Question 6.4.2: Raoult’s Law and Henry’s Law Use either Raoult’s law or Henr......

1. Hydrocarbons normally are relatively insoluble in water, so that the solution of ethane is probably extremely dilute. Let us therefore apply Henry’s law. Page 2 130 of Perry’s Chemical Engineers’ Handbook (see Footnote 1) gives the Henry’s law constant for ethane in water at 25°C as 2.67 × 10^{4} atm/mole fraction.^{5}From
Equation 6.4-2

Henry’s Law:             p_{A} \equiv \boxed{y_{A} P = x_{A} H_{A}(T)}               (6.4-2)

x_{C_{2}H_{6}} = \frac{y_{C_{2}H_{6}}  P}{H_{C_{2}H_{6}}} = \frac{(0.0100)(20.0  atm)}{2.67 \times 10^{4}  atm/mole  fraction} = \boxed{7.49 \times 10^{-6}  \frac{mol  C_{2}H_{6}}{mol}}

2. Since benzene and toluene are structurally similar compounds, we may apply Raoult’s law. From Table B.4,

\log_{10} p^{*}_{B} = 6.906  –  \frac{1211}{T+220.8} \overset{T= 30°C}{\Longrightarrow } p^{*}_{B} = 119  mm  Hg \\ \log_{10} p^{*}_{T} = 6.9533  –  \frac{1343.9}{T + 219.38}\overset{T= 30°C}{\Longrightarrow } p^{*}_{T} = 36.7  mm  Hg

(The values of p^{*}_{B} and p^{*}_{T} could have been obtained using the AntoineP function of APEx.)

Using Equation 6.4-1,

Raoult’s Law:             p_{A}\equiv \boxed{y_{A} P = x_{A}p^{*}_{A}(T)}           (6.4-1)

p_{B}= x_{B}p^{*}_{B}= (0.500 )(119  mm  Hg) = 59.5  mm  Hg \\ p_{T} = x_{T}p^{*}_{T}= (0.500 )(36.7  mm  Hg) = 18.35  mm  Hg \\ P= p_{B} +p_{T}= \boxed{77.9  mm  Hg} \\ y_{B} = p_{B}/P = \boxed{0.764  mole  benzene/mole} \\ y_{T} = p_{T}/P = \boxed{0.236  mole  toluene/mole}

^{5} The uncertainty associated with Henry’s law constants is illustrated by the fact that on page 2-130 of Perry’s Chemical Engineers’ Handbook (see Footnote 1), two different values for ethane in water at 25°C are given: Table 2-123 gives H = 2.94 × 10^{4} atm/mole fraction while Table 2-124 gives H = 2.67 × 10^{4} atm/mole fraction, a 9% difference.