Question

Is the yield of \(\mathrm{SO}_{3}\) at equilibrium favored by a higher or lower pressure? By a higher or lower temperature? $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) \quad \Delta H=-197 \mathrm{~kJ} / \mathrm{mol} $$

Short answer

Higher pressure and lower temperature favor the yield of \( \mathrm{SO}_3 \).

Step-by-step solution

Identify the effect of pressure on the equilibrium

Consider the balanced equation for the synthesis of \( \mathrm{SO}_3 \) from \( \mathrm{SO}_2 \) and \( \mathrm{O}_2 \):\[ 2 \mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftarrows 2 \mathrm{SO}_3(g) \]. Notice that there are 3 moles of gaseous reactants and 2 moles of gaseous products. According to Le Chatelier's principle, increasing the pressure shifts the equilibrium towards the side with fewer moles of gas. Therefore, the yield of \( \mathrm{SO}_3 \) is favored by an increase in pressure.

Analyze the effect of temperature on the equilibrium

The reaction is exothermic (\( \Delta H = -197 \text{ kJ/mol} \)), meaning heat is released during the forward reaction. According to Le Chatelier's principle, increasing the temperature will shift the equilibrium towards the endothermic direction (reactants) to absorb the extra heat. Conversely, lowering the temperature shifts the equilibrium towards the exothermic direction (products), thus favoring the formation of \( \mathrm{SO}_3 \).

Key concepts

Equilibrium

In chemical reactions, equilibrium refers to the state where the rate of the forward reaction equals the rate of the reverse reaction. This balance means the concentrations of reactants and products remain constant over time, although they are not necessarily equal. For the synthesis of \(\mathrm{SO}_3\) from \(\mathrm{SO}_2\) and \(\mathrm{O}_2\), reaching equilibrium implies a specific ratio of reactants and products that stabilizes.

• Equilibrium is dynamic; molecules are continuously reacting, yet overall composition remains constant.
• Le Chatelier's Principle predicts how changes in conditions affect equilibrium.
• If a system at equilibrium is disturbed, it will adjust to counteract the disturbance and establish a new equilibrium.

Understanding equilibrium is key to predicting the yields of desired products in industrial chemical processes. For example, adjusting conditions such as pressure and temperature can significantly impact the amounts of products formed.

Pressure Effects

Pressure has a significant influence on the equilibrium of reactions involving gases. According to Le Chatelier's Principle, if you increase the pressure on a reaction mixture, the equilibrium position will shift towards the side with the fewer moles of gas. This adjustment helps the system relieve some of the increased pressure.

• In the \(\mathrm{SO}_3\) production: \(2 \mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftarrows 2 \mathrm{SO}_3(g)\).
• There are 3 moles of reactants (2 \(\mathrm{SO}_2\) + 1 \(\mathrm{O}_2\)) and 2 moles of products (\(\mathrm{SO}_3\)).

This means that increasing pressure will favor the formation of \(\mathrm{SO}_3\), thus maximizing its yield. Conversely, reducing pressure would shift the equilibrium toward the reactants, decreasing the production of \(\mathrm{SO}_3\).

Temperature Effects

Temperature changes can impact equilibrium in reactions, significantly altering product yields. Le Chatelier's Principle guides us here too, by stating that if temperature increases, the system will shift in the direction that absorbs heat.

• Consider the reaction: \(2 \mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftarrows 2 \mathrm{SO}_3(g)\).
• With \(\Delta H = -197 \text{ kJ/mol}\), the reaction releases heat (exothermic reaction).

Thus, increasing the temperature pushes the equilibrium towards the reactants (endo direction), reducing \(\mathrm{SO}_3\) yield. Lowering the temperature will favor forward reactions, as the system seeks to produce more heat, thereby increasing \(\mathrm{SO}_3\) yield.

Exothermic Reactions

Exothermic reactions are characterized by the release of heat energy as products form. This is an important factor in determining reaction direction and conditions for maximum yield.

• In an exothermic reaction, such as \(2 \mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftarrows 2 \mathrm{SO}_3(g)\) with \(\Delta H = -197 \text{ kJ/mol}\), heat is a product of the reaction.
• The negative sign of \(\Delta H\) indicates that the reaction releases energy.

So, if the temperature is lowered, the reaction "wants" to shift towards more heat production, which means moving forward to produce \(\mathrm{SO}_3\). On the other hand, if the temperature is raised, the reaction will shift in the reverse direction to absorb excess heat by favoring reactants. Understanding these dynamics is crucial for optimizing reactions in industrial applications where maximizing product yield is essential.