Question

Consider the exothermic reaction: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g)$$ If you were trying to maximize the amount of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) produced, which tactic might you try? Assume that the reaction mixture reaches equilibrium. \begin{equation}\begin{array}{l}{\text { a. increasing the reaction volume }} \\\ {\text { b. removing } \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2} \text { from the reaction mixture as it forms }} \\ {\text { c. lowering the reaction temperature }} \\ {\text { d. adding } \mathrm{Cl}_{2}}\end{array}\end{equation}

Short answer

To maximize the production of \(\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{Cl}_{2}\), removing \(\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{Cl}_{2}\) from the reaction mixture as it forms would be the most effective tactic.

Step-by-step solution

Understand Le Chatelier's Principle

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In the case of an exothermic reaction, increasing temperature shifts the equilibrium to the left (endothermic direction), and decreasing temperature shifts it to the right (exothermic direction). Removing products or adding reactants also shifts the equilibrium to the right, favoring product formation.

Evaluate the Given Options

Option a (increasing reaction volume) does not shift the equilibrium in a predictable way because the reaction involves equal moles of gas on both sides. Option b (removing \(\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{Cl}_{2}\) from the reaction mixture) would shift the equilibrium to the right, increasing product formation. Option c (lowering the reaction temperature) would also shift the equilibrium to the right, as it is an exothermic reaction. Option d (adding \(\mathrm{Cl}_{2}\)) shifts the equilibrium to the right by increasing the concentration of a reactant.

Choose the Best Tactic

To maximize the amount of \(\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{Cl}_{2}\) produced, removing the product from the reaction mixture (Option b) would be the most effective tactic as it continuously shifts the equilibrium towards the products. Lowering the reaction temperature (Option c) would also help to some extent, but it also slows down the rate of reaction.

Key concepts

Chemical Equilibrium

In chemistry, chemical equilibrium refers to a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. As a result, the concentrations of the reactants and products remain unchanged over time, although both reactions continue to occur.

This balance can be represented with the expression \( K_c = \frac{[Products]}{[Reactants]} \) where \( K_c \) is the equilibrium constant, and the square brackets denote concentrations. At equilibrium, regardless of the initial amounts of reactants and products, the ratio of concentration of products to reactants always converges to a particular value, given constant temperature.

It's crucial to understand that equilibrium does not mean the reactants and products are in equal concentrations; rather, it's their rates of formation and consumption that are equal. The position of the equilibrium, as described by the equilibrium constant, tells us the extent to which a reaction proceeds to form products under those specific conditions.

Exothermic Reactions

A reaction that releases heat to its surroundings is termed an exothermic reaction. This release of heat is often observed as an increase in temperature in the reaction mixture.

Exothermic reactions are characterized by negative enthalpy change (\(\Delta H < 0\)), indicating that energy is exiting the system. In the context of the provided exercise, the reaction \( \mathrm{C}_{2} \mathrm{H}_{4}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g) \) is exothermic. When this reaction moves in the forward direction, it releases heat, signifying that heat can be considered as a 'product' in a metaphorical sense.

Therefore, when the temperature is lowered, this type of reaction will shift to the right according to Le Chatelier's Principle, to produce more heat, favouring the formation of \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2} \) in this case. This aligns with the principle's statement that a system at equilibrium will try to counteract the changes imposed on it.

Equilibrium Shift

When the conditions of a system at chemical equilibrium are altered, the system responds to counteract the imbalance, a concept known as an equilibrium shift. Le Chatelier's Principle guides us in predicting the direction of this shift.

Changing concentrations of reactants or products, pressure changes (involving gases), volume changes, or temperature alterations can cause this response. For instance, if more reactants are added, the equilibrium will shift to the right to form more products, aiming to reduce the concentration of added reactants. Conversely, removing products as they are formed, as discussed in the exercise, causes the system to produce more products to maintain the equilibrium constant.

In conclusion, manipulating various conditions of a reaction can strategically shift the equilibrium. In the exercise, removing the product, \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2} \), as well as lowering temperature, are effective tactics to drive the reaction forward, thereby increasing the yield of the desired product due to the equilibrium shift to the right.