Shifting Equilibria: Le Châtelier’s Principle Quiz

PDF page 745: A system at equilibrium is in a state of dynamic balance, with forward and reverse reactions taking place at equal rates. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. The system will subsequently experience a net reaction in the direction of greater rate (a shift) that will re-establish the equilibrium. This phenomenon is summarized by Le Châtelier’s principle: if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium. Reaction rates are affected primarily by concentrations, as described by the reaction’s rate law, and temperature, as described by the Arrhenius equation. Consequently, changes in concentration and temperature are the two stresses that can shift an equilibrium.

PDF page 745: A system at equilibrium is in a state of dynamic balance, with forward and reverse reactions taking place at equal rates. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. The system will subsequently experience a net reaction in the direction of greater rate (a shift) that will re-establish the equilibrium. This phenomenon is summarized by Le Châtelier’s principle: if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium. Reaction rates are affected primarily by concentrations, as described by the reaction’s rate law, and temperature, as described by the Arrhenius equation. Consequently, changes in concentration and temperature are the two stresses that can shift an equilibrium.

PDF page 746: H (g) + I (g) ⇌ 2HI(g) Kc = 50.0 at 400 °C 2 2 The rate laws for the forward and reverse reactions are m n f orward H (g) + I (g) → 2HI(g) rate = k [H ] [I ] 2 2 2 f f 2 x reverse 2HI(g) → H (g) + I (g) rater = kr[HI] 2 2 When this system is at equilibrium, the forward and reverse reaction rates are equal. rate = rater f If the system is stressed by adding reactant, either H2 or I2, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction: rate > rater f The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates. The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction rate < rater f and a temporary net reaction in the reverse direction to re-establish equilibrium. As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium, 2 [HI] Qc = = Kc [H ][I ] 2 2 If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the new equilibrium mixture will have greater concentrations of H2 and I2 and a lesser concentration of HI. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left. For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure P of an ideal gas is proportional to its molar concentration M, n P

PDF page 746: H (g) + I (g) ⇌ 2HI(g) Kc = 50.0 at 400 °C 2 2 The rate laws for the forward and reverse reactions are m n f orward H (g) + I (g) → 2HI(g) rate = k [H ] [I ] 2 2 2 f f 2 x reverse 2HI(g) → H (g) + I (g) rater = kr[HI] 2 2 When this system is at equilibrium, the forward and reverse reaction rates are equal. rate = rater f If the system is stressed by adding reactant, either H2 or I2, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction: rate > rater f The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates. The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction rate < rater f and a temporary net reaction in the reverse direction to re-establish equilibrium. As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium, 2 [HI] Qc = = Kc [H ][I ] 2 2 If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the new equilibrium mixture will have greater concentrations of H2 and I2 and a lesser concentration of HI. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left. For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure P of an ideal gas is proportional to its molar concentration M, n P

PDF page 746: H (g) + I (g) ⇌ 2HI(g) Kc = 50.0 at 400 °C 2 2 The rate laws for the forward and reverse reactions are m n f orward H (g) + I (g) → 2HI(g) rate = k [H ] [I ] 2 2 2 f f 2 x reverse 2HI(g) → H (g) + I (g) rater = kr[HI] 2 2 When this system is at equilibrium, the forward and reverse reaction rates are equal. rate = rater f If the system is stressed by adding reactant, either H2 or I2, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction: rate > rater f The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates. The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction rate < rater f and a temporary net reaction in the reverse direction to re-establish equilibrium. As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium, 2 [HI] Qc = = Kc [H ][I ] 2 2 If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the new equilibrium mixture will have greater concentrations of H2 and I2 and a lesser concentration of HI. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left. For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure P of an ideal gas is proportional to its molar concentration M, n P

PDF page 746: H (g) + I (g) ⇌ 2HI(g) Kc = 50.0 at 400 °C 2 2 The rate laws for the forward and reverse reactions are m n f orward H (g) + I (g) → 2HI(g) rate = k [H ] [I ] 2 2 2 f f 2 x reverse 2HI(g) → H (g) + I (g) rater = kr[HI] 2 2 When this system is at equilibrium, the forward and reverse reaction rates are equal. rate = rater f If the system is stressed by adding reactant, either H2 or I2, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction: rate > rater f The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates. The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction rate < rater f and a temporary net reaction in the reverse direction to re-establish equilibrium. As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium, 2 [HI] Qc = = Kc [H ][I ] 2 2 If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the new equilibrium mixture will have greater concentrations of H2 and I2 and a lesser concentration of HI. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left. For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure P of an ideal gas is proportional to its molar concentration M, n P

PDF page 746: H (g) + I (g) ⇌ 2HI(g) Kc = 50.0 at 400 °C 2 2 The rate laws for the forward and reverse reactions are m n f orward H (g) + I (g) → 2HI(g) rate = k [H ] [I ] 2 2 2 f f 2 x reverse 2HI(g) → H (g) + I (g) rater = kr[HI] 2 2 When this system is at equilibrium, the forward and reverse reaction rates are equal. rate = rater f If the system is stressed by adding reactant, either H2 or I2, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction: rate > rater f The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates. The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction rate < rater f and a temporary net reaction in the reverse direction to re-establish equilibrium. As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium, 2 [HI] Qc = = Kc [H ][I ] 2 2 If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the new equilibrium mixture will have greater concentrations of H2 and I2 and a lesser concentration of HI. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left. For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure P of an ideal gas is proportional to its molar concentration M, n P

PDF page 746: V RT and so changes in the partial pressures of any reactant or product are essentially changes in concentrations and thus yield the same effects on equilibria. Aside from adding or removing reactant or product, the pressures (concentrations) of species in a gas-phase equilibrium can also be changed by changing the volume occupied by the system. Since all species of a gas-phase equilibrium occupy the same volume, a given change in volume will cause the same change in concentration for both reactants and products. In order to discern what shift, if any, this type of stress will induce the stoichiometry of the reaction must be considered. At equilibrium, the reaction H2(g) + I2(g) ⇌ 2HI(g) is described by the reaction quotient

PDF page 747: 2 P HI Q p = = K p P P H2 I2 If the volume occupied by an equilibrium mixture of these species is decreased by a factor of 3, the partial pressures of all three species will be increased by a factor of 3:

PDF page 747: And so, changing the volume of this gas-phase equilibrium mixture does not results in a shift of the equilibrium. A similar treatment of a different system, 2NO2(g) ⇌ 2 NO(g) + O2(g), however, yields a different result: P 2 P NO O2 Q p = = K p P 2 NO2 2 P 3P 9P 27P 2 (3P ) 3P 2 O2 O2 ' NO NO O2 NO = = Q p = = 3Q p > K p 2 9P 9P 2 2 (3P ) NO2 NO2 NO2 ' Q p = 3Q p > K p In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left. These results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better “accommodates” the volume change. In the second example, two moles of reactant (NO2) yield three moles of product (2NO + O2), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas (2 mol) than the forward reaction (3 mol). Conversely, increasing the volume of this equilibrium system would result in a shift towards products.

PDF page 747: And so, changing the volume of this gas-phase equilibrium mixture does not results in a shift of the equilibrium. A similar treatment of a different system, 2NO2(g) ⇌ 2 NO(g) + O2(g), however, yields a different result: P 2 P NO O2 Q p = = K p P 2 NO2 2 P 3P 9P 27P 2 (3P ) 3P 2 O2 O2 ' NO NO O2 NO = = Q p = = 3Q p > K p 2 9P 9P 2 2 (3P ) NO2 NO2 NO2 ' Q p = 3Q p > K p In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left. These results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better “accommodates” the volume change. In the second example, two moles of reactant (NO2) yield three moles of product (2NO + O2), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas (2 mol) than the forward reaction (3 mol). Conversely, increasing the volume of this equilibrium system would result in a shift towards products.

PDF page 747: And so, changing the volume of this gas-phase equilibrium mixture does not results in a shift of the equilibrium. A similar treatment of a different system, 2NO2(g) ⇌ 2 NO(g) + O2(g), however, yields a different result: P 2 P NO O2 Q p = = K p P 2 NO2 2 P 3P 9P 27P 2 (3P ) 3P 2 O2 O2 ' NO NO O2 NO = = Q p = = 3Q p > K p 2 9P 9P 2 2 (3P ) NO2 NO2 NO2 ' Q p = 3Q p > K p In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left. These results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better “accommodates” the volume change. In the second example, two moles of reactant (NO2) yield three moles of product (2NO + O2), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas (2 mol) than the forward reaction (3 mol). Conversely, increasing the volume of this equilibrium system would result in a shift towards products.

PDF page 748: 3 2 3 These same equilibrium reactions are the basis of today’s soft-drink carbonation process. Beverages are exposed to a high pressure of gaseous carbon dioxide during the process to shift the first equilibrium above to the right, resulting in desirably high concentrations of dissolved carbon dioxide and, per similar shifts in the other two equilibria, its hydrolysis and ionization products. A bottle or can is then nearly filled with the carbonated beverage, leaving a relatively small volume of air in the container above the beverage surface (the headspace) before it is sealed. The pressure of carbon dioxide in the container headspace is very low immediately after sealing, but it rises as the dissolution equilibrium is re-established by shifting to the left. Since the volume of the beverage is significantly greater than the volume of the headspace, only a relatively small amount of dissolved carbon dioxide is lost to the headspace. When a carbonated beverage container is opened, a hissing sound is heard as pressurized CO2 escapes from the headspace. This causes the dissolution equilibrium to shift left, resulting in a decrease in the concentration of dissolved CO2 and subsequent left-shifts of the hydrolysis and ionization equilibria. Fortunately for the consumer, the dissolution equilibrium is usually re-established slowly, and so the beverage may be enjoyed while its dissolved carbon dioxide concentration remains palatably high. Once the equilibria are re-established, the CO2(aq) concentration will be significantly lowered, and the beverage acquires a characteristic taste referred to as “flat.”

PDF page 748: Consistent with the law of mass action, an equilibrium stressed by a change in concentration will shift to re-establish equilibrium without any change in the value of the equilibrium constant, K. When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition that exhibits a different value for the equilibrium constant.