Question

Carbon dioxide is commonly produced through the reaction \(\mathrm{C}+\mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\). Determine the yield of carbon dioxide (mole fraction) when this is done in a reactor maintained at 1 atm and 3800 K. The natural logarithm of the equilibrium constant for the reaction \(\mathrm{C}+\mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\) at \(3800 \mathrm{K}\) is \(-0.461 .\)

Short answer

Answer: The yield of carbon dioxide (mole fraction) in the reactor is approximately 0.205.

Step-by-step solution

Write the balanced reaction equation

The given reaction can be written as follows: C + O2 <-> CO2

Calculate the equilibrium constant K

We are given the natural logarithm of the equilibrium constant: ln(K) = -0.461 To find the equilibrium constant K, we need to take the exponential of ln(K): K = e^(-0.461) ≈ 0.630

Write the expression for K using concentrations

We can write the expression for K using the concentrations of reactants and products: K = [CO2] / ([C] * [O2])

Assume initial and equilibrium concentrations

Since the system is in equilibrium, we can assume initial concentrations and calculate equilibrium concentrations by introducing a variable x. Let x be the amount of C that reacts with O2. Initial concentrations: [C] = 1 - x, [O2] = 1 - x, [CO2] = x Equilibrium concentrations: [C] = 1 - x, [O2] = 1 - x, [CO2] = x

Substitute equilibrium concentrations into the K expression

Now we can substitute the equilibrium concentrations in terms of x into the K expression: 0.630 = x / ((1 - x) * (1 - x))

Solve for x

We'll rearrange the equation and solve for x: x = 0.630 * (1 - x)² x = 0.630 - 1.26x + 0.630x² 0.630x² - 0.890x + 0.630 = 0 Using a quadratic formula or a solver, we find x ≈ 0.614

Calculate the mole fraction of CO2

The mole fraction of CO2 is the ratio of moles of CO2 to the total moles in the system: Mole fraction of CO2 = x / (x + (1 - x) + (1 - x)) Mole fraction of CO2 ≈ 0.614 / (0.614 + (1 - 0.614) + (1 - 0.614)) ≈ 0.614 / 3 Mole fraction of CO2 ≈ 0.205 Therefore, the yield of carbon dioxide (mole fraction) in the reactor is approximately 0.205.