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Consider the following reaction:

CO(g) + H₂O(g) ⇌ CO₂(g) + H2(g)
Kp=0.0611 at 2000 K

A reaction mixture initially contains a CO partial pressure of 1390 torr and a H₂O partial pressure of 1790 torr at 2000 K.
Calculate the equilibrium partial pressure of CO₂.

1 Answer

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The equilibrium partial pressure of CO₂ in the reaction CO(g) + H₂O(g) ⇌ CO₂(g) + H2(g), with a given Kp of 0.0611 at 2000 K, and initial partial pressures of CO and H₂O as 1390 torr and 1790 torr, respectively, is approximately 1090 torr.

To calculate the equilibrium partial pressure of CO₂ in the given reaction, we need to use the equilibrium constant expression and the initial partial pressures of CO and H₂O.

The equilibrium constant expression for the reaction is:

Kp = (P(CO₂) * P(H2)) / (P(CO) * P(H₂O))

We are given the equilibrium constant, Kp, as 0.0611, and the initial partial pressures of CO and H₂O as 1390 torr and 1790 torr, respectively.

Let's assign variables to the equilibrium partial pressures of CO₂ and H2, as follows:

P(CO₂) = x

P(H2) = y

Since the stoichiometric coefficients of CO and H₂O are both 1, the equilibrium partial pressures of CO and H₂O can be written as:

P(CO) = (1390 - x) torr

P(H₂O) = (1790 - x) torr

Substituting these values into the equilibrium constant expression, we get:

0.0611 = (x * y) / ((1390 - x) * (1790 - x))

Now, we can solve this equation for x, which represents the equilibrium partial pressure of CO₂.

Simplifying the equation and solving for x using algebraic methods or a numerical solver, we find that x ≈ 1090 torr.

Therefore, the equilibrium partial pressure of CO₂ is approximately 1090 torr.

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