Question

ethylene (ch₂ch₂) is the starting point for a wide array of industrial chemical syntheses. for example, worldwide about 8.0×10¹⁰ kg of polyethylene are made from ethylene each year, for use in everything from household plumbing to artificial joints. natural sources of ethylene are entirely inadequate to meet world demand, so ethane (ch₃ch₃) from natural gas is \cracked\ in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas.
suppose an engineer studying ethane cracking fills a 60.0 l reaction tank with 40.0 atm of ethane gas and raises the temperature to 900. °c. she believes kₚ = 0.050 at this temperature.
calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture. round your answer to 2 significant digits.
note for advanced students: the engineer may be mistaken about the correct value of kₚ, and the mass percent of ethylene you calculate may not be what she actually observes.

Explanation:

Step1: Write the balanced chemical equation

The cracking of ethane \(CH_3CH_3(g)
ightarrow CH_2CH_2(g)+H_2(g)\)
Let the change in partial pressure of \(CH_3CH_3\) be \(-x\) atm. Then, at equilibrium, \(P_{CH_3CH_3}=(40 - x)\) atm, \(P_{CH_2CH_2}=x\) atm, and \(P_{H_2}=x\) atm.

Step2: Write the expression for \(K_p\)

\(K_p=\frac{P_{CH_2CH_2}\times P_{H_2}}{P_{CH_3CH_3}}\)
Substitute the equilibrium partial pressures into the \(K_p\) expression: \(0.050=\frac{x\times x}{40 - x}\)

Step3: Solve the quadratic equation

\(0.050(40 - x)=x^{2}\)
\(2-0.05x=x^{2}\)
\(x^{2}+0.05x - 2=0\)
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 1\), \(b=0.05\), and \(c=- 2\)
\(x=\frac{-0.05\pm\sqrt{(0.05)^{2}-4\times1\times(-2)}}{2\times1}=\frac{-0.05\pm\sqrt{0.0025 + 8}}{2}=\frac{-0.05\pm\sqrt{8.0025}}{2}=\frac{-0.05\pm2.83}{2}\)
We take the positive root \(x=\frac{- 0.05+2.83}{2}=\frac{2.78}{2}=1.39\) atm (we discard the negative root because pressure cannot be negative in this context)

Step4: Calculate the moles of each gas using the ideal gas law (\(PV = nRT\), \(T=(900 + 273)K=1173K\), \(R = 0.0821\frac{L\cdot atm}{mol\cdot K}\))

For \(CH_3CH_3\): \(n_{CH_3CH_3}=\frac{P_{CH_3CH_3}V}{RT}=\frac{(40 - 1.39)\times60}{0.0821\times1173}=\frac{38.61\times60}{0.0821\times1173}\)
\(n_{CH_3CH_3}=\frac{2316.6}{96.3}\approx24.06\) mol
Molar mass of \(CH_3CH_3,M_{CH_3CH_3}=(2\times12 + 6\times1)g/mol = 30g/mol\), mass of \(CH_3CH_3,m_{CH_3CH_3}=n_{CH_3CH_3}\times M_{CH_3CH_3}=24.06\times30 = 721.8g\)
For \(CH_2CH_2\): \(n_{CH_2CH_2}=\frac{P_{CH_2CH_2}V}{RT}=\frac{1.39\times60}{0.0821\times1173}=\frac{83.4}{96.3}\approx0.866\) mol
Molar mass of \(CH_2CH_2,M_{CH_2CH_2}=(2\times12+4\times1)g/mol = 28g/mol\), mass of \(CH_2CH_2,m_{CH_2CH_2}=n_{CH_2CH_2}\times M_{CH_2CH_2}=0.866\times28 = 24.25g\)
For \(H_2\): \(n_{H_2}=\frac{P_{H_2}V}{RT}=\frac{1.39\times60}{0.0821\times1173}\approx0.866\) mol, molar mass of \(H_2,M_{H_2}=2g/mol\), mass of \(H_2,m_{H_2}=0.866\times2 = 1.73g\)

Step5: Calculate the mass percent of \(CH_2CH_2\)

Total mass \(m_{total}=m_{CH_3CH_3}+m_{CH_2CH_2}+m_{H_2}=721.8 + 24.25+1.73=747.78g\)
Mass percent of \(CH_2CH_2=\frac{m_{CH_2CH_2}}{m_{total}}\times100=\frac{24.25}{747.78}\times100\approx3.2\%\)

Answer:

\(3.2\%\)