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 200. ml flask with 4.7 atm of carbon monoxide gas and 1.2 atm of water vapor. when the mixture has come to equilibrium she determines that it contains 4.01 atm of carbon monoxide gas, 0.51 atm of water vapor and 0.69 atm of carbon dioxide. the engineer then adds another 2.4 atm of carbon monoxide, and allows the mixture to come to equilibrium again. calculate the pressure of hydrogen after equilibrium is reached the second time. round your answer to 2 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The reaction is \(CO(g)+H_2O(g)
ightleftharpoons CO_2(g)+H_2(g)\)

Step2: Calculate \(K_p\) for the first - equilibrium

Using the formula \(K_p=\frac{P_{CO_2}\times P_{H_2}}{P_{CO}\times P_{H_2O}}\). At the first - equilibrium, \(P_{CO}=4.01\ atm\), \(P_{H_2O}=0.51\ atm\), \(P_{CO_2}=0.69\ atm\), and \(P_{H_2}\) (from the first - equilibrium) can be found from the stoichiometry of the reaction.
Since for the reaction \(CO(g)+H_2O(g)
ightleftharpoons CO_2(g)+H_2(g)\), the change in pressure of \(CO\) is \(\Delta P_{CO}=4.7 - 4.01=0.69\ atm\). By stoichiometry, \(P_{H_2}=0.69\ atm\) (because the mole ratio of \(CO\) to \(H_2\) is \(1:1\) in the reaction).
Then \(K_p=\frac{0.69\times0.69}{4.01\times0.51}=\frac{0.4761}{2.0451}\approx0.233\)

Step3: Set up the second - equilibrium calculation

After adding \(2.4\ atm\) of \(CO\), the initial pressure of \(CO\) is \(P_{CO}^{initial}=4.01 + 2.4=6.41\ atm\), \(P_{H_2O}=0.51\ atm\), \(P_{CO_2}=0.69\ atm\), and let the change in pressure of \(CO\) be \(-x\). Then the equilibrium pressures are \(P_{CO}=6.41 - x\), \(P_{H_2O}=0.51 - x\), \(P_{CO_2}=0.69 + x\), \(P_{H_2}=0.69 + x\)
Substitute into the \(K_p\) formula: \(K_p = 0.233=\frac{(0.69 + x)(0.69 + x)}{(6.41 - x)(0.51 - x)}\)
\(0.233=\frac{(0.69 + x)^2}{(6.41 - x)(0.51 - x)}\)
Expand the denominator: \((6.41 - x)(0.51 - x)=6.41\times0.51-6.41x-0.51x + x^{2}=3.2691-6.92x + x^{2}\)
Expand the numerator: \((0.69 + x)^2=0.4761 + 1.38x+x^{2}\)
So, \(0.233(3.2691-6.92x + x^{2})=0.4761 + 1.38x+x^{2}\)
\(0.7617 - 1.612x+0.233x^{2}=0.4761 + 1.38x+x^{2}\)
\(x^{2}-0.233x^{2}+1.38x + 1.612x+0.4761 - 0.7617 = 0\)
\(0.767x^{2}+2.992x - 0.2856 = 0\)
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 0.767\), \(b = 2.992\), \(c=- 0.2856\)
\(x=\frac{-2.992\pm\sqrt{(2.992)^{2}-4\times0.767\times(-0.2856)}}{2\times0.767}=\frac{-2.992\pm\sqrt{8.952+0.877}}{1.534}=\frac{-2.992\pm\sqrt{9.829}}{1.534}=\frac{-2.992\pm3.135}{1.534}\)
We take the positive root \(x=\frac{-2.992 + 3.135}{1.534}=\frac{0.143}{1.534}\approx0.0932\)

Step4: Calculate the pressure of \(H_2\) at the second - equilibrium

\(P_{H_2}=0.69+x\)
\(P_{H_2}=0.69 + 0.0932=0.7832\approx0.78\ atm\)

Answer:

\(0.78\ atm\)