Question

while ethanol (ch₃ch₂oh) is produced naturally by fermentation, e.g. in beer - and wine - making, industrially it is synthesized by reacting ethylene (ch₂ch₂) with water vapor at elevated temperatures.
a chemical engineer studying this reaction fills a 5.0 l flask with 3.1 atm of ethylene gas and 3.7 atm of water vapor. when the mixture has come to equilibrium she determines that it contains 1.8 atm of ethylene gas and 2.4 atm of water vapor.
the engineer then adds another 0.78 atm of ethylene, and allows the mixture to come to equilibrium again. calculate the pressure of ethanol after equilibrium is reached the second time. round your answer to 2 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The reaction is \(CH_{2}=CH_{2}(g)+H_{2}O(g)
ightleftharpoons CH_{3}CH_{2}OH(g)\)

Step2: Calculate \(K_{p}\) for the first - equilibrium

At the first equilibrium, \(P_{C_{2}H_{4}} = 1.8\ atm\), \(P_{H_{2}O}=2.4\ atm\), and \(P_{C_{2}H_{5}OH}=(3.1 - 1.8)=(3.7 - 2.4)=1.3\ atm\)
Using the formula \(K_{p}=\frac{P_{C_{2}H_{5}OH}}{P_{C_{2}H_{4}}\times P_{H_{2}O}}\)
\(K_{p}=\frac{1.3}{1.8\times2.4}\)
\(K_{p}=\frac{1.3}{4.32}\approx0.301\)

Step3: Set up the ICE table for the second - equilibrium

After adding \(0.78\ atm\) of \(C_{2}H_{4}\), the initial pressures are \(P_{C_{2}H_{4}}^{initial}=(1.8 + 0.78)=2.58\ atm\), \(P_{H_{2}O}^{initial}=2.4\ atm\), and \(P_{C_{2}H_{5}OH}^{initial}=1.3\ atm\)
Let \(x\) be the change in pressure of \(C_{2}H_{4}\) (and also of \(H_{2}O\)) as the reaction shifts to re - establish equilibrium.
The equilibrium pressures are \(P_{C_{2}H_{4}}=(2.58 - x)\ atm\), \(P_{H_{2}O}=(2.4 - x)\ atm\), and \(P_{C_{2}H_{5}OH}=(1.3 + x)\ atm\)

Step4: Substitute into the \(K_{p}\) expression

Since \(K_{p}=\frac{P_{C_{2}H_{5}OH}}{P_{C_{2}H_{4}}\times P_{H_{2}O}}\) and \(K_{p}=0.301\)
\(0.301=\frac{1.3 + x}{(2.58 - x)(2.4 - x)}\)
Expand the denominator: \((2.58 - x)(2.4 - x)=6.192-2.58x - 2.4x+x^{2}=6.192 - 4.98x+x^{2}\)
The equation becomes \(0.301(6.192 - 4.98x+x^{2})=1.3 + x\)
\(1.863792-1.49898x + 0.301x^{2}=1.3 + x\)
\(0.301x^{2}-2.49898x + 0.563792 = 0\)
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 0.301\), \(b=-2.49898\), \(c = 0.563792\)
\(x=\frac{2.49898\pm\sqrt{(-2.49898)^{2}-4\times0.301\times0.563792}}{2\times0.301}\)
\(x=\frac{2.49898\pm\sqrt{6.2449 - 0.677}}{0.602}\)
\(x=\frac{2.49898\pm\sqrt{5.5679}}{0.602}\)
\(x=\frac{2.49898\pm2.36}{0.602}\)
We take the smaller root (because if we take the larger root, the pressures of \(C_{2}H_{4}\) and \(H_{2}O\) will be negative).
\(x=\frac{2.49898 - 2.36}{0.602}=\frac{0.13898}{0.602}\approx0.231\)

Step5: Calculate the pressure of ethanol at the second - equilibrium

\(P_{C_{2}H_{5}OH}=(1.3 + x)\)
Substitute \(x = 0.231\)
\(P_{C_{2}H_{5}OH}=1.3+0.231 = 1.531\approx1.5\ atm\)

Answer:

\(1.5\ atm\)