We are given the reaction:

$\text{C(s)} + 2\text{H}_2(g) \rightleftharpoons \text{CH}_4(g)$

with an equilibrium constant $K_p = 0.643$ and the equilibrium partial pressure of methane, $P_{\text{CH}_4} = 6.73 \, \text{atm}$. We need to find the equilibrium partial pressure of hydrogen, $P_{\text{H}_2}$.

The equilibrium expression for the reaction is given by:

$K_p = \frac{P_{\text{CH}_4}}{P_{\text{H}_2}^2}$

Substitute the known values into the equation:

$0.643 = \frac{6.73}{P_{\text{H}_2}^2}$

Now, solve for $P_{\text{H}_2}$:

$P_{\text{H}_2}^2 = \frac{6.73}{0.643}$

$P_{\text{H}_2}^2 = 10.466$

Take the square root of both sides:

$P_{\text{H}_2} = \sqrt{10.466}$

$P_{\text{H}_2} \approx 3.24 \, \text{atm}$

So, the equilibrium partial pressure of $\text{H}_2$ is approximately $3.24 \, \text{atm}$.

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Would you like any further details or clarifications?

Here are 5 related questions to expand your understanding:

1. How does the value of $K_p$ change with temperature?
2. What would happen to the equilibrium if more hydrogen gas was added?
3. How is $K_p$ related to $K_c$ for this reaction?
4. Why is the pressure of solid carbon not included in the equilibrium expression?
5. How would the equilibrium shift if the volume of the container was decreased?

Tip: Always remember to check the units of pressure and ensure they are consistent with the equilibrium constant $K_p$ used in your calculations.