A2 Vapour-liquid equilibrium behaviour for mixtures of 1-propanol and water carn

Question

A2 Vapour-liquid equilibrium behaviour for mixtures of 1-propanol and water carn be roughly approximated using the Margules activity model with the Margules parameters A12 2.248 and A21 1.019, where '1 is propanol and '2' is water. Antoine coefficient data for water and 1-propanol are provided in Table A2. It is known that an azeotrope exists for this binary system at atmospheric pressure Show that this azeotrope occurs at a 1-propanol mole fraction of 0.436 and determine the temperature at which the azeotrope occurs. [16 marks] Table A2 - Antoine coefficient data (pressure in kPa, temperature in oC_logarithm to base e) Water 1-propanol 16.3872 16.1154 3885.7 3483.67 230.170 205.807

Explanation / Answer

Margules equation

ln ?1=x22(A12+2(A21-A12)*x1)

ln ?2=x12(A21+2(A12-A21)*x2)

So

x1=0.436

x2=0.564

ln ?1=0.5642*(2.248+2(1.019-2.248)0.436)

?1=1.453

ln ?2=0.4362(1.019+2(2.248-1.019)0.564)

?2=1.58

Now we know that relative volatility

K1=y1/x1=?1P1sat/P

K2=y2/x2=?2P2sat/P

Now for azeotrope

y=x

K=1

So

?1P1sat/P=?2P2sat/P

So

?1*P1sat=?2*P2sat

So from antoine equation

ln P1sat=16.1154 - 3483.67/(T+205.807)

ln P2sat=16.3872-3885.7/(T+230.170)

So

ln ?1+ln P1sat=ln ?2+ln P2sat

ln 1.453+16.1154 - 3483.67/(T+205.807)=ln 1.58 +16.3872-3885.7/(T+230.170)

3885.7/(T+230.170)- 3483.67/(T+205.807)=0.36

0.36T2-245.08T+1726885.48=0

Solving this equation we get

T=340.38 0C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.