Problem 9 - Chemistry and Chemical Reactivity

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 \(\mathrm{CO}, 0.20 \mathrm{mol}\) of \(\mathrm{CO}_{2}\) and 0.40 mol of \(\mathrm{C}\) at equilibrium. (a) Calculate \(K_{c}\) for the reaction at \(700^{\circ} \mathrm{C}\) (b) Calculate \(K_{c}\) for the reaction, also at \(700^{\circ} \mathrm{C},\) if the amounts at equilibrium in the 2.0 -I. flask are 0.10 mol of \(\mathrm{CO}, 0.20 \mathrm{mol}\) of \(\mathrm{CO}_{2},\) and \(0.80 \mathrm{mol}\) of C. (c) Compare the results of (a) and (b). Does the quantity of carbon affect the value of \(K_{c} ?\) Explain.

Step-by-Step Solution

Answer

(a) \(K_c = 0.025\); (b) \(K_c = 0.025\); (c) No, solid carbon does not affect \(K_c\).

Step 1: Determine Initial Concentrations for Part (a)

First, we calculate the concentrations of each component in the flask. Since the given volume of the flask is 2.0 L, the concentrations are:- Concentration of CO: \( [\mathrm{CO}] = \frac{0.10 \text{ mol}}{2.0 \text{ L}} = 0.05 \text{ M} \).- Concentration of CO₂: \( [\mathrm{CO_2}] = \frac{0.20 \text{ mol}}{2.0 \text{ L}} = 0.10 \text{ M} \).

Step 2: Expression for Equilibrium Constant (Kc)

The equilibrium constant expression for the reaction \(\mathrm{C(s)} + \mathrm{CO_2(g)} \rightleftarrows 2 \mathrm{CO(g)}\) is given by:\[ K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO_2}]} \].

Step 3: Calculate Kc for Part (a)

Substitute the equilibrium concentrations into the expression:\[ K_c = \frac{(0.05)^2}{0.10} = \frac{0.0025}{0.10} = 0.025 \].

Step 4: Determine Initial Concentrations for Part (b)

The concentrations for this scenario are the same as in part (a) because the molar amounts of CO and CO₂ remain unchanged:- \( [\mathrm{CO}] = 0.05 \text{ M} \),- \( [\mathrm{CO_2}] = 0.10 \text{ M} \).

Step 5: Calculate Kc for Part (b)

Since the equilibrium concentrations of CO and CO₂ have not changed, and carbon is a solid and does not appear in the equilibrium expression, the value of \(K_c\) remains the same:\[ K_c = 0.025 \].

Step 6: Comparison and Impact of Carbon Quantity

In both scenarios (a) and (b), the value of \(K_c\) is 0.025. This demonstrates that the amount of solid carbon does not affect the value of \(K_c\), as \(K_c\) depends only on the concentrations of gaseous species in equilibrium for this reaction.

Key Concepts

Chemical Equilibrium

Chemical equilibrium represents a state in which the forward and reverse reactions occur at the same rate. It means that the concentrations of the reactants and products do not change over time. In the reaction involving carbon, carbon dioxide, and carbon monoxide, equilibrium is reached at a high temperature.

At equilibrium, the system is balanced. This balance is described quantitatively by the equilibrium constant, \(K_c\). The constant \(K_c\) gives us an idea of the position of the equilibrium. A high value of \(K_c\) indicates more products at equilibrium, while a low \(K_c\) shows more reactants.

In this problem, even though there is carbon, it is not included in the equilibrium expression because it is a solid and does not affect equilibrium concentrations of gaseous components.

• Equilibrium shows the exact balance point between product and reactant concentration.
• The reaction given maintains its balance regardless of the solid carbon present.

Reaction Stoichiometry

Stoichiometry of a reaction helps us understand the relationship between reactants and products in a balanced chemical equation. The stoichiometry for the given reaction is derived from its balanced form, where one mole of \(\mathrm{C}(\mathrm{s})\) reacts with one mole of \(\mathrm{CO_2}(\mathrm{g})\) to form two moles of \(\mathrm{CO}(\mathrm{g})\).

Understanding stoichiometry is vital to calculate the concentration of products and reactants. Here, it indicates that for every one mole of \(\mathrm{CO_2}\) consumed, two moles of \(\mathrm{CO}\) are produced.

The stoichiometric coefficients thus directly impact the calculation of \(K_c\) because \(\mathrm{CO}\) appears as a factor squared in the \(K_c\) expression due to its coefficient of 2 in the balanced equation:

• Stoichiometric coefficients dictate how reactants are transformed into products.
• A solid component, like carbon, remains unchanged in these calculations.

Concentration Calculations

Concentration calculations are crucial for determining the equilibrium constant \(K_c\). It involves calculating how much of each substance is present in a given volume at equilibrium.

For the reaction, you calculate the molarity of the gases:

• \([\mathrm{CO}] = \frac{0.10 \, \mathrm{mol}}{2.0 \, \mathrm{L}} = 0.05 \, \mathrm{M}\)
• \([\mathrm{CO_2}] = \frac{0.20 \, \mathrm{mol}}{2.0 \, \mathrm{L}} = 0.10 \, \mathrm{M}\)

These concentrations are plugged into the \(K_c\) expression:

• \(K_c = \frac{(\mathrm{0.05})^2}{\mathrm{0.10}} = \mathrm{0.025}\)

Thus, solid carbon, while part of the reaction, does not directly influence the \(K_c\) calculation as its concentration is not factored into the molarity calculations. The constancy of \(K_c\) in both scenarios of the problem showcases how concentrations of gases dictate the position of equilibrium.