In the previous post, we talked about Le Châtelier’s principle that helps evaluate the effect of changing the concentrations, volume, pressure, and temperature on the equilibrium. In short, you need to remember that if a system at equilibrium is disturbed, it will try to minimize the effect of external stress.
Practice
Consider the following equilibrium process
2SO3(g) ⇆ 2SO2(g) + O2(g) ΔH° = 2198 kJ/mol
How will the concentrations of SO2, O2, and SO3 be affected in each scenario?
(a) the temperature is increased
(b) the pressure is decreased by increasing the volume of the container
(c) the concentration of O2 is increased
(d) a is added catalyst
(e) an inert gas is added at a constant volume
(a) The reaction will shift to the right.
(b) The reaction will shift to the right.
(c) The reaction will shift left.
(d) It has no effect on the equilibrium position.
(e) It has no effect on the equilibrium composition.
(a) the temperature is increased: We need to first realize that this is an endothermic reaction (ΔHo positive). This means the reaction consumes heat, or, in other words, heat is a reactant. According to Le Châtelier’s principle, if we increase the temperature, the reaction will go in the direction that absorbs heat. Therefore, the reaction will shift right.
(b) the pressure on the gases is decreased: The number of gas molecules is different, so the reaction will respond in a way that brings the pressure up again. The products contain more gas molecules, therefore, the reaction will shift to the right.
(c) the concentration of O2 is increased: The reaction will shift left to use up some of the added oxygen gas and decrease its concentration.
(d) a is added catalyst: The catalyst is used to speed up the reaction so it reaches equilibrium more quickly. However, it has no effect on the equilibrium position.
(e) an inert gas is added at constant volume: Remember, changing the pressure of a gaseous reaction only has an effect on the equilibrium position if it results in changes of partial pressures of gases in the chemical equation. And although adding an inert gas to the system, increases the total pressure, the partial pressures of gases do not change. Therefore, it has no effect on the equilibrium composition.
Suppose you need to increase the amount of C3H6Cl2 produced in the following exothermic reaction:
C3H6(g) + Cl2(g) ⇆ C3H6Cl2(g)
Which of the following strategies will work once the reaction mixture reaches equilibrium?
a) decreasing the reaction volume
b) removing C3H6Cl2 from the reaction mixture as it forms
c) adding a catalyst
d) adding Cl2
b) removing C3H6Cl2 from the reaction mixture as it forms
d) adding Cl2
a) decreasing the reaction volume – Has no effect since the number of gas molecules is equal on both sides.
b) removing C3H6Cl2 from the reaction mixture as it forms – Will increase the yield of the reaction since the equilibrium will shift to the right to reinstate the amount of product removed from the system.
c) adding a catalyst – Has no effect on the equilibrium position
d) adding Cl2 – Will shift the reaction right and thus, increase the amount of product.
Predict the shift in the equilibrium position that will occur for each of the following reactions at equilibrium when the volume of the reaction container is increased.
1) PCl3(g) + Cl2(g) ⇆ PCl5(g)
2) 2NBr3(g) ⇆ N2(g) + 3Br2(g)
3) CO(g) + Cl2(g) ⇆ COCl2(g)
4) H2(g) + B2(g) ⇆ 2HBr(g)
5) MgCO3(s) ⇆ MgO(s) + CO2(g)
1) The reaction will shift left
2) The reaction will shift right
3) The equilibrium will not change
4) The equilibrium will not change
5) The reaction will shift right.
If we increase the volume of a reaction container, the reaction itself will shift to the side that contains more gas molecules since this increases its own volume. If there are an equal number of gas molecules on both sides of the reaction, then the reaction will remain at equilibrium regardless of how we change the volume of the container.
1) The reaction will shift left producing more reactants because there are 2 molecules of gas reactants compared with 1 molecule of gas on the product side.
2) The reaction will shift right producing more products because there are 4 molecules of gas products compared with 2 molecules of gas on the reactants side.
3) The equilibrium will not change because both sides contain an equal number of molecules in gas state.
4) The equilibrium will not change because both sides contain an equal number of molecules in gas state.
5) Only the products contain a gas molecule, therefore, the reaction will shift right. Remember, we ignore the solids and liquids and only concentrate on the gases because of their extremely large volume compared with solids and liquids.
Consider the following equilibrium process for the commercial production of hydrogen:
CO(g) + H2O(g) ⇆ CO2(g) + H2(g) ΔH° = +42 kJ/mol
Predict the direction of the shift in equilibrium when
(a) the temperature is raised.
(b) more CO gas is added to the reaction mixture.
(c) some CO2 is removed from the mixture.
(d) the pressure on the gases is increased by changing the volume of the container.
(e) a catalyst is added to the reaction mixture
(a) The reaction will shift right.
(b) The reaction will shift right.
(c) The reaction will shift right.
(d) No effect on the equilibrium.
(e) It has no effect on the equilibrium position.
(a) the temperature is raised: We need to first realize that this is an endothermic reaction (ΔHo positive). This means the reaction consumes heat, or, in other words, heat is a reactant. According to Le Châtelier’s principle, if we increase the temperature, the reaction will go in the direction that absorbs heat. Therefore, the reaction will shift right.
(b) more CO gas is added to the reaction mixture: The reaction will shift right to consume some of the added CO gas and decrease its concentration.
(c) some CO2 is removed from the mixture: The reaction will shift right to reinstate the original concentration of CO2.
(d) the pressure on the gases is increased: If the number of gas molecules was different, the reaction would respond in a way that brings the pressure back down. However, the number of gas molecules is equal here, therefore, there will be no effect on the equilibrium.
(e) a catalyst is added to the reaction mixture: The catalyst is used to speed up the reaction so it reaches equilibrium more quickly. However, it has no effect on the equilibrium position.
Consider this reaction at equilibrium:
N2 + O2(g) ⇆ 2NO(g)
Predict whether the reaction will shift left, shift right, or remain unchanged after each disturbance.
a) NO is added to the reaction mixture.
b) N2 is added to the reaction mixture.
c) NO is removed from the reaction mixture.
a) The reaction will shift left.
b) The reaction will shift right.
c) The reaction will shift right.
According to the Le Châtelier’s principle, when a system in chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the system shifts in the direction that counteracts this change.
a) NO is added: The reaction will shift left to use up some of the added nitrogen monoxide and decrease its concentration.
b) N2 is added: The reaction will shift right to consume some of the added nitrogen gas and decrease its concentration.
c) NO is removed: The reaction will shift right to reinstate the original concentration of NO.
Check Also
- Chemical Equilibrium
- Equilibrium Constant
- Kp and Partial Pressure
- Kp and KcRelationship
- K Changes with Chemical Equation
- Equilibrium Constant K from Two Reactions
- Reaction Quotient – Q
- ICE Table – Calculating Equilibrium Concentrations
- ICE Table Practice Problems
- Le Châtelier’s principle
- Chemical Equilibrium Practice Problems