Question

The reaction $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}(\mathrm{g}) $$ occurs at high temperatures. At \(700^{\circ} \mathrm{C},\) a \(2.0-\mathrm{L}\). flask contains 0.10 mol of \(\mathrm{CO}, 0.20\) mol of \(\mathrm{CO}_{2},\) and 0.40 mol of \(\mathrm{C}\) at equilibrium. (a) Calculate \(K\) for the reaction at \(700^{\circ} \mathrm{C}\) (b) Calculate \(K\) for the reaction, also at \(700^{\circ} \mathrm{C},\) if the amounts at equilibrium in the \(2.0-\mathrm{L}\). flask are \(0.10 \mathrm{mol}\) of \(\mathrm{CO}, 0.20 \mathrm{mol}\) of \(\mathrm{CO}_{2},\) and \(0.80 \mathrm{mol}\) of \(\mathrm{C}\) (c) Compare the results of (a) and (b). Does the quantity of carbon affect the value of \(K\) ? Explain.

Short answer

(a) \( K = 0.025 \); (b) \( K = 0.025 \); the amount of carbon does not affect \( K \).

Step-by-step solution

Understand the Reaction and Equilibrium Expression

The reaction is \( \mathrm{C}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}(\mathrm{g}) \). Since C(s) is a solid, it does not appear in the expression for the equilibrium constant \( K \). The equilibrium constant expression is \( K = \frac{[\mathrm{CO}]^2}{[\mathrm{CO_2}]} \), where \([\mathrm{CO}]\) and \([\mathrm{CO_2}]\) are the concentrations of CO and CO_2, respectively, in mol/L.

Calculate Equilibrium Concentrations for (a)

Given the amounts in a 2.0 L flask, the concentrations are: \([\mathrm{CO}] = \frac{0.10}{2.0} = 0.05\ mol/L\) and \([\mathrm{CO_2}] = \frac{0.20}{2.0} = 0.10\ mol/L\).

Calculate \( K \) for (a)

Using the equilibrium expression, \( K = \frac{(0.05)^2}{0.10} = \frac{0.0025}{0.10} = 0.025 \).

Calculate Equilibrium Concentrations for (b)

In the second scenario, the concentrations remain the same as in (a) because they depend only on the amounts of gases, not the solid. Thus, \([\mathrm{CO}] = 0.05\ mol/L\) and \([\mathrm{CO_2}] = 0.10\ mol/L\).

Calculate \( K \) for (b)

Since the equilibrium concentrations are the same as in part (a), \( K = 0.025 \).

Compare Results and Effect of Carbon Amount

The value of \( K \) in both (a) and (b) is 0.025. The amount of carbon does not affect the equilibrium constant because \( K \) depends only on the concentrations of gases at equilibrium, and solids are not included in the equilibrium expression.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a crucial concept in chemical equilibrium. It provides a numerical value that reflects the ratio of concentrations of products to reactants at equilibrium. This ratio remains constant for a given reaction at a specific temperature.
For the reaction \( \mathrm{C}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}(\mathrm{g}) \), the equilibrium constant \( K \) is expressed as \( K = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \). Here, \([\mathrm{CO}]\) is the concentration of CO, and \([\mathrm{CO}_2]\) is the concentration of CO2. Notice that the solid carbon, \( \mathrm{C}(\mathrm{s}) \), does not appear in the expression for \( K \). This is because solids and pure liquids are not included in equilibrium constant expressions.
The value of \( K \) can help determine the extent to which a reaction occurs. A large \( K \) indicates a reaction favoring products, whereas a small \( K \) favors reactants.

Reaction Quotient

The reaction quotient, \( Q \), is a factor used to predict the direction in which a reaction will proceed. It is calculated using the same formula as the equilibrium constant, but at non-equilibrium conditions.
Like \( K \), \( Q \) is expressed using the concentrations of products divided by reactants raised to their stoichiometric coefficients. For example, for the reaction \( \mathrm{C}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}(\mathrm{g}) \), the reaction quotient is \( Q = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \).
When \( Q < K \), the system shifts towards the products to reach equilibrium. Conversely, if \( Q > K \), the reaction will shift towards the reactants. If \( Q = K \), the system is already at equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle describes how a system at equilibrium responds to disturbances. It will adjust to counteract any changes in concentration, pressure, temperature, or volume.
For the given reaction, Le Chatelier's Principle suggests how changes can affect the reaction direction. For instance:

• If more \( \mathrm{CO}_2 \) is added, the system will shift towards the products, forming more \( \mathrm{CO} \).
• If \( \mathrm{CO} \) is removed, the system will shift towards the products to replenish the lost \( \mathrm{CO} \).
• If the temperature is increased, depending on whether the reaction is exothermic or endothermic, the equilibrium will shift to absorb or emit heat.

Le Chatelier's Principle helps predict the qualitative changes in a system, but it does not quantify how much the reaction shifts.

Heterogeneous Equilibrium

Heterogeneous equilibrium involves reactions with substances in different phases, such as solids, liquids, and gases. In the given reaction, \( \mathrm{C}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}(\mathrm{g}) \), we observe a heterogeneous system because solid carbon and gaseous components are involved.
One significant aspect of heterogeneous equilibrium is that the concentration of pure solids and liquids remains constant. Therefore, they do not appear in the equilibrium constant expression. This is why solid carbon is excluded from the \( K \) expression for this reaction.
Understanding heterogeneous equilibria is important because it simplifies the expressions used for equilibrium constants by ignoring the phases that do not change in concentration.