Chem 73 Third Hour Exam

Answers

1. What is E∘ for the following unbalanced reaction?

Fe(s) + O2(g) + 2H2O(l) −→Fe3++ 4OH−

(The ultimate products are FeO(OH) and H2O, but they are formed by a non redox reaction.The hydrated FeO(OH) is what we know as rust.)

Soln: With the help of the table:

R: O2(g) + 2H2O(l) + 4e−−→4OH− E∘ = +0.401

L: Fe3++ 3e−−→ Fe(s) E∘ = −0.037

E∘R − E∘

L = 0.438V

2. Given the non-standard concentration of the following reactions, calculate the instantaneous Ecell ofthe Daniel cell.

Zn(s) + Cu2+(aq, 0.0333M) −→ Zn2+(aq, 0.0444M) + Cu(s)

Soln: Q =MZn2+/M∘

MCu2+/M∘ = [Zn2+][Cu2+] = 0.0444

0.0333 = 0.133

The voltage under standard conditions, E∘ = 1.104V

Ecell = E∘ − (8.314 J/mol⋅K)(298K)(2mol e−)(96485C/mol e−) ln(0.133)

Ecell = 1.104V − (−0.0259) = 1.130V

3. For the cell reaction,

Hg2Cl2(s) + H2(1atm) −→2Hg(l ) + 2H+(a = 1) + 2Cl−(a = 1)

E∘ = +0.2676V ;(

∂E∘

∂T

)P

= −3.19× 10−4V/K

Calculate △G∘,△S∘, and△H∘.

Soln: n = 2△G∘ = −nFE∘ = −(2mol e−)(96484C/mol e−)(0.2676V )10−3kJ/J) = −51.64 kJ/mol

△S∘ = nF(

∂E∘

∂T

)P

= 2(96484C/mol e−)(−3.19× 10−4V/K) = −61.6 J/mol

△H∘ = ΔG∘ + TΔS∘

= −51.64 kJmol + 298.15K (−61.6 J

mol ) = −69.99 kJmol

4. The molar enthalpy of vaporization of water is 40.65 kJ/mol at its normal boiling point. Use theClausius-Clapeyron equation to calculate the vapour pressure of water at 100∘C. (Assuming ΔvapHremains constant with respect to T over the ten-degree range).

Soln: Assuming ΔvapH remains constant with respect to T over the ten-degree range.

lnP2

P1=

ΔvapHR

[T2−T1

T1T2

]= 40650 J⋅mol

8.3145 J⋅mol⋅K−1

[10K

(373.15)(383.15)

]= 0.342

ln P2

1atm = 0.342 or P2 = 1.408 atm = 1070Torr

5. A solution containing 1.470 g of dichlorobenzene in 50.00 g of benzene boils at 80.60∘C at a pressureof 1.00 bar. The boiling point of pure benzene is 80.09∘C and the molar enthalpy of vaporization ofpure benzene is 32.0 kJ/mol. Determine the molecular mass of dichlorobenzene from these data. \

Soln:

ΔvapT = 80.60∘C − 80.09∘C = 0.51K

Kb =[

M1

1000g⋅kg−1

]R(Tvap)2

ΔvapH= (78.108g/mol)(8.314J/mol⋅K)(353.2K)2

(1000g/kg)(32.0×103J/mol) = 2.53K ⋅ kg/mol

The molality is given by : m =ΔvapTKb

= 0.51K2.53K⋅kg/mol = 0.20mol/kg

Therefore: 1.470 g C6H4Cl2= 50.0 g C6H6

29.4 g C6H4Cl2 =1000 g C6H6= 0.20molSo the molar mass is 147

1

6. Verify the ionic strength of the following:

Type of Salt Example Im

1-1 KCl ?1-2 CaCl2 ?2-1 K2SO4 ?2-2 MgSO4 ?1-3 LaCl3 ?3-1 Na3PO4 ?

7. Estimate the mean ionic activity coe�cient for CaCl2 in a solution that is 0.010 mol/kg CaCl2(aq)and 0.030 mol/kg NaF (aq).

Soln. Im = 12 [(4× 0.010) + (1× 0.020) + (1× 0.030) + (1× 0.030)] = 0.060

log ± = −2× 1× 0.509× (0.060)1/2 = −0.249̄4

± = 0.56

8. Given the following reactions:

2BrCl(g) ⇌ Cl2(g) + Br2(g) Kp = 0.169

IBr(g) ⇌ Br2(g) + I2(g) Kp = 0.0149

Determine KP for the reaction : BrCl(g) + 12 I2(g) ⇌ IBr(g) + 1

2Cl2(g)

Soln: Eqn(3) = 12Eqn(1) -

12Eqn(2)

ΔrG∘3 = 1

2ΔrG∘1-

12ΔrG

∘2

K3 =K

1/21

K1/22

= (0.169)1/2

(0.0149)1/2= 3.37

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