91Ó°ÊÓ

Le Chatelier's Principle helps us understand how a chemical equilibrium responds to changes in conditions like temperature, pressure, and concentration. It posits that if a dynamic equilibrium is disturbed by changing conditions, the position of the equilibrium moves to counteract the change.

For example, in the reaction given in the exercise, $$ \text{Ag}_2 \text{CO}_3(\text{s}) \rightleftharpoons \text{Ag}_2 \text{O}(\text{s}) + \text{CO}_2(\text{g}) $$
if we increase the temperature, we disturb the equilibrium. According to Le Chatelier's Principle, the system will try to counteract this change by favoring the reaction that absorbs heat, which in our case is the endothermic reaction. This shifts the equilibrium position to the left, favoring the formation of reactants.

In summary, Le Chatelier's Principle allows us to predict the behavior of a system in equilibrium when external conditions change.

The Equilibrium Constant (\text{K}_p) quantifies the ratio of the concentrations (or partial pressures for gases) of products to reactants at equilibrium.

In our context, the introduced reaction has an equilibrium constant defined by:
$$ \text{K}_p = \frac{[\text{CO}_2]}{[\text{Ag}_2 \text{CO}_3]} $$
Here, since \text{Ag}_2 \text{CO}_3 and \text{Ag}_2 \text{O} are solids, their activities are considered to be 1 and do not appear in the equation. Therefore, the expression simplifies to \text{K}_p = [\text{CO}_2].

The values of \text{K}_p given are:
$$ \text{K}_p (298\text{K}) = 3.16 \times 10^3 \text{ atm} $$ $$ \text{K}_p (500\text{K}) = 1.48 \text{ atm} $$
These values clearly indicate the extent to which the reaction favors the formation of \text{CO}_2 at different temperatures. \text{K}_p is a crucial indicator of the position of equilibrium.

An endothermic reaction absorbs heat from its surroundings. The reaction requires energy to proceed.

For the given reaction: $$ \text{Ag}_2 \text{CO}_3(\text{s}) \rightleftharpoons \text{Ag}_2 \text{O}(\text{s}) + \text{CO}_2(\text{g}) $$
we observed that the equilibrium constant decreases as the temperature increases, which is a characteristic behavior of endothermic reactions.

When the temperature increases, the equilibrium shifts towards the reactants to absorb the additional heat, reducing the yield of \text{CO}_2.

Thus, identifying the reaction as endothermic explains why \text{K}_p is lower at higher temperatures.

The reaction's dependence on temperature is a crucial aspect of understanding chemical equilibrium.

In the given reaction: $$ \text{Ag}_2 \text{CO}_3(\text{s}) \rightleftharpoons \text{Ag}_2 \text{O}(\text{s}) + \text{CO}_2(\text{g}) $$
We see that \text{K}_p values change with temperature:
$$ \text{K}_p (298\text{K}) = 3.16 \times 10^3 \text{ atm} $$ $$ \text{K}_p (500\text{K}) = 1.48 \text{ atm} $$
This indicates a significant shift in equilibrium due to temperature change.

According to Le Chatelier's Principle, for an endothermic reaction, which absorbs heat, increasing temperature will shift the equilibrium towards the reactants, thereby decreasing the value of \text{K}_p.

Hence, temperature plays a vital role in determining the direction and extent of the reaction, impacting the yield of \text{CO}_2}.