Answer:
1.6 × 10⁻⁴ atm
Explanation:
Let's consider the following reaction at equilibrium.
CO(g) + 2 H₂(g) ⇄ CH₃OH(l)
The pressure equilibrium constant (Kp) is the product of the equilibrium pressures of the products raised to their stoichiometric coefficients divided by the product of the equilibrium pressures of the reactants raised to their stoichiometric coefficients.
Kp = 1 / (pCo)eq × (pH₂)eq²
(pCo)eq = 1 / Kp × (pH₂)eq²
(pCo)eq = 1 / 2.25 × 10⁴ × 0.52²
(pCo)eq = 1.6 × 10⁻⁴ atm
Answer:
[tex]1.6x10^{-4}atm[/tex]
Explanation:
Hello,
In this case, to clarify the situation, the undergoing chemical reaction is:
[tex]CO(g)+2H_2(g)\leftrightarrow CH_3OH(l)[/tex]
Thus, since both the pressure equilibrium constant and the partial pressure of hydrogen are given, one writes the law of mass action excluding methanol because it is liquid as shown below:
[tex]Kp=\frac{1}{p_{H_2}^2p_{CO}}=2.25x10^{4}[/tex]
Finally, solving for the carbon monoxide's partial pressure at equilibrium, one obtains:
[tex]p_{CO}=\frac{1}{p_{H_2}^2*Kp}=\frac{1}{(0.52atm)^2*2.25x10^4/atm^3} =1.6x10^{-4}atm[/tex]
Best regards.
