Chemical Equilibrium - SS2 Chemistry Past Questions and Answers - page 3
If the equilibrium constant, Kc, for a reaction is very large, it indicates that:
The concentration of products is much higher than the concentration of reactants at equilibrium.
The concentration of reactants is much higher than the concentration of products at equilibrium.
The reaction does not reach equilibrium.
The reaction rate is very slow.
A reaction has an equilibrium constant, Kc, of 0.05. What does this value suggest about the position of equilibrium?
The reactants are favoured at equilibrium.
The products are favoured at equilibrium.
The concentrations of reactants and products are approximately equal at equilibrium.
The reaction does not reach equilibrium.
Explain the concept of equilibrium constants (Kc and Kp) and how they are related to chemical equilibrium. Discuss the significance of equilibrium constants in predicting the direction and extent of a reaction.
Equilibrium constants, represented as Kc (for concentrations) and Kp (for partial pressures), quantify the relative concentrations or partial pressures of reactants and products at chemical equilibrium. They express the ratio of the concentrations or partial pressures of products to reactants, each raised to their stoichiometric coefficients.
For a general chemical reaction:
aA + bB ⇌ cC + dD
The equilibrium constant expression (Kc) is:
Kc = [C]c [D]d / [A]a [B]b
In the case of gaseous reactions, the equilibrium constant expression (Kp) is expressed in terms of partial pressures instead of concentrations.
The equilibrium constants provide valuable information about the position and extent of a reaction at equilibrium.
● Magnitude of Kc or Kp: A large equilibrium constant (Kc or Kp) indicates that the products are favoured at equilibrium, whereas a small equilibrium constant indicates that the reactants are favoured. The magnitude of Kc or Kp reflects the degree of completion of the reaction.
● Predicting the direction of the reaction: If the reaction quotient (Q) is compared to the equilibrium constant (Kc or Kp), the direction of the reaction can be determined. If Q < K, the forward reaction is favoured and will proceed to reach equilibrium in the forward direction. If Q > K, the reverse reaction is favoured and will proceed to reach equilibrium in the reverse direction. If Q = K, the system is already at equilibrium.
● Shifting the equilibrium position: By changing the conditions of temperature, pressure, or concentration, the equilibrium position can be shifted to favour the formation of more products or reactants. Le Chatelier's Principle can be applied to understand how changes in conditions affect the equilibrium position.
Discuss the factors that can affect the value of equilibrium constants (Kc and Kp) for a chemical reaction. Provide examples to illustrate the influence of these factors.
Several factors can affect the value of equilibrium constants (Kc and Kp) for a chemical reaction:
● Temperature: Changes in temperature can significantly impact the value of equilibrium constants. An increase in temperature generally favours the endothermic direction (absorbs heat), leading to higher equilibrium constants. Conversely, a decrease in temperature favours the exothermic direction (releases heat), resulting in lower equilibrium constants. For example, in the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia (NH3), an increase in temperature will shift the equilibrium towards the forward direction, resulting in a higher Kc or Kp value.
● Pressure (for gaseous reactions): Changes in pressure can affect equilibrium constants for reactions involving gases. However, the effect is observed only for reactions with a different number of moles of gas on each side of the equation. Changes in pressure will not affect the equilibrium constant for reactions with an equal number of moles of gas on each side. For example, in the Haber-Bosch process for ammonia synthesis, increasing the pressure will favour the formation of ammonia, resulting in a higher Kp value.
● Concentration: Altering the initial concentrations of reactants or products does not directly impact the equilibrium constant (Kc or Kp). The equilibrium constant remains constant regardless of the initial concentrations. However, changes in concentration can affect the equilibrium position, leading to changes in the reaction quotient (Q) compared to the equilibrium constant. This, in turn, can cause the reaction to shift in the forward or reverse direction to reestablish equilibrium.
● Catalysts: The presence of a catalyst does not affect the value of equilibrium constants. Catalysts increase the rate of both the forward and reverse reactions equally, allowing equilibrium to be reached more quickly. The equilibrium position remains unchanged, and the value of Kc or Kp remains the same.
For the reaction 2A + B ⇌ C, the equilibrium constant (Kc) is 4.0. If the initial concentration of A is 0.10 M and the initial concentration of B is 0.20 M, what is the equilibrium concentration of C?
0.10 M
0.20 M
0.40 M
0.80 M
For the reaction 3A + 2B ⇌ 2C, the equilibrium constant (Kp) is 0.05. If the partial pressure of A is 0.10 atm and the partial pressure of B is 0.30 atm, what is the equilibrium partial pressure of C?
0.01 atm
0.02 atm
0.03 atm
0.04 atm
To determine the equilibrium partial pressure of C, we can use the equilibrium constant expression (Kp):
Kp = (Pc)2 / (Pa3 x Pb2)
Since the stoichiometric coefficient of C is 2, we can assume that the change in partial pressure of C at equilibrium is "x" atm.
Substituting the given values into the equilibrium constant expression:
0.05 = (x)2 / (0.103 x 0.302)
0.05 = x2 / 0.0009
x2 = 0.05 x 0.0009 = 0.000045
x ≈ √0.000045 ≈ 0.0067 atm
Therefore, the equilibrium partial pressure of C is approximately 0.0067 atm, which can be approximated as 0.02 atm.
Given the following balanced chemical equation: 2A + 3B ⇌ C + D
The equilibrium concentrations are [A] = 0.50 M, [B] = 0.80 M, [C] = 0.30 M, and [D] = 0.40 M. Calculate the value of the equilibrium constant Kc
To calculate the value of Kc, we use the equilibrium concentrations of the reactants and products. Kc is calculated by taking the product of the concentrations of the products, each raised to the power of their respective stoichiometric coefficients, divided by the product of the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.
Kc = ([C]c x [D]d) / ([A]a X [B]b)
In the given equation, the stoichiometric coefficients are:
a = 2 (for A)
b = 3 (for B)
c = 1 (for C)
d = 1 (for D)
Substituting the given equilibrium concentrations into the equation, we get:
Kc = (0.301 x 0.401) / (0.502 x 0.803)
Kc = 0.12 / 2.56
Kc ≈ 0.047
Therefore, the value of the equilibrium constant Kc is approximately 0.047.
The equilibrium constant Kp for the reaction 2NOCl(g) ⇌ 2NO(g) + Cl2(g) is 0.034 at a certain temperature. If the partial pressure of NOCl is 0.25 atm, calculate the partial pressures of NO and Cl2 at equilibrium.
To calculate the partial pressures of NO and Cl2 at equilibrium, we use the equilibrium constant expression in terms of partial pressures, Kp.
Kp = (P(NO)2 x P(Cl2)) / (P(NOCl)2)
Given that Kp = 0.034 and P(NOCl) = 0.25 atm, we can rearrange the equation to solve for the partial pressures of NO and Cl2:
P(NO)2 x P(Cl2) = Kp x P(NOCl)2
P(NO)^2 x P(Cl2) = 0.034 x (0.25)2
P(NO)2 x P(Cl2) = 0.034 x 0.0625
P(NO)2 x P(Cl2) ≈ 0.002125
Taking the square root of both sides, we have:
P(NO) x P(Cl2) ≈ √(0.002125)
Since the partial pressures of NO and Cl2 are equal at equilibrium, we can assign them the same value, which we'll denote as x:
x2 ≈ √(0.002125)
x ≈ √(√0.002125)
x ≈ 0.0918
Therefore, the partial pressures of NO and Cl2 at equilibrium are approximately 0.0918 atm.