Question

What relative conditions (reactant concentrations, pressure, and temperature) would favor a high equilibrium concentration of the underlined substance in each of the following equilibrium systems? a. $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \underline{\mathrm{CO}_{2}(g)}+167 \mathrm{kJ}$$ b. $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftarrows \underline{\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)}+42 \mathrm{kJ}$$ c. $$2 \mathrm{HI}(g)+12.6 \mathrm{kJ} \rightleftarrows \mathrm{H}_{2}(g)+ \underline{\mathrm{I}_{2}(g)}$$ d. $$4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(g)+ \underline{2 \mathrm{Cl}_{2}(g)}+113 \mathrm{kJ}$$ e. $$\mathrm{PCl}_{5}(g)+88 \mathrm{kJ} \rightleftarrows \mathrm{PCl}_{3}(g)+ \underline{\mathrm{Cl}_{2}(g)}$$

Short answer

a. Low Temp, High Pressure, High CO & O2; b. Low Temp, High Cu2+ & NH3; c. High Temp, High HI; d. Low Temp, High Pressure, High HCl & O2; e. High Temp, High PCl5.

Step-by-step solution

- Analyze Reaction (a)

Consider the reaction: \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \underline{\mathrm{CO}_{2}(g)}+167 \mathrm{kJ}\). Since the formation of \(\mathrm{CO}_{2}(g)\) releases heat (exothermic), higher concentrations of \(\mathrm{CO}_{2}(g)\) are favored by low temperatures. Additionally, increasing the pressure will favor the side with fewer gas molecules (right side in this case). Higher reactant concentrations (\(\mathrm{CO}(g)\) and \(\mathrm{O}_{2}(g)\)) will also push the equilibrium towards more \(\mathrm{CO}_{2}(g)\).

- Analyze Reaction (b)

Consider the reaction: \( \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftarrows \underline{\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)}+42 \mathrm{kJ} \). This is also exothermic, thus a high concentration of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) is favored by low temperature. Increasing the concentrations of the reactants (\(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_{3}\)) will shift the equilibrium to the right, increasing \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\).

- Analyze Reaction (c)

For the reaction: \(2 \mathrm{HI}(g)+12.6 \mathrm{kJ} \rightleftarrows \mathrm{H}_{2}(g)+ \underline{\mathrm{I}_{2}(g)}\), the formation of \(\mathrm{I}_{2}(g)\) is endothermic, favoring high temperatures. Increasing the pressure will favor the side with fewer gas molecules, although neither side has different moles of gases, so pressure changes do not significantly affect the equilibrium. Increasing the concentration of \(\mathrm{HI}(g)\) favors the formation of \(\mathrm{I}_{2}(g)\).

- Analyze Reaction (d)

For the reaction: \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(g)+ \underline{2 \mathrm{Cl}_{2}(g)}+113 \mathrm{kJ}\), high yields of \(\mathrm{Cl}_{2}(g)\) are favored by low temperatures because the reaction is exothermic. Increasing the pressure will favor the side with fewer gas molecules. Higher reactant concentrations will push the equilibrium to the right, forming more \(\mathrm{Cl}_{2}(g)\).

- Analyze Reaction (e)

In the reaction: \(\mathrm{PCl}_{5}(g)+88 \mathrm{kJ} \rightleftarrows \mathrm{PCl}_{3}(g)+ \underline{\mathrm{Cl}_{2}(g)}\), the formation of \(\mathrm{Cl}_{2}(g)\) is endothermic, favoring high temperatures. Increasing the pressure favors the side with fewer gas molecules, though this has no different in the case. Higher concentrations of \(\mathrm{PCl}_{5}(g)\) will shift the equilibrium to the right, increasing \(\mathrm{Cl}_{2}(g)\).

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is essential when analyzing equilibrium reactions. This principle states that if a system at equilibrium experiences a change in concentration, temperature, or pressure, the equilibrium will shift to counteract the change and restore a new balance.
For instance, if we increase the concentration of a reactant, the system will shift towards forming more products to reduce the added reactant.
Conversely, if we decrease the concentration of a reactant, the system will shift towards forming more reactants.
This concept is particularly useful in predicting how different conditions will affect the concentration of substances in a reaction, as seen in the original exercise.

Endothermic and Exothermic Reactions

Reactions can either absorb heat (endothermic) or release heat (exothermic).
Understanding this helps us predict the effect of temperature changes on equilibrium.
In endothermic reactions, heat is a reactant, so increasing temperature shifts the equilibrium to the right, favoring the formation of products. For example, the formation of \(\mathrm{Cl}_2(g)\) in reaction e is endothermic.
In exothermic reactions, heat is a product, so increasing temperature shifts the equilibrium to the left, favoring the formation of reactants. For example, the formation of \(\mathrm{CO}_2(g)\) in reaction a is exothermic.
Recognizing the nature of the reaction can help manipulate conditions to obtain more or less of the desired substance.

Effects of Temperature and Pressure on Equilibrium

Pressure and temperature changes can significantly affect the equilibrium state of a reaction, especially those involving gases.
When a system involves gases, changes in pressure will affect the side of the reaction with more or fewer gas molecules.
Increasing pressure favors the side with fewer gas molecules to reduce the imposed pressure.

• For instance, in reaction a, increasing pressure favors the formation of \(\mathrm{CO}_2(g)\), which is on the side with fewer gas molecules.
• In reaction steps where the number of moles of gas is balanced on both sides, pressure changes will have minimal impact, such as in reaction c.

Temperature changes, as discussed, will shift the equilibrium depending on whether the reaction is endothermic or exothermic. Combining all these factors allows us to predict and control the equilibrium concentrations of various substances effectively.