At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass C (s) + CO2 (g) = 2CO (g) Calculate Kc for this reaction at the above temperature.

To calculate the equilibrium constant \( K_c \) for the reaction \( C (s) + CO_2 (g) \rightleftharpoons 2CO (g) \) at a temperature of 1127 K and a pressure of 1 atm, we need to start by analyzing the information given about the gaseous mixture and the equilibrium condition.

Understanding the Reaction

This reaction involves solid carbon reacting with carbon dioxide to produce carbon monoxide. The equilibrium constant \( K_c \) is defined in terms of the concentrations of the gaseous products and reactants at equilibrium. Since solid carbon is a pure substance, it does not appear in the expression for \( K_c \).

Setting Up the Equilibrium Expression

The equilibrium expression for this reaction is given by:

Kc = \frac{[CO]^2}{[CO_2]}

Where \([CO]\) and \([CO_2]\) are the molar concentrations of carbon monoxide and carbon dioxide, respectively.

Calculating Molar Masses

First, we need to determine the molar masses of the gases involved:

• Molar mass of \( CO \) = 12.01 (C) + 16.00 (O) = 28.01 g/mol
• Molar mass of \( CO_2 \) = 12.01 (C) + 2 × 16.00 (O) = 44.01 g/mol

Finding Masses and Moles

Given that the gaseous mixture consists of 90.55% CO by mass, we can assume we have 100 g of the mixture for simplicity. This means:

• Mass of \( CO \) = 90.55 g
• Mass of \( CO_2 \) = 100 g - 90.55 g = 9.45 g

Calculating Moles of Each Gas

Next, we convert these masses into moles:

• Moles of \( CO \) = \( \frac{90.55 \, g}{28.01 \, g/mol} \approx 3.23 \, mol \)
• Moles of \( CO_2 \) = \( \frac{9.45 \, g}{44.01 \, g/mol} \approx 0.215 \, mol \)

Calculating Concentrations

At 1 atm and 1127 K, we can use the ideal gas law \( PV = nRT \) to find the volume of the gas mixture. Rearranging gives us:

V = \frac{nRT}{P}

Using \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \):

V = \frac{(3.23 + 0.215) \cdot 0.0821 \cdot 1127}{1} \approx 25.1 \, L

Calculating Concentrations

Now we can find the concentrations:

• \([CO] = \frac{3.23 \, mol}{25.1 \, L} \approx 0.128 \, mol/L\)
• \([CO_2] = \frac{0.215 \, mol}{25.1 \, L} \approx 0.00855 \, mol/L\)

Calculating Kc

Now we can substitute these values into the equilibrium expression:

Kc = \frac{(0.128)^2}{0.00855} \approx 1.93 \, (unitless)

Final Result

Thus, the equilibrium constant \( K_c \) for the reaction at 1127 K and 1 atm is approximately 1.93. This value indicates the extent to which the reaction favors the formation of carbon monoxide over carbon dioxide at this temperature and pressure.