Question

For the reaction \(\mathrm{Ag}_{2} \mathrm{CO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) the value of the equilibrium constant \(\mathrm{K}_{\mathrm{p}}\) is \(3.16 \times 10^{3}\) atm at \(298 \mathrm{~K}\) and is \(1.48 \mathrm{~atm}\) at \(500 \mathrm{~K}\); then which one of the following statements is correct? (1) The yield of \(\mathrm{CO}_{2}\) is independent of temperature. (2) The yield of \(\mathrm{CO}_{2}\) will increase at high pressure. (3) The value of \(\Delta H\) for the forward reaction is negligible. (4) The forward reaction is endothermic.

Short answer

The forward reaction is endothermic.

Step-by-step solution

Write the reaction

Consider the reaction: \[ \mathrm{Ag}_{2} \mathrm{CO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \]

Identify given values

The equilibrium constant \( \mathrm{K}_{\mathrm{p}} \) is given as: \[ \mathrm{K}_{\mathrm{p}}(298 \mathrm{~K}) = 3.16 \times 10^{3} \: \mathrm{atm} \] \[ \mathrm{K}_{\mathrm{p}}(500 \mathrm{~K}) = 1.48 \: \mathrm{atm} \]

Understand the implications of \( \mathrm{K}_{\mathrm{p}} \)

The equilibrium constant \( \mathrm{K}_{\mathrm{p}} \) represents the partial pressure of \( \mathrm{CO}_{2} \) in the reaction as only gases contribute to \( \mathrm{K}_{\mathrm{p}} \).

Determine the nature of the reaction

Since \( \mathrm{K}_{\mathrm{p}} \) decreases with an increase in temperature from 298 K to 500 K, it indicates a shift in equilibrium towards the reactants. For the forward reaction (production of \( \mathrm{CO}_{2} \)) to be less favored at higher temperatures, the reaction must be endothermic. According to Le Chatelier's principle, an endothermic reaction's equilibrium constant will decrease with an increase in temperature.

Analyze the options

From the analysis: (1) This option is incorrect since the yield of \( \mathrm{CO}_{2} \) does depend on temperature.(2) High pressure would not affect the equilibrium in significant amounts because solids and gases are involved.(3) The value of \( \Delta H \) cannot be negligible because we observe a change in \( \mathrm{K}_{\mathrm{p}} \) with temperature.(4) The forward reaction being endothermic is correct since \( \mathrm{K}_{\mathrm{p}} \) decreases with an increase in temperature.

Key concepts

Le Chatelier's Principle

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.

Equilibrium Constant

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.

Endothermic Reaction

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.

Reaction Temperature Dependence

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}.