test

(1)
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(2)
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(3) Styrene is hydrogenated to ethyl benzene in the gas phase at 600 K and 1 bar with a reactor feed ratio of 3 moles H2 per 1 mole styrene. What is the equilibrium composition of the three components?

C6H5CH=CH2 + H2 ( C6H5CH2CH3

ΔGf° = -19.84 kcal/mol ΔHf° = -29.5 kcal/mol R = 1.987 kcal / (mol K)

ln(K298) = -ΔGf°/RT ( K298 = exp(-ΔGf°/RT) = exp(19.84 /(1.987x10-3)/298) = 3.562x1014

K600 = K298 (-ΔΗf°(1/T – 1/298)/R) = 3.562x1014 (29.5(1/600 – 1/298)/(1.987x10-3)) = 4583

νetb=1, νsty=-1, νh2=-1, ν=-1 nh20 = 3 nsty0 = 1

ysty = (1 – ε) / (4 – ε) yh2 = (3 – ε) / (4 – ε) yetb = ε / (4 – ε)

K = yetb / (yh2 ysty) = [ ε / (4 – ε)] / [(1 – ε) / (4 – ε) *(3 – ε) / (4 – ε)] = [ε(4–ε)]/[(1–ε) (3–ε)]

3K -4Kε + Κε2 -4ε + ε2 = 0

(K+1)ε2 – 4(K+1)ε + 3K = 0 ( (4584)ε2 – 4(4584)ε + 3(4583) = 0

ε ’ 0.9997

ysty = 0.0001 0.162 0.160 yh2 = 0.6667 OR 0.762 OR 0.160 yetb = 0.3332 0.162 0.160 if you used incorrect R-value printed on exam if you switched DG and DH (since DH not labeled correctly)

(4) In the absence of a selective catalyst, the synthesis of methanol can be accompanied by a side reaction in which coke (solid carbon) is formed (see below). If 1 mol CO and 2 mol H2 are fed to a reactor at 298 K and 0.5 bar, approximately how many moles of coke will form?

CO(g) + 2H2(g) ( CH3OH(g) 2CO(g) ( CO2(g) + C(s)

Compound ΔGf° ΔΗf° (kJ/mol) CO -137.2 -110.53 MeOH -162.2 -200.9 CO2 -394.4 -393.5 H2O -228.6 -241.9

Free energies of formation of pure elements are taken as the reference state, hence ΔGf°H2 = 0 and ΔGf°C(s) = 0

ΔG1 = ΔGf°MeOH – 2ΔGf°H2 – ΔGf°CO = -162.2 -2(0) –(-137.2) = -25 kJ/mol ΔG2 = ΔGf°C + ΔGf°CO2 – 2ΔGf°CO = 0 + (-394.4) -2 (-137.2) = -120 kJ/mol

| | ΔG | K |
| Reaction 1 |-25 |24113.43 |
| Reaction 2 |-120 |1.08E+21 |

I did give a lot of partial credit to those who set up the full calculation correctly, but didn’t reach the answer. But a detailed calculation is not needed to obtain the conversions. The point of this problem is that you need to think carefully about the K values before beginning to try and solve.

These reactions occur in parallel and they both have large K values. Even though K2 is much larger than K1, each of these reactions is going to go essentially to completion. For the initial 1 mole CO, ½ mole is converted in reaction 1 and ½ mole is converted in reaction 2. For reaction 2:

1/2 mole CO * (1 mole C / 2 mole CO) = 0.25 mole C produced
(5) Ethylene is produced from ethanol by the reaction below. A colleague has measured the compositions (mole fractions) of the reactants and products at two temperatures for the reaction carried out at 1 bar. Using this information, estimate the enthalpy of the reaction, ΔH°.

C2H5OH ( C2H4 + H20

T (K) 298 423 yC2H4 0.333 0.489 yH2O 0.333 0.489

Enthalpy can be estimated from van’t Hoff relationship if the K values are known at two temperatures.

Find the K values first.

K298 = (yH20 yC2H4) / yEtOH = (0.333)(0.333)/0.333 = 0.333

K423 = (yH20 yC2H4) / yEtOH = (0.489)(0.489)/(1-0.489-0.489) = 10.9

ln(K / K’) = - ΔH (1/T – 1/T’)/R ΔH = -Rln(K/K’)/(1/T – 1/T’)= -Rln(0.333 / 10.9)(8.314x10-3) / (1/298 -1/423) =

ΔH = 29.2 kJ/mol