_2NO2

  • The total mass of the system changes constantly

View Answer & Explanation
Correct Answer: Option C -

The concentrations of both N2O4\text{N}_2\text{O}_4N2O4 and NO2\text{NO}_2NO2

Explanation:

Once a system reaches equilibrium, the concentrations of all reactants and products remain constant over time, although they are not necessarily equal.

Which of the following conditions is not necessary for a system to reach chemical equilibrium?

  • The system must be closed.

  • The reaction must be reversible.

  • The temperature must remain constant.

  • The initial concentrations of reactants must be equal to the initial concentrations of products.

View Answer & Explanation
Correct Answer: Option D -

The initial concentrations of reactants must be equal to the initial concentrations of products.

Explanation:

Equilibrium can be approached from any starting concentrations. It is not required that initial concentrations be equal.

Graphically, how is the state of equilibrium identified in a Concentration vs. Time plot?

  • When the reactant and product curves intersect.

  • When the reactant and product curves become horizontal (slope = 0).

  • When the reactant curve hits zero.

  • When the product curve reaches its maximum theoretical yield.

View Answer & Explanation
Correct Answer: Option B -

When the reactant and product curves become horizontal (slope = 0).

Explanation:

When the curves become horizontal, the concentrations are no longer changing, indicating the rates of forward and reverse reactions are equal.

Why are radioactive isotopes often used to demonstrate the dynamic nature of chemical equilibrium?

  • They change the position of equilibrium.

  • They stop the reverse reaction allowing observation of the forward reaction.

  • Mixing a radioactive product into an equilibrium mixture results in radioactive reactants over time, proving the reverse reaction is occurring.

  • They increase the rate of reaction significantly.

View Answer & Explanation
Correct Answer: Option C -

Mixing a radioactive product into an equilibrium mixture results in radioactive reactants over time, proving the reverse reaction is occurring.

Explanation:

If a radioactive isotope is introduced to an equilibrated system, it distributes between reactants and products, proving that the molecular interchange (forward and reverse reactions) continues despite constant macroscopic concentrations.

Write the equilibrium constant expression (KcK_cKc) for the following heterogeneous reaction:\nCaCO3(s)CaO(s)+CO2(g)\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)CaCO3(s)CaO(s)+CO2(g)

  • Kc=[CaO][CO2][CaCO3]K_c = \frac{[\text{CaO}][\text{CO}_2]}{[\text{CaCO}_3]}Kc=[CaCO3][CaO][CO2]

  • Kc=[CO2]K_c = [\text{CO}_2]Kc=[CO2]

  • Kc=[CaCO3][CO2]K_c = \frac{[\text{CaCO}_3]}{[\text{CO}_2]}Kc=[CO2][CaCO3]

  • Kc=[CaO][CO2]K_c = [\text{CaO}][\text{CO}_2]Kc=[CaO][CO2]

View Answer & Explanation
Correct Answer: Option B -

Kc=[CO2]K_c = [\text{CO}_2]Kc=[CO2]

Explanation:

The concentrations of pure solids and pure liquids are constant and are omitted from the equilibrium expression. Thus, Kc=[CO2]K_c = [\text{CO}_2]Kc=[CO2].

For the reaction 2SO2(g)+O2(g)2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)2SO2(g)+O2(g)2SO3(g), what is the relationship between KpK_pKp and KcK_cKc?

  • Kp=Kc(RT)K_p = K_c(RT)Kp=Kc(RT)

  • Kp=Kc(RT)1K_p = K_c(RT)^{-1}Kp=Kc(RT)1

  • Kp=Kc(RT)2K_p = K_c(RT)^{2}Kp=Kc(RT)2

  • Kp=KcK_p = K_cKp=Kc

View Answer & Explanation
Correct Answer: Option B -

Kp=Kc(RT)1K_p = K_c(RT)^{-1}Kp=Kc(RT)1

Explanation:

The relationship is Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}Kp=Kc(RT)Δng. Here, Δng=(moles product gas)(moles reactant gas)=2(2+1)=1\Delta n_g = (\text{moles product gas}) - (\text{moles reactant gas}) = 2 - (2+1) = -1Δng=(moles product gas)(moles reactant gas)=2(2+1)=1. Therefore, Kp=Kc(RT)1K_p = K_c(RT)^{-1}Kp=Kc(RT)1.

If the equilibrium constant for ABA \rightleftharpoons BAB is K1=2K_1 = 2K1=2, what is the equilibrium constant for the reaction 2B2A2B \rightleftharpoons 2A2B2A?

  • 4

  • 0.5

  • 0.25

  • -2

View Answer & Explanation
Correct Answer: Option C -

0.25

Explanation:

Reversing the reaction inverts KK​K (1/K1=1/21/K_1 = 1/21/K1=1/2). Multiplying coefficients by 2 raises the new constant to the power of 2. New K=(1/2)2=1/4=0.25K = (1/2)^2 = 1/4 = 0.25K=(1/2)2=1/4=0.25.

Calculate KcK_cKc for the reaction 2NO(g)N2(g)+O2(g)2\text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g)2NO(g)N2(g)+O2(g) given the equilibrium concentrations: [NO]=0.2M[\text{NO}] = 0.2\,\text{M}[NO]=0.2M, [N2]=0.1M[\text{N}_2] = 0.1\,\text{M}[N2]=0.1M, and [O2]=0.1M[\text{O}_2] = 0.1\,\text{M}[O2]=0.1M.

  • 0.25

  • 2.5

  • 0.5

  • 1.0

View Answer & Explanation
Correct Answer: Option A -

0.25

Explanation:

Kc=[N2][O2][NO]2=(0.1)(0.1)(0.2)2=0.010.04=0.25K_c = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2} = \frac{(0.1)(0.1)}{(0.2)^2} = \frac{0.01}{0.04} = 0.25Kc=[NO]2[N2][O2]=(0.2)2(0.1)(0.1)=0.040.01=0.25

Which statement about the value of the equilibrium constant KK​K is true?

  • A large KK​K (K1K \gg 1K1) indicates that reactants are favored at equilibrium.

  • A small KK​K (K1K \ll 1K1) indicates that products are favored at equilibrium.

  • The value of KK​K is independent of temperature.

  • A large KK​K (K1K \gg 1K1) indicates that products are favored at equilibrium.

View Answer & Explanation
Correct Answer: Option D -

A large KK​K (K1K \gg 1K1) indicates that products are favored at equilibrium.

Explanation:

If KK​K is very large, the numerator (products) is much larger than the denominator (reactants), meaning the equilibrium favors the products.

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