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To find the equilibrium constants \( K_c \) and \( K_p \) for the given reaction:

\[
\text{CH}_4(g) + \text{H}_2\text{O}(g) \leftrightarrow \text{CO}(g) + 3\text{H}_2(g)
\]

we'll proceed as follows:

1. **Convert the masses of each substance to moles** using their molar masses:
   - Molar mass of \( \text{CO} \) = 28.01 g/mol
   - Molar mass of \( \text{H}_2 \) = 2.02 g/mol
   - Molar mass of \( \text{CH}_4 \) = 16.04 g/mol
   - Molar mass of \( \text{H}_2\text{O} \) = 18.02 g/mol

2. **Calculate the molar concentrations** by dividing the number of moles by the volume of the flask (5.00 L).

3. **Determine \( K_c \)** using the expression for the equilibrium constant based on concentrations:

   \[
   K_c = \frac{[\text{CO}][\text{H}_2]^3}{[\text{CH}_4][\text{H}_2\text{O}]}
   \]

4. **Calculate \( K_p \)** using the relationship between \( K_c \) and \( K_p \):

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

   where \( R \) is the gas constant (0.0821 L·atm/(mol·K)), \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in the number of moles of gas (\( 3+1-1-1=2 \)).

### Step 1: Convert masses to moles
- Moles of \( \text{CO} = \frac{8.62 \text{g}}{28.01 \text{g/mol}} \approx 0.3078 \text{mol} \)
- Moles of \( \text{H}_2 = \frac{2.60 \text{g}}{2.02 \text{g/mol}} \approx 1.2871 \text{mol} \)
- Moles of \( \text{CH}_4 = \frac{43.0 \text{g}}{16.04 \text{g/mol}} \approx 2.6808 \text{mol} \)
- Moles of \( \text{H}_2\text{O} = \frac{48.4 \text{g}}{18.02 \text{g/mol}} \approx 2.6864 \text{mol} \)

### Step 2: Calculate molar concentrations
- \( [\text{CO}] = \frac{0.3078 \text{mol}}{5.00 \text{L}} = 0.0616 \text{M} \)
- \( [\text{H}_2] = \frac{1.2871 \text{mol}}{5.00 \text{L}} = 0.2574 \text{M} \)
- \( [\text{CH}_4] = \frac{2.6808 \text{mol}}{5.00 \text{L}} = 0.5362 \text{M} \)
- \( [\text{H}_2\text{O}] = \frac{2.6864 \text{mol}}{5.00 \text{L}} = 0.5373 \text{M} \)

### Step 3: Calculate \( K_c \)
\[
K_c = \frac{(0.0616)(0.2574)^3}{(0.5362)(0.5373)}
\]
Calculating this step-by-step:

- \( (0.2574)^3 = 0.0171 \)
- Numerator: \( 0.0616 \times 0.0171 = 0.001053 \)
- Denominator: \( 0.5362 \times 0.5373 = 0.2881 \)

Thus, \( K_c = \frac{0.001053}{0.2881} \approx 0.00365 \).

### Step 4: Calculate \( K_p \)
First, convert temperature to Kelvin: \( T = 320 + 273.15 = 593.15 \text{K} \).

Now calculate \( K_p \):
\[
K_p = 0.00365 \times (0.0821 \times 593.15)^2
\]
- \( 0.0821 \times 593.15 = 48.69 \)
- \( 48.69^2 = 2370.38 \)

Therefore, \( K_p = 0.00365 \times 2370.38 \approx 8.65 \).

### Summary
- \( K_c \approx 0.00365 \)
- \( K_p \approx 8.65 \)

Would you like further details, or do you have any questions?

### Related Questions:
1. How do changes in temperature affect \( K_c \) and \( K_p \)?
2. What would happen to the equilibrium position if the volume of the flask was increased?
3. How would adding an inert gas at constant volume affect \( K_p \) and \( K_c \)?
4. Why do we use Kelvin for temperature in the \( K_p \) calculation?
5. How does the value of \( \Delta n \) influence the relationship between \( K_c \) and \( K_p \)?

### Tip:
Always ensure that temperature is converted to Kelvin when using gas laws or equilibrium expressions involving gases.

Consider the equilibrium system described by the chemical reaction below. At equilibrium, a sample of gas from the system is collected into 5.00 L flask. The flask is found to contain 8.62 g of CO, 2.60 g of H2, 43.0 g of CH4, and 48.4 g of H2O at 320.0 °C. What are the values of Kc and Kp for this reaction?

Find the equilibrium constants Kc and Kp for the reaction CH4 + H2O ⇌ CO + 3H2 at 320 °C. This detailed solution covers stoichiometry, chemical equilibrium, and gas laws, suitable for college-level chemistry.

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