The given question is incomplete, the complete question is:
Calculating an equilibrium constant from a partial equilibrium... Steam reforming of methane (CH) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 25.0L tank with 8.0 mol of methane gas and 1.9 mol of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be 1.5 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits.
Answer:
The correct answer is 2.47.
Step-by-step explanation:
Based on the given information, the equation for the synthesis gas is,
CH₄ (g) + H₂O (g) ⇔ CO (g) + 3H₂ (g)
Based on the given information, 25.0 L is the volume of the tank, the concentration of CH₄ is 8.0 mol, the concentration of water vapor is 1.9 mol, and the concentration of CO gas is 1.5 mol.
Therefore, 25 L of the solution comprise 8.0 mole of CH₄. So, 1 L of the solution will comprise 8.0 / 25 mole CH₄,
= 0.32 mole of CH₄
Thus, the concentration of CH₄ or [CH₄] will be 0.32 mole/L or 0.32 M.
Similarly, the concentration of H₂O or [H₂O] will be 1.9/25 = 0.076 M
and [CO] is 1.5/25 = 0.06 M
The concentration equilibrium constant for the steam will be,
Kc = [CO] pH₂ / [CH₄] [H₂O] (Here pH₂ is the partial pressure of H₂)
Now lets us assume that the reaction has taken place in a constant atmospheric pressure, therefore, pH₂ will be equal to 1.
= 0.06 M/0.32 M × 0.076 M
= 2.47