Question

"Synthesis gas" is a mixture of carbon monoxide and water vapor. At high temperature synthesis gas will form carbon dioxide and hydrogen, and in fact this reaction is one of the ways hydrogen is made industrially. A chemical engineer studying this reaction fills a 1.5 L flask at 8.3 °C with 2.6 atm of carbon monoxide gas and 3.4 atm of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 1.88 atm of carbon monoxide gas, 2.68 atm of water vapor and 0.72 atm of hydrogen gas The engineer then adds another 1.1 atm of water, and allows the mixture to come to equilibrium again. Calculate the pressure of carbon dioxide after equilibrium is reached the second time. Round your answer to 2 significant digits. atm x10

Answer 1

Answer:

8.2x10⁻² atm

Explanation:

For the reversible reaction betwenn carbon monoxide and water vapor, let's do an equilibrium chart:

CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)

2.6 atm 3.4 atm 0 0 Initial

1.88 atm 2.68 atm x 0.72 atm Equilibrium

So, 0.72 atm of CO and H₂O had reacted. Because of the stoichiometry of the reaction is 1:1:1:1, x = 0.72 atm.

The constant of equilibrium based on pressure (Kp) can be calculated by the partial pressure of the products elevated by their coefficients, divided by the partial pressure of the reactants elevated by their coefficients.

Kp = (pCO₂ * pH₂)/(pCO * pH₂O)

Kp = (0.72 * 0.72)/(1.88 * 2.68)

Kp = 0.1030

Kp only change with a change in temperature. The, when more water vapor is added, the equilibrium will shift for the right, to compensate the disturb. The new equilibrium chart will be:

CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)

1.88 2.68 0.72 0.72 Initial

+ 1.1 Added

1.88 3.78 0.72 0.72 New partial pressures

-x -x +x +x React

1.88-x 3.78-x 0.72+x 0.72+x Equilibrium

Kp = [(0.72+x)*(0.72+x)]/[(1.88-x)*(3.78-x)]

0.1030 = (0.5184 + 1.44x + x²)/(7.1064 - 5.66x + x²)

0.5184 + 1.44x + x² = 0.1030*(7.1064 - 5.66x + x²)

0.5184 + 1.44x + x² = 0.1030x² - 0.58298x + 0.7319592

0.8970x² + 2.02298x - 0.2135592 = 0

Solving by a graphic calculator, and 0 < x < 1.88, x = 0.10104

pCO₂ = 0.72 + 0.10104 = 0.82104 atm

pCO₂ = 8.2x10⁻² atm

Answer 2

The pressure of carbon dioxide after equilibrium is reached the second time is 8.2x10⁻² atm.

What is "Synthesis gas"?

Syngas, sometimes known as synthesis gas, is a fuel gas mixture mostly composed of hydrogen, carbon monoxide, and, in rare cases, carbon dioxide.

The reaction is

$\rm CO(g) + H_2O(g) <---> CO_2(g) + H_2(g)$

Calculating the Kp

The constant of equilibrium based on pressure (Kp) is calculated by the partial pressure of the products elevated by their coefficients divided by the partial pressure of the reactants increased by their coefficients yields.

The partial pressure of the product is 0.72 atm

The partial pressure of the reactants are:

1.88 atm of carbon monoxide gas

2.68 atm of water vapor

$Kp =\dfrac{0.72 \times0.72}{1.88 \times 2.68} = 0.1030$

Kp = 0.1030

Kp changes solely when the temperature changes. The equilibrium will shift to the right when more water vapor is introduced, compensating for the disturbance.

Before the water vapor is 2.68 atm.

When 1.1 atm. of water added

The new water vapor is 3.7 atm.

The new equilibrium diagram will look like this:

$1.88-x 3.78-x 0.72+x 0.72+x\;\;Equilibrium$

$Kp = [(0.72+x)\times \dfrac{(0.72+x)}{[(1.88-x)*(3.78-x)]} ]$

$0.1030 = (0.5184 + 1.44x + x^2)/(7.1064 - 5.66x + x^2)$

$0.5184 + 1.44x + x^2 = 0.1030\times(7.1064 - 5.66x + x^2)$

$0.5184 + 1.44x + x^2= 0.1030x^2 - 0.58298x + 0.7319592$

$0.8970x^2 + 2.02298x - 0.2135592 = 0$

$\rm pCO_2 = 0.72 + 0.10104 = 0.82104\; atm$

pCO₂ = 8.2x10⁻² atm.

Thus, the pressure of carbon dioxide after equilibrium is reached the second time is 8.2x10⁻² atm.

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