For the following reaction, KP = 0.455 at 945°C: At equilibrium, is 1.78 atm. What is the equilibrium partial pressure of CH4 in atm?

See explanation below

Explanation:

The question is incomplete. However, here's the missing part of the question:

"For the following reaction, Kp = 0.455 at 945 °C:

C (s) + 2H2 (g) CH4 (g).

At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4 (g) ?"

With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.

The expression of Kp for this reaction is:

Kp = PpCH4 / (PpH2) ²

We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:

PpCH4 = Kp * PpH2²

*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.

Now solving for PpCH4:

PpCH4 = 0.455 * (1.78) ²

PpCH4 = 1.44 atm