Question

For the reaction $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ \(K=168\) at \(1273 \mathrm{~K}\). If one starts with \(0.3 \mathrm{~atm}\) of \(\mathrm{CO}_{2}\) and \(12.0 \mathrm{~g}\) of \(\mathrm{C}\) at \(1273 \mathrm{~K}\), will the equilibrium mixture contain (a) mostly \(\mathrm{CO}_{2} ?\) (b) mostly CO? (c) roughly equal amounts of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) ? (d) only C?

Short answer

Question: At a certain temperature, the equilibrium constant (K) for the reaction C (graphite) + CO₂ (g) ⇌ 2CO (g) is 168. A container holds 12 g of C (graphite) and 0.3 atm of CO₂. At equilibrium, the mixture will contain mostly: (a) CO₂ (b) CO (c) roughly equal amounts of CO₂ and CO Answer: (b) mostly CO

Step-by-step solution

Identify the balanced chemical equation and equilibrium constant

From the exercise, we have the balanced chemical equation: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ The equilibrium constant, K, is given as \(168\).

Analyze the initial conditions

From the exercise, we are given the initial conditions: - C (graphite) = 12.0 grams - CO₂ = 0.3 atm Since we have grams of C, let's convert it to moles: $$ \mathrm{moles~of~C} = \frac{12.0\mathrm{~g}}{12.01\mathrm{~g/mol}} \approx 1\mathrm{~mol} $$ In this problem, C(s) is a solid, and its concentration won't change during the reaction. So, we will focus on the concentration changes of CO₂ and CO.

Formulate the reaction table

A reaction table or ICE table helps us to keep track of the concentration changes during the reaction. In this case, we will use pressures instead of concentrations due to given initial pressure values. | | CO₂ (g) | C (s) | 2CO (g) | |---------|---------|-------|---------| | Initial | 0.3 atm | N/A | 0 atm | | Change | -x atm | N/A | +2x atm | | Equilibrium | 0.3-x atm | N/A | 2x atm | We can use the value of K to solve for x and find the equilibrium pressures of CO₂ and CO.

Calculate the value of x using equilibrium constant (K)

We know that the equilibrium constant expression for this reaction is: $$ K = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} $$ As we are using pressures in this problem, we can rewrite it as: $$ K = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}} $$ Since we have established that the initial pressures result in equilibrium pressures of \(P_{\mathrm{CO}}=2x\) atm and \(P_{\mathrm{CO}_2} = 0.3-x\) atm, we can plug these into the expression above as: $$ 168 = \frac{(2x)^2}{(0.3-x)} $$ Solving the quadratic equation for x, we get: $$ x \approx 0.258 $$

Find the equilibrium pressures for CO and CO₂

Now we can use the value of x to calculate the equilibrium pressures of CO and CO₂: - For CO: \(P_{\mathrm{CO}} = 2x = 2 (0.258) \approx 0.516\) atm - For CO₂: \(P_{\mathrm{CO}_2} = 0.3 - x = 0.3 - 0.258 \approx 0.042\) atm

Determine the dominant component

Comparing the values we got for the equilibrium pressures of CO and CO₂: - \(P_{\mathrm{CO}} \approx 0.516\) atm - \(P_{\mathrm{CO}_2} \approx 0.042\) atm Since the equilibrium pressure of CO is significantly higher than that of CO₂, the equilibrium mixture will contain mostly CO. Therefore, the correct answer is: (b) mostly CO.