Chemistry 360 Dr. Jean M. Standard

Problem Set 8 Solutions 1. Consider the reaction between zinc metal and hydrochloric acid,

€

Zn s( ) + 2 HCl aq( )

€

H2 g( ) + ZnCl2 aq( ) . If the initial amounts present are 100.0 g Zn and 150.0 mL of 2.25 M HCl, determine the minimum and maximum values of the extent of reaction

€

ξ that are possible for this reaction. To determine the extent of reaction, we need to figure out the numbers of moles of each of the reactants,

€

nZn = 100.0 g65.39 g/mol

nZn = 1.529 mol.

€

nHCl = 0.150 L( ) 2.25mol/L( )nHCl = 0.3375 mol.

The minimum extent of reaction occurs when there is no reaction; that is, the moles of reactants at the end of the reaction are the same as the initial moles,

€

ξmin = ni − ni0vi

= ni0 − ni0vi

ξmin = 0 . The maximum extent of reaction occurs when all of the limiting reagent is consumed and converted to products. For this reaction, the limiting reagent is HCl. Calculating the moles, we have

€

Zn s( ) + 2 HCl aq( )

€

H2 g( ) + ZnCl2 aq( ) .

€

moles init. 1.529 0.3375 0 0moles reacted − 0.169 − 0.3375 0.169 0.169

moles final 1.361 0 0.169 0.169 Using the final number of moles of any of the species, we can calculate the maximum extent of reaction. Calculating this value using the moles of HCl, we have

€

ξmax = nHCl − nHCl,0

vHCl

= 0− 0.3375mol−2

ξmax = 0 .169 mol.

2

2. The equilibrium constant for the isomerization of cis-2-butene to trans-2-butene is

€

Keq = 2.07 at 400 K. Determine the standard molar Gibbs free energy, ΔGR

! , for the reaction at 400 K.

€

ΔGr! = − RT ln Keq

= − 8.314 J mol−1K−1( ) 400 K( ) ln 2.07( )

ΔGr! = − 2420 J mol−1 .

3. Buckminsterfullerene, C60, is an icosohedral (soccer-ball) shaped molecule composed of hexagons and

pentagons of carbon atoms. For C60, the standard molar Gibbs free energy of formation,

€

ΔG f! , is 23.98

kJ/mol at 25.0ºC. Write the balanced formation reaction for C60 and determine the equilibrium constant

€

Keq for the formation of C60.

The formation reaction of C60 is

The molar Gibbs free energy of this reaction is the same as the Gibbs free energy of formation of C60, so the equilibrium constant is

€

ΔGr! = − RT ln Keq ,

or Keq = exp −ΔGr!

RT

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪ .

Substituting,

€

Keq = exp −ΔGr!

RT

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

= exp −23980 J/mol( )

8.314 J mol−1K−1( ) 298 K( )

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪

= e−9.6740

Keq = 6.29×10−5 .

60 C (s, graphite) C60 (s) .

3

4. For the reaction

determine an expression for the equilibrium constant in terms of the extent of reaction

€

ξ when one mole of CO gas is mixed with one mole of hydrogen gas.

The initial and equilibrium moles, as well as the mole fractions can be determined in terms of the extent of reaction,

€

moles init. 1 1 0 0moles equil. 1−ξ 1− 3ξ ξ ξ

xi 1−ξ2− 2ξ

1− 3ξ2− 2ξ

ξ2− 2ξ

ξ2− 2ξ

The equilibrium constant is then

€

Keq =

PCH4

P!⎛

⎝ ⎜

⎞

⎠ ⎟ PH2O

P!⎛

⎝ ⎜

⎞

⎠ ⎟

PCOP!

⎛

⎝ ⎜

⎞

⎠ ⎟ PH2

P!⎛

⎝ ⎜

⎞

⎠ ⎟

3

=

xCH4 P

P!⎛

⎝ ⎜

⎞

⎠ ⎟ xH2OP

P!⎛

⎝ ⎜

⎞

⎠ ⎟

xCOPP!

⎛

⎝ ⎜

⎞

⎠ ⎟ xH2P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

3

=

ξ2− 2ξ⎛

⎝ ⎜

⎞

⎠ ⎟

2 PP!⎛

⎝ ⎜

⎞

⎠ ⎟ 2

1−ξ2− 2ξ⎛

⎝ ⎜

⎞

⎠ ⎟

1− 3ξ2− 2ξ⎛

⎝ ⎜

⎞

⎠ ⎟

3 PP!⎛

⎝ ⎜

⎞

⎠ ⎟ 4

Keq = 4ξ 2 1−ξ( )

1− 3ξ( )3P!

P

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

.

CO (g) + 3 H2 (g) CH4 (g) + H2O (g) ,

CO (g) + 3 H2 (g) CH4 (g) + H2O (g)

4

5. Nitrogen dioxide, NO2, dimerizes to form nitrogen tetroxide according to the reaction

€

2NO2 g( )

€

N2O4 g( ) .

(a.) Using data from the appendix of your textbook, calculate the standard molar Gibbs free energy of reaction, ΔGR

! , and the equilibrium constant,

€

Keq , at 298 K.

From the appendix, the standard molar Gibbs free energy of formation of NO2 is 51.31 kJ/mol and for N2O4 it is 97.89 kJ/mol at 298 K. The standard molar Gibbs free energy of reaction is therefore

€

ΔGr! = ΔG f

! N2O4( ) − 2ΔG f! NO2( )

= 97.89 kJ/mol − 2 51.31 kJ/mol( )ΔGr

! = − 4.73 kJ/mol.

Then, using the equation

€

ΔGr! = − RT ln Keq , the equilibrium constant is

€

ln Keq = − ΔGr!

RT,

or Keq = exp −ΔGr!

RT

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪ .

Substituting, the equilibrium constant is

€

Keq = exp −ΔGr!

RT

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

= exp −−4.73×103 J/mol( )

8.314 J mol−1K−1( ) 298.15K( )

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪

Keq = 6.74 . Note that the temperature of 298.15 K was used since that is the temperature at which the data in the appendix is tabulated.

(b.) Calculate the extent of reaction

€

ξ for this reaction if 1.0 mole of NO2 and 0.0 moles of N2O4 are present initially. The equilibrium pressure is 1.5 bar.

The equilibrium constant can be expressed in terms of the extent of reaction,

€

2NO2 g( )

€

N2O4 g( ) .

€

moles init. 1 0moles equil. 1− 2ξ ξ

xi 1− 2ξ1−ξ

ξ1−ξ

5

5 b.) Continued The equilibrium constant is

€

Keq =

PN2O4

P!⎛

⎝ ⎜

⎞

⎠ ⎟

PNO2

P!⎛

⎝ ⎜

⎞

⎠ ⎟

2

Keq =

xN2O4 P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

xNO2P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

2 .

The equilibrium constant can be written terms of the extent of reaction,

€

Keq =

xN2O4 P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

xNO2P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

2

=

ξ1−ξ⎛

⎝ ⎜

⎞

⎠ ⎟ PP!⎛

⎝ ⎜

⎞

⎠ ⎟

1− 2ξ1−ξ

⎛

⎝ ⎜

⎞

⎠ ⎟

2 PP!⎛

⎝ ⎜

⎞

⎠ ⎟ 2

Keq = ξ 1−ξ( )1− 2ξ( )2

P!

P

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ .

Substituting the value for the equilibrium constant , equilibrium pressure (P=1.5 bar), and

€

P! = 1bar yields

€

Keq = ξ 1−ξ( )1− 2ξ( )2

P !

P

⎛

⎝ ⎜

⎞

⎠ ⎟

6.74 = ξ 1−ξ( )1− 2ξ( )2

1bar1.5 bar⎛

⎝ ⎜

⎞

⎠ ⎟ .

6

5 b.) Continued

Solving for the extent of reaction,

€

6.74 = ξ 1−ξ( )1− 2ξ( )2

1bar1.5 bar

⎛

⎝ ⎜

⎞

⎠ ⎟

10.11 = ξ 1−ξ( )1− 2ξ( )2

10.11 1− 2ξ( )2 = ξ 1−ξ( )10.11 1− 4ξ + 4ξ 2( ) = ξ −ξ 2

40.44ξ 2 − 40.44ξ +10.11 = ξ −ξ 2

41.44ξ 2 − 41.44ξ +10.11 = 0

ξ 2 −ξ + 0.244 = 0ξ = 0.422 or 0.578.

Since the moles at equilibrium of NO2 equal

€

1− 2ξ , plugging

€

ξ = 0.578 into this would yield a negative number. Thus, the only correct physical answer is

€

ξ = 0.422.

7

6. A mixture of diatomic nitrogen and diatomic hydrogen gas in a 1:3 molar ratio was passed over a catalyst to form ammonia at 450°C. At a pressure of 10.13 atm, there was 2.04% by volume of ammonia gas formed. Determine the equilibrium constant for the reaction

€

12 N2 g( ) + 3

2 H2 g( )

€

NH3 g( ) . In terms of partial pressures (assuming that the gases are ideal), the equilibrium constant is

€

Keq =

PNH3

P!⎛

⎝ ⎜

⎞

⎠ ⎟

PN2

P!⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2 PH2

P!⎛

⎝ ⎜

⎞

⎠ ⎟

3/ 2 .

Since the amount of NH3 is 2.04% by volume, the mole fraction of NH3 is therefore 0.0204 (assuming ideal gas behavior). Since the total pressure is 10.13 atm = 10.26 bar, the partial pressure of NH3 is

€

PNH3 = xNH3P

= 0.0204( ) 10.13atm( )PNH3 = 0.207 atm = 0.210 bar.

Since the ratio of N2 to H2 is 1:3,

€

PH2 = 3PN2 . The total pressure is

€

P = PNH3 + PH2 + PN2

= PNH3 + 3PN2 + PN2

P = PNH3 + 4PN2 .

Substituting P=10.13 atm and

€

PNH3 = 0.207 atm,

€

P = PNH3 + 4PN2

10.13atm = 0.207 atm + 4PN2

PN2 = 2.48 atm = 2.51 bar . And, the partial pressure of hydrogen is therefore

€

PH2 = 3PN2

= 3 2.48 atm( )PH2 = 7.44 atm = 7.54 bar .

8

6. Continued Substituting into the expression for the equilibrium constant, using

€

P! = 1bar and P=10.26 bar, we have

€

Keq =

PNH3

P!⎛

⎝ ⎜

⎞

⎠ ⎟

PN2

P!⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2 PH2

P!⎛

⎝ ⎜

⎞

⎠ ⎟

3/ 2

=

0.210 bar1bar

⎛

⎝ ⎜

⎞

⎠ ⎟

2.51bar1bar

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 27.54 bar

1bar⎛

⎝ ⎜

⎞

⎠ ⎟

3/ 2

Keq = 0.00640.

9

7. A 1.0 L reaction vessel containing 0.233 moles of N2 and 0.341 moles of PCl5 is heated to 250°C. The total pressure once the system reaches equilibrium is 29.33 atm. Calculate the equilibrium constant for the reaction

In terms of partial pressures, assuming that the gases behave ideally, the equilibrium constant is

€

Keq =

PPCl3P!

⎛

⎝ ⎜

⎞

⎠ ⎟ PCl2P!

⎛

⎝ ⎜

⎞

⎠ ⎟

PPCl5P!

⎛

⎝ ⎜

⎞

⎠ ⎟

.

In terms of partial pressures, the equilibrium constant becomes

€

Keq =

xPCl3P

P!⎛

⎝ ⎜

⎞

⎠ ⎟ xCl2P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

xPCl5P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

.

To determine the equilibrium constant, we need to express the mole fractions in terms of the extent of reaction.

€

moles init. 0.341 0 0moles equil. 0.341−ξ ξ ξ

xi 0.341−ξ0.574 +ξ

ξ0.574 +ξ

ξ0.574 +ξ

Note that the total number of moles when calculating the mole fractions must include the 0.233 moles of N2 gas as well as the products and reactants. In terms of the extent of reaction, the equilibrium constant is

€

Keq =

xPCl3P

P!⎛

⎝ ⎜

⎞

⎠ ⎟ xCl2P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

xPCl5P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

=

ξ0.574 +ξ

⎛

⎝ ⎜

⎞

⎠ ⎟ PP!⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

ξ0.574 +ξ

⎛

⎝ ⎜

⎞

⎠ ⎟ PP!⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

0.341−ξ0.574 +ξ

⎛

⎝ ⎜

⎞

⎠ ⎟ PP!⎛

⎝ ⎜

⎞

⎠ ⎟

=

ξ0.574 +ξ

⎛

⎝ ⎜

⎞

⎠ ⎟

2 PP!⎛

⎝ ⎜

⎞

⎠ ⎟

0.341−ξ0.574 +ξ

⎛

⎝ ⎜

⎞

⎠ ⎟

Keq = ξ 2P0.574 +ξ( ) 0.341−ξ( )P!

.

PCl5 (g) PCl3 (g) + Cl2 (g) .

PCl5 (g) PCl3 (g) + Cl2 (g) .

10

7. Continued To determine the extent of reaction, we can use the ideal gas law,

€

ntot = PVRT

= 29.33atm( ) 1.0 L( )

0.08206 L atmmol−1K−1( ) 523.15K( )ntot = 0.683mol.

Then, in terms of the extent of reaction,

€

ntot = 0.574 + ξ0.683mol = 0.574 + ξ

ξ = 0.109 mol.

Finally, the equilibrium constant may be calculated (using P = 29.33 atm = 29.72 bar),

€

Keq = ξ 2P0.574 +ξ( ) 0.341−ξ( )P!

= 0.109 mol( )2 29.72 bar( )

0.574 + 0.109 mol( ) 0.341− 0.109 mol( ) 1bar( )Keq = 2.23.

11

8. Determine the equilibrium constant and extent of reaction for the dissociation of hydrogen gas at 3000 K and 1 atm,

The standard molar Gibbs free energies of formation,

€

ΔG f! , of H2 and H at 3000 K are 0 and 46.01

kJ/mol, respectively.

The equilibrium constant can be expressed in terms of the extent of reaction,

€

moles init. 1 0moles equil. 1−ξ 2ξ

xi 1−ξ1+ξ

2ξ1+ξ

The equilibrium constant is

€

Keq =

PHP!

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

PH2

P!⎛

⎝ ⎜

⎞

⎠ ⎟

or Keq =

xH PP!

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

xH2P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

.

Writing the equilibrium constant in terms of the extent of reaction yields

€

Keq =

xH PP!

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

xH2P

P!⎛

⎝ ⎜

⎞

⎠ ⎟

=

2ξ1+ξ

⎛

⎝ ⎜

⎞

⎠ ⎟

2 PP!⎛

⎝ ⎜

⎞

⎠ ⎟ 2

1−ξ1+ξ

⎛

⎝ ⎜

⎞

⎠ ⎟ PP!⎛

⎝ ⎜

⎞

⎠ ⎟

Keq = 4ξ 2

1−ξ 2( )PP!⎛

⎝ ⎜

⎞

⎠ ⎟ .

H2 (g) 2 H (g) .

H2 (g) 2 H (g) .

12

8. Continued To determine the extent of reaction, we need to calculate the equilibrium constant. We can determine the equilibrium constant from the standard molar Gibbs free energy for the reaction. Using the values given for the standard molar Gibbs free energies of formation at 3000 K, we have

€

ΔGr! = 2 ΔG f

! H( ) − ΔG f! H2( )

= 2 46.01kJ/mol( ) − 0

ΔGr! = 92.02 kJ/mol.

Then, using the equation

€

ΔGr! = − RT ln Keq , the equilibrium constant is

€

ln Keq = − ΔGr!

RT,

or Keq = exp − ΔGr

!

RT

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

= exp −92.02×103 J/mol( )

8.314 J mol−1K−1( ) 3000 K( )

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪

Keq = 0.0250. The equilibrium constant in terms of extent of reaction is then

€

Keq = 4ξ 2

1−ξ 2( )PP!⎛

⎝ ⎜

⎞

⎠ ⎟

0.0250 = 4ξ 2

1−ξ 2( )1.013bar

1bar⎛

⎝ ⎜

⎞

⎠ ⎟ ,

where the relation P = 1 atm = 1.013 bar has been used. Solving the equation for the extent of reaction,

€

0.0250 = 4ξ 2

1−ξ 2( )1.013bar

1bar⎛

⎝ ⎜

⎞

⎠ ⎟

0.0247 1−ξ 2( ) = 4ξ 2

ξ 2 = 0.02474.0247

ξ = 0.0783. Thus, from the extent of reaction, diatomic hydrogen gas is 7.8% dissociated at 3000 K and 1 atm pressure.