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, given that the gaseous mixture contains \( 90.55\% \) CO by mass, we can follow these steps:
### Step 1: Determine the Mass of CO and CO2
Given that the mixture is \( 90.55\% \) CO by mass, if we assume a total mass of the mixture to be \( 100 \, g \):
- Mass of CO = \( 90.55 \, g \)
- Mass of CO2 = \( 100 - 90.55 = 9.45 \, g \)
### Step 2: Calculate the Molar Mass of CO and CO2
- Molar mass of CO = \( 12 \, g/mol \, (C) + 16 \, g/mol \, (O) = 28 \, g/mol \)
- Molar mass of CO2 = \( 12 \, g/mol \, (C) + 32 \, g/mol \, (O_2) = 44 \, g/mol \)
### Step 3: Calculate the Number of Moles of CO and CO2
Using the formula \( \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} \):
- Moles of CO = \( \frac{90.55 \, g}{28 \, g/mol} = 3.234 \, mol \)
- Moles of CO2 = \( \frac{9.45 \, g}{44 \, g/mol} = 0.215 \, mol \)
### Step 4: Calculate the Total Number of Moles
Total moles = Moles of CO + Moles of CO2
\[
\text{Total moles} = 3.234 + 0.215 = 3.449 \, mol
\]
### Step 5: Calculate the Mole Fraction of CO and CO2
- Mole fraction of CO:
\[
X_{CO} = \frac{3.234}{3.449} \approx 0.938
\]
- Mole fraction of CO2:
\[
X_{CO2} = \frac{0.215}{3.449} \approx 0.062
\]
### Step 6: Calculate the Partial Pressures of CO and CO2
Using the total pressure of \( 1 \, atm \):
- Partial pressure of CO:
\[
P_{CO} = X_{CO} \times P_{total} = 0.938 \times 1 \, atm = 0.938 \, atm
\]
- Partial pressure of CO2:
\[
P_{CO2} = X_{CO2} \times P_{total} = 0.062 \times 1 \, atm = 0.062 \, atm
\]
### Step 7: Calculate \( K_p \)
Using the formula for \( K_p \):
\[
K_p = \frac{(P_{CO})^2}{P_{CO2}} = \frac{(0.938)^2}{0.062} = \frac{0.879}{0.062} \approx 14.19
\]
### Step 8: Calculate \( K_c \)
Using the relationship between \( K_p \) and \( K_c \):
\[
K_c = K_p \cdot R^{\Delta n} \cdot T
\]
Where:
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- \( T = 1127 \, K \)
- \( \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 1 = 1 \)
Calculating \( K_c \):
\[
K_c = 14.19 \cdot 0.0821^{1} \cdot 1127
\]
Calculating \( K_c \):
\[
K_c = 14.19 \cdot 0.0821 \cdot 1127 \approx 0.1533
\]
### Final Answer
The value of \( K_c \) at \( 1127 \, K \) is approximately \( 0.1533 \).
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