Problem 15 - Chemistry Matter and Change

Question

Decide whether higher or lower temperatures will produce more \(\mathrm{CH}_{3} \mathrm{CHO}\) in the following equilibrium. \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \Delta H^{\circ}=-151 \mathrm{k}\)

Step-by-Step Solution

Answer

In the given equilibrium reaction, \(C_2H_2(g) + H_2O(g) \rightleftharpoons CH_3CHO(g)\), with a negative enthalpy change (\(\Delta H^{\circ}=-151 \mathrm{k}\)), indicating it is exothermic. According to Le Chatelier's Principle, lower temperatures will produce more CH3CHO, as the system will favor the exothermic forward reaction to counteract the decrease in temperature.

Step 1: Identify the reaction's direction

The given equilibrium reaction is: \[C_2H_2(g) + H_2O(g) \rightleftharpoons CH_3CHO(g)\] And the enthalpy change for the reaction is \(\Delta H^{\circ}=-151 \mathrm{k}\), which is negative, indicating the reaction is exothermic (releases heat) in the forward direction (formation of CH3CHO). Conversely, the reverse reaction (breaking down of CH3CHO) would be endothermic (absorbs heat).

Step 2: Apply Le Chatelier's Principle

According to Le Chatelier's Principle, when a change in temperature is applied to a system at equilibrium, the system will adjust to counteract the change. In this case, we need to determine whether higher or lower temperatures will favor the formation of more CH3CHO.

Step 3: Analyze the impact of temperature changes

If the temperature is increased, the system will try to counteract the increase by favoring the endothermic (heat-absorbing) reaction, which is the reverse reaction (breaking down CH3CHO). Therefore, higher temperatures will produce less CH3CHO. Conversely, if the temperature is decreased, the system will try to counteract the decrease in temperature by favoring the exothermic (heat-releasing) reaction, which is the forward reaction (formation of CH3CHO). Therefore, lower temperatures will produce more CH3CHO.

Step 4: Conclusion

In conclusion, lower temperatures will produce more CH3CHO in the given equilibrium reaction, as lowering the temperature favors the exothermic forward reaction due to Le Chatelier's Principle.

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is essential for students studying chemistry. It refers to the state where the concentrations of reactants and products in a reversible chemical reaction do not change over time. Despite appearances, this doesn't imply that the reaction has stopped; instead, it continues with reactants turning into products and vice versa at an equal rate, establishing a balance of the two.

Imagine watching two teams of equally skilled basketball players passing the ball among themselves without either side scoring; this movement of the ball is akin to a chemical reaction at equilibrium. It's not that the players are motionless; they are actively playing, but the score remains the same, analogous to the concentrations of reactants and products remaining constant in a chemically balanced system.

For the given exercise, \(C_2H_2(g) + H_2O(g) \rightleftharpoons CH_3CHO(g)\), students should note that the reaction can proceed in both the forward and backward directions, and equilibrium is reached when the rate of the forward reaction (formation of acetone) equals the rate of the backward reaction (breakdown of acetone).

Temperature Effects on Equilibrium

A critical aspect of chemical equilibrium is how it responds to changes in temperature. Le Chatelier's Principle helps us understand that a system at equilibrium will adjust to negate any imposed changes, including temperature. Think of it like a thermostat in your home that adjusts the heat to maintain a set temperature.

When the temperature of a system is increased, the reaction will shift in favor of the endothermic direction, absorbing extra heat and partially restoring the temperature balance. On the other hand, a decrease in temperature prompts the system to offset the loss of heat by favoring the exothermic reaction, which releases heat.

Referring to our previous example, when the equilibrium of \(C_2H_2(g) + H_2O(g) \rightleftharpoons CH_3CHO(g)\) is subject to a temperature change, the system shifts to counteract that change. You can visualize this as either turning on the air conditioning when it gets too warm or cranking up the furnace when it gets too cold, each action restoring comfort—or in chemical terms, equilibrium.

Enthalpy Change

Enthalpy change, denoted by \(Delta H\), is a measurement of heat change during a reaction at constant pressure. A negative \(Delta H\) value indicates an exothermic reaction, where heat is released into the surroundings, and it feels warm to the touch. In contrast, a positive \(Delta H\) signifies an endothermic reaction, which absorbs heat, making the surroundings feel cooler.

Considering our original problem where \(Delta H^{circ}=-151 k\), this negative enthalpy change corresponds to the exothermic forward reaction. It's like releasing a warm burst of air into a room every time the reaction progresses toward the product \(CH_3CHO(g)\). Thus, by lowering the temperature, we are effectively encouraging the production of more \(CH_3CHO\) as the reaction seeks to release heat and balance the temperature decrease. Learning to associate the sign of \(Delta H\) with either heat release or absorption will vastly simplify the comprehension of reaction energetics.