Unit 4: Chemical Equilibrium

Understanding the balance in reversible reactions and how to manipulate it.

4.9 Reversible Reactions & Dynamic Equilibrium

Many chemical reactions are reversible, meaning they can proceed in both the forward (reactants to products) and reverse (products to reactants) directions. This is represented by the equilibrium symbol ($\rightleftharpoons$).

In a closed system, as reactants form products, the forward reaction rate slows down. Simultaneously, as product concentration increases, the reverse reaction rate speeds up. Eventually, the system reaches a state of dynamic equilibrium.

Dynamic Equilibrium: The state reached in a reversible reaction where the rate of the forward reaction is equal to the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant, even though both reactions are still occurring.
Solved Examples:
  1. What does the symbol $\rightleftharpoons$ indicate in a chemical equation?
    Solution: It indicates that the reaction is reversible and can proceed in both the forward and reverse directions.
  2. Define 'dynamic equilibrium'.
    Solution: It is the state where the rates of the forward and reverse reactions are equal, and the net concentrations of reactants and products do not change.
  3. Why is the equilibrium described as 'dynamic'?
    Solution: Because both the forward and reverse reactions are still occurring continuously, even though there is no overall change in concentrations. The system is active, not static.
  4. What condition is necessary for a reversible reaction to reach equilibrium?
    Solution: The reaction must take place in a closed system, where no reactants or products can escape.
  5. At equilibrium, are the concentrations of reactants and products equal?
    Solution: Not necessarily. They are simply constant. The relative concentrations depend on the specific reaction.
  6. Why can't an open system, like a puddle evaporating, reach equilibrium?
    Solution: Because the product (water vapor) escapes. This prevents the reverse reaction (condensation) from occurring at a rate equal to the forward reaction (evaporation).
  7. Initially, which rate is faster in a reversible reaction, forward or reverse?
    Solution: The forward rate is initially faster because the concentration of reactants is at its maximum and the concentration of products is zero.
  8. What happens to the rate of the reverse reaction as a system approaches equilibrium?
    Solution: The rate of the reverse reaction increases because the concentration of products is increasing.
  9. When equilibrium is reached, what can be said about the macroscopic properties of the system (e.g., color, pressure)?
    Solution: The macroscopic properties become constant because the concentrations of all substances are no longer changing.
  10. Give a real-world example of a system at dynamic equilibrium.
    Solution: A sealed bottle of soda. COâ‚‚(g) is in equilibrium with dissolved COâ‚‚(aq). The rate at which COâ‚‚ dissolves into the liquid equals the rate at which it escapes from the liquid into the space above.

4.10 Equilibrium Position & Equilibrium Constant ($K_c$)

The position of equilibrium describes the relative amounts of reactants and products in a system at equilibrium. If there are mostly products, we say the equilibrium "lies to the right." If there are mostly reactants, it "lies to theleft." This position is quantified by the equilibrium constant ($K_c$).

For a general reversible reaction: $aA + bB \rightleftharpoons cC + dD$, the expression for the equilibrium constant is:

Equilibrium Constant Expression: $$ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} $$ Where $[X]$ represents the molar concentration of substance X at equilibrium. The value of $K_c$ is constant for a given reaction at a constant temperature.
  • If $K_c > 1$, the equilibrium lies to the right (products are favored).
  • If $K_c < 1$, the equilibrium lies to the left (reactants are favored).
Solved Examples:
  1. Write the $K_c$ expression for $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
    Solution: $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$.
  2. For the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, $K_c = 55.3$ at a certain temperature. Are reactants or products favored?
    Solution: Since $K_c > 1$, the products are favored at equilibrium. The equilibrium lies to the right.
  3. Write the $K_c$ expression for $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$.
    Solution: $K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$.
  4. What does a very small $K_c$ value (e.g., $1 \times 10^{-5}$) indicate?
    Solution: It indicates that the denominator (reactants) is much larger than the numerator (products) at equilibrium. The equilibrium lies far to the left.
  5. Why are solids and pure liquids omitted from the $K_c$ expression?
    Solution: Their concentrations (or densities) are essentially constant and are incorporated into the value of $K_c$.
  6. For the reaction $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$, the equilibrium concentrations are $[PCl_5]=1.0 M$, $[PCl_3]=0.2 M$, and $[Cl_2]=0.2 M$. Calculate $K_c$.
    Solution: $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.2)(0.2)}{1.0} = 0.04$.
  7. What is the only factor that can change the value of $K_c$ for a given reaction?
    Solution: Temperature.
  8. Write the $K_c$ expression for the dissociation of ethanoic acid: $CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)$.
    Solution: $K_c = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$.
  9. If $K_c=1$, what can be said about the position of equilibrium?
    Solution: The concentrations of products (raised to their powers) are roughly equal to the concentrations of reactants (raised to their powers). The equilibrium position is intermediate.
  10. Write the $K_c$ expression for $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$.
    Solution: Since $CaCO_3$ and $CaO$ are solids, they are omitted. The expression is $K_c = [CO_2]$.

4.11 Le Chatelier's Principle

Le Chatelier's Principle provides a way to predict how a system at equilibrium will respond to a change in conditions. It states: "If a change of condition is applied to a system in equilibrium, the system will shift in a direction that opposes the change." This allows chemists to manipulate reactions to maximize the yield of desired products.

Effects of Changes on Equilibrium:
  • Concentration: If a substance is added, the equilibrium shifts to consume it. If a substance is removed, the equilibrium shifts to produce more of it.
  • Pressure (gases only): If pressure is increased, the equilibrium shifts to the side with fewer moles of gas to reduce the pressure. If pressure is decreased, it shifts to the side with more moles of gas.
  • Temperature: If temperature is increased, the equilibrium shifts in the endothermic direction to absorb the added heat. If temperature is decreased, it shifts in the exothermic direction to release heat. (Note: A change in temperature is the only factor that also changes the value of $K_c$).
  • Catalyst: A catalyst has no effect on the position of equilibrium. It speeds up both the forward and reverse reactions equally, so equilibrium is reached faster.
Solved Examples:
  1. For $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, what happens if more $H_2$ is added?
    Solution: The system will shift to the right to consume the added $H_2$, producing more $NH_3$.
  2. For the same reaction, what is the effect of increasing the pressure?
    Solution: The left side has 4 moles of gas, and the right has 2. To oppose the increased pressure, the equilibrium will shift to the right, the side with fewer moles of gas.
  3. The forward reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ is exothermic. What happens if the temperature is increased?
    Solution: To oppose the increase in temperature, the equilibrium will shift in the endothermic (reverse) direction. The position shifts to the left.
  4. For $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, what is the effect of changing the pressure?
    Solution: There are 2 moles of gas on the left and 2 moles of gas on the right. Since there is no change in the number of moles of gas, changing the pressure has no effect on the position of equilibrium.
  5. What happens if a catalyst is added to a system at equilibrium?
    Solution: There is no change to the position of equilibrium. The system simply reaches equilibrium faster.
  6. In the reaction $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$, what happens if some $Cl_2$ is removed?
    Solution: The system will shift to the right to produce more $Cl_2$ and restore equilibrium.
  7. For the exothermic reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, what conditions of temperature and pressure would maximize the yield of $NH_3$?
    Solution: To favor the forward (exothermic) reaction, a low temperature is needed. To favor the side with fewer moles of gas, a high pressure is needed.
  8. If increasing the temperature of a system at equilibrium causes $K_c$ to increase, is the forward reaction exothermic or endothermic?
    Solution: Endothermic. An increase in temperature favors the endothermic direction. If this causes $K_c$ to increase, the forward reaction must be the endothermic one.
  9. Consider $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$. How does adding more $CaCO_3(s)$ affect the equilibrium?
    Solution: It has no effect. Since $CaCO_3$ is a solid, its concentration is constant, and adding more does not shift the position of equilibrium.
  10. For $2NO_2(g) \rightleftharpoons N_2O_4(g)$, what is the effect of decreasing the volume of the container?
    Solution: Decreasing the volume increases the pressure. The equilibrium will shift to the side with fewer moles of gas to reduce the pressure. It will shift to the right.

End of Topic Questions

Answer: A reaction that can proceed in both the forward and reverse directions.

Answer: The rate of the forward reaction and the rate of the reverse reaction.

Answer: $K_c = \frac{[C]^3}{[A]^2[B]}$.

Answer: If a change is applied to a system at equilibrium, the system will shift to oppose that change.

Answer: No effect, because there are 2 moles of gas on both sides.

Answer: To the right, in the endothermic direction, to absorb the added heat.

Answer: Temperature.

Answer: The equilibrium shifts to the right to produce more $NH_3$.

Answer: Reactants are heavily favored at equilibrium; the equilibrium lies to the left.

Answer: It has no effect on the position of equilibrium; it only speeds up the rate at which equilibrium is reached.

Answer: A system from which no matter (reactants or products) can escape.

Answer: $K_c$ decreases, because the equilibrium shifts to the left (favoring reactants).

Answer: Constant.

Answer: The equilibrium shifts to the side with more moles of gas, so it shifts to the left.

Answer: $K_c = \frac{[H_2][I_2]}{[HI]^2}$.

Answer: To the right (products are favored).

Answer: To the left, to consume the added C.

Answer: High temperature.

Answer: Because the forward and reverse reactions are still occurring, even though there is no net change.

Answer: No effect. While total pressure increases, the partial pressures (concentrations) of the reacting gases do not change.
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