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.