At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2)` in equilibrium with solid carbon has `90.55% CO` by mass:
`C_((s))+CO_(2(g))hArr2CO_((g))`
Calculate `K_(c)` for the reaction at the above temperature.

Explanation

To calculate the equilibrium constant \( K_c \) for the reaction:

\[ C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)} \]

at \( 1127 \, K \) and \( 1 \, atm \) pressure, we can follow these steps:

### Step 1: Determine the mass of CO and CO2

Given that the gaseous mixture has \( 90.55\% \) CO by mass, we can assume a total mass of the gas mixture to be \( 100 \, g \).

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

### Step 2: Calculate the number of moles of CO and CO2

Using the molar masses:
- Molar mass of \( CO = 28 \, g/mol \)
- Molar mass of \( CO_2 = 44 \, g/mol \)

Calculating the moles:

\[ \text{Moles of } CO = \frac{90.55 \, g}{28 \, g/mol} \approx 3.23 \, mol \]

\[ \text{Moles of } CO_2 = \frac{9.45 \, g}{44 \, g/mol} \approx 0.21 \, mol \]

### Step 3: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) for the reaction is given by:

\[ K_p = \frac{(P_{CO})^2}{P_{CO_2}} \]

Where \( P_{CO} \) and \( P_{CO_2} \) are the partial pressures of \( CO \) and \( CO_2 \) respectively.

### Step 4: Calculate total moles and partial pressures

Total moles \( n_{total} = n_{CO} + n_{CO_2} = 3.23 + 0.21 = 3.44 \, mol \)

Now, we can find the partial pressures using the ideal gas law \( P = \frac{nRT}{V} \). Since we are given \( 1 \, atm \) total pressure, we can find the partial pressures directly as:

\[ P_{CO} = \frac{n_{CO}}{n_{total}} \times P_{total} = \frac{3.23}{3.44} \times 1 \, atm \approx 0.94 \, atm \]

\[ P_{CO_2} = \frac{n_{CO_2}}{n_{total}} \times P_{total} = \frac{0.21}{3.44} \times 1 \, atm \approx 0.061 \, atm \]

### Step 5: Substitute into the \( K_p \) expression

Now substituting into the \( K_p \) expression:

\[ K_p = \frac{(0.94)^2}{0.061} \approx \frac{0.8836}{0.061} \approx 14.44 \]

### Step 6: Convert \( K_p \) to \( K_c \)

Using the relationship between \( K_p \) and \( K_c \):

\[ K_c = \frac{K_p}{RT^{\Delta n}} \]

Where \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \), \( T = 1127 \, K \), and \( \Delta n = 2 - 1 = 1 \).

Substituting the values:

\[ K_c = \frac{14.44}{(0.0821)(1127)^1} \approx \frac{14.44}{92.71} \approx 0.156 \, mol/L \]

### Final Answer

Thus, the equilibrium constant \( K_c \) for the reaction at \( 1127 \, K \) is approximately:

\[ K_c \approx 0.156 \, mol/L \]

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