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.

Options

A. a. 1.53
B. b. 0.153
C. c. 0.53
D. d. 0.76

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, 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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