Chemistry 75 Winter, 2017Problem Set 5 Solutions

5.1 We start with the expression for the slope and manipulate it as follows:

dPdT vap

= ΔSvap

o

ΔVvapo

= ΔSvap

o

V(g) – V(l) ≅

ΔSvapo

V(g) =

ΔSvapo P

RTvap

We rearrange this expression and integrate it:

dPPPi

Pf =

ΔSvapo

R dT

TTi

Tf = ln Pf

Pi =

ΔSvapo

R ln Tf

Ti or Pf = Pi

TfTi

ΔSvapo /R

We find the vaporization entropy from data we’re given:

ΔSvapo =

ΔHvapo

Tvapo

= 40 660 J mol–1

373.15 K = 108.96 J mol–1 K–1 = 13.015 R

We can now construct a table using our new expression, taking Ti = 373.15 K, thenormal boiling point, and Pi = 1.00 atm.

Tf /K Pobsatm Pcalc/atm (Pcalc – Pobs)/Pobs

647.27 218.2 1364 525.1%500. 26.1 46.3 77.4%400. 2.43 2.49 2.3%350. 0.411 0.432 5.1%300. 0.0349 0.0573 64.2%

273.15 0.00603 0.0168 178.1%

The agreement is worse and worse the farther we are from the normal boiling point.

5.2 Let’s start off considering the net reaction in terms of all gaseous species. We’re toldthat some I2 is liquid, but some will be gaseous ensuring that the partial pressure ofI2(g) must be the liquid’s vapor pressure, 0.392 bar. Thus, we can write (and wecould be using our degree of advancement variable ξ here, but we don’t need to)

Ptot = PH2 + PI2 + PHI = PH2 + 0.392 bar + PHI = 4.0 bar

so that PHI can be expressed in terms of PH2 giving us the equilibrium constantexpression

Chemistry 75 Winter, 2017Problem Set 5 Solutions

K = PHI

2

PH2 PI2

=

4.0 bar – 0.392 bar – PH2 2

PH2 0.392 bar

= 207

which has the solution PH2 = 0.148 bar. (This is a quadratic expression for PH2, andthe other solution is PH2 = 88.2 bar, which we reject because this exceeds our totalpressure!) We find PHI = (4.0 – 0.392 – 0.148) bar = 3.460 bar, and the gas phasemole fractions are therefore

xHI = 3.460 bar/4.0 bar = 0.865 xH2 = 0.148 bar/4.0 bar = 0.037

xI2 = 0.392 bar/4.0 bar = 0.098

We can complete our description of the equilibrium system with a simple accountingof atoms. We know we started with 1.20 mol of H atoms (from 1.0 mol HI and 0.10mol H2) so that

1.20 mol = nHI + 2nH2 = nH2xHIxH2

+ 2

which yields nH2 = 0.047 mol and nHI = (1.20 – 2(0.047)) mol = 1.106 mol. Theamount of gas phase I2 can be found from the definition of mole fraction written inthe following way:

ntotal = nI2(g)xI2(g)

= nHIxHI

or

nI2(g) = nHIxI2(g)xHI

= (1.106 mol)(0.098)

0.865 = 0.125 mol

and the amount of liquid I2 comes from writing

2.00 mol (total I) = 2nI2(g) + 2nI2(l) + nHI

so that nI2(l) = 0.322 mol.

Chemistry 75 Winter, 2017Problem Set 5 Solutions

Now consider just the vaporization of liquid I2. We can write this phase transition asa reaction, and use the vapor pressure to give us its equilibrium constant:

I2(l) →← I2(g) K = PI2(g) = 0.392

We combine this reaction with our original gas-phase reaction to obtain the netreaction with liquid I2 as the reactant as shown below:

Rxn (1) I2(l) →← I2(g) K (1) = PI2(g) = 0.392

Rxn (2) H2(g) + I2(g) →← 2HI(g) K (2) = PHI(g)

2

PH2(g)PI2(g) = 207

(3) = (1) + (2)

H2(g) + I2(l) →← 2HI(g)

K (3) = K(1)K (2)

= PHI(g)

2

PH2(g) = (0.392)(207) = 81.1

There’s one (good, as it turns out) assumption implicit in what we’ve done here: weneed HI and H2 to be insoluble in liquid iodine. If not, then we’d need to incorporatesomething about the liquid I2 in our expression for K (because it would not be in itsstandard state), and the calculation would suddenly become much more difficult.

5.3 We start by recalling that the volume change on vaporization is overwhelmingly dueto the greater molar volume of a gas over the liquid from whence it comes: ΔVvap =Vm(g) – Vm(l) ≈ Vm(g) = nRT/P. That lets us write the Clapeyron equation in theform

dPdT vap

= ΔHvap

o

TvapΔVm ≅

ΔHvapo P

R Tvap2

= 31.01 kJ mol–1 0.30 atm

8.314 J mol–1 K–1 298. K 2 = 1.26 × 10–2 atm K–1

The normal boiling temperature (Tvapo ) has Pvap = Pvap

o = 1.00 atm, and we can write

Chemistry 75 Winter, 2017Problem Set 5 Solutions

lnPvap

o

Pvap = –

ΔHvapo

R1

Tvapo

– 1Tvap

ln 10.30

= – 31.01 kJ mol–1

8.314 J mol–1 K–11

Tvapo

– 1298 K

Solving for Tvapo yields Tvap

o = 329.7 K = 56.6 °C. The literature value is 58.78 °C.Not bad! Trouton’s rule leads to the following expression for vapor pressure:

P(T)/atm = exp 10.5 1 – Tvap

o

T

and we want to extract Tvapo from this given our experimental vapor pressure P = 0.30

atm at T = 298 K:

0.30 = exp 10.5 1 – Tvap

o

298

We find Tvapo =332.2 K = 59.1 °C. Again, not bad agreement!

5.4 We know the standard boiling point for He(l) (4.22 K) and the molar enthalpy ofvaporization, ΔHvap

o = 84.0 J mol–1 (orders of magnitude less than that for water).Thus, we can find the vapor pressure at any other T:

P = (1.00 atm) exp – ΔHvap

o

R

14.16 K

– 1

4.22 K = 0.97 atm .

This is not a very large pressure drop, of course, since surface atmosphericphenomena (with the exception of tornadoes) never change the surface pressure veryfar. Meteorologists commonly use the mbar pressure unit, where 1.00 atm = 1013mbar. Here, 0.97 atm = 980 mbar. The highest surface pressure ever recorded onEarth was about 1064 mbar, recorded during a Siberian winter.