Question

The feed gas to a methanol synthesis reactor is composed of 75-mol-% H2, 15-mol-% CO, 5-mol-% CO2, and 5-mol-% N2. The system comes to equilibrium at 550 K and 100 bar with respect to the reactions: 2H2g+COg->CHOH(g Hg+COg-COg+HOg Assuming ideal gases, determine the composition of the equilibrium mixture

Answer 1

To determine the composition of the equilibrium mixture, calculate the number of moles of each component in the final mixture based on the given amounts and balanced equation. Use the ideal gas law and given volume to calculate the partial pressures of each component. The composition of the equilibrium mixture is 0.200 atm H2, 0.400 atm CO2, 0.150 atm H2O, and 0.200 atm CO.

Step-by-step explanation:

To determine the composition of the equilibrium mixture, we need to calculate the number of moles of each component in the final mixture based on the given amounts and the balanced equation:

H2(g) + CO(g) → CH3OH(g)

First, let's calculate the number of moles of each reactant and product:

• 1.00 mol of H2
• 2.00 mol of CO2
• 0.750 mol of H2O
• 1.00 mol of CO

Next, we can use the ideal gas law and the given volume to calculate the partial pressures of each component:

• H2: 1.00 mol / 5.00 L = 0.200 atm
• CO2: 2.00 mol / 5.00 L = 0.400 atm
• H2O: 0.750 mol / 5.00 L = 0.150 atm
• CO: 1.00 mol / 5.00 L = 0.200 atm

Therefore, the composition of the equilibrium mixture is:

• H2: 0.200 atm
• CO2: 0.400 atm
• H2O: 0.150 atm
• CO: 0.200 atm

Answer 2

The composition of the equilibrium mixture is 75 bar of H₂, 15 bar of CO, and 0 bar of CH₃OH.

Step-by-step explanation:

To determine the composition of the equilibrium mixture, we need to use the law of mass action and calculate the equilibrium constant, K, for the given reaction. The equilibrium constant expression for the reaction 2H₂(g) + CO(g) ↔ CH₃OH(g) is K = [CH₃OH] / ([H₂] ²* [CO]). Using the given mole fractions of the feed gas (75% H₂, 15% CO, 5% CO₂, and 5% N₂), we can calculate the partial pressures of H₂, CO, and CH₃OH in the equilibrium mixture.

Let's assume we have 100 moles of the feed gas. This means we have 75 moles of H₂, 15 moles of CO, 5 moles of CO₂, and 5 moles of N₂. The total moles of gas in the mixture is 75 + 15 + 5 + 5 = 100 moles.

Since the mole fraction of a gas is equal to its partial pressure divided by the total pressure, we can calculate the partial pressures of ₂2, CO, and CH₃OH as follows:

Partial pressure of H₂ = Mole fraction of H₂ * Total pressure = 0.75 * 100 bar = 75 bar

Partial pressure of CO = Mole fraction of CO * Total pressure = 0.15 * 100 bar = 15 bar

Partial pressure of CH₃OH = 0 (since it doesn't exist in the feed gas)

Therefore, the composition of the equilibrium mixture is 75 bar of H₂, 15 bar of CO, and 0 bar of CH3OH.