chemistry:solved problems 6

1.   Using the values of DH_f° and S° provided, calculate the quantities listed below for the following phase change in ethanol,

CH₃CH₂OH.

CH₃CH₂OH(l)     ⇄     CH₃CH₂OH(g)

DH_f° [CH₃CH₂OH(l)] = -277.69 kJ/mol                   DH_f° [CH₃CH₂OH(g)] = -235.10 kJ/mol

S°[CH₃CH₂OH(l)] = 160.7 J/mol·K                        S°[CH₃CH₂OH(g)] = 282.7 J/mol·K

The sign of DS°_total will inform us as to whether the evaporation of ethanol is spontaneous under standard          conditions (P = 1 bar and T = 25°C). If DS°_total is positive then the process is spontaneous under standard     conditions; if DS°_total is negative then the process is non-spontaneous under standard conditions.

(a) What intermolecular attractive forces are being overcome in this phase change?

The following intermolecular attractive forces are present between liquid phase ethanol molecules.

Dispersion forces

Dipole-dipole interactions

Hydrogen “bonds”

(b) Calculate DS°_system at 25.0°C.

DS°_system = [(1)(282.7 J/mol·K)] – [(1)(160.7 J/mol·K)] = 122.0 J/mol·K

(c) Calculate DS°_surroundings at 25.0°C.

DS°_surroundings = – DH°_system/T

DH°_system = [(1)(-235.10 kJ/mol)] – [(1)(-277.69 kJ/mol)] = 42.59 kJ/mol

DH°_system = 42590 J/mol (4 sig. fig.)

DS°_surroundings = – (42590 J/mol)/298.15 K = -142.8 J/mol·K

(d) Calculate DS°_otal at 25.0°C.

DS°_total = DS°_system + DS°_surroundings

DS°_total = 122.0 J/mol·K  +  (-142.8 J/mol·K)  = – 20.8 J/mol·K

(e) Is this phase change spontaneous under standard conditions? Explain your reasoning.

Since DS°_total < 0 the evaporation of ethanol, according to the data provided, is not spontaneous under                 standard conditions.

2.   Calculate the following quantities for the evaporation of ethanol.

(a) Using the values of DH_f° and S° provided in problem 1, calculate DG_rxn° at 25.0°C.

DG_rxn° = DH_rxn° – TDS_rxn°

DH_rxn° = 42.59 kJ/mol (from 1c)

DS_rxn° = 122.0 J/mol·K = 0.1220 kJ/ mol·K (from 1b)

DG_rxn° = 42.59 kJ/mol – [(298.15 K)(0.1220 kJ/ mol·K)]

DG_rxn° = 42.59 kJ/mol – 36.37 kJ/mol = 6.22 kJ/mol

(b) Calculate the value of the equilibrium constant at 25.0°C.

K = e^{–DG° / RT} = e ^{-6.22 kJ/mol / [(0.008314 kJ/KCmol)(298.15 K)]}

K = e^-2.51 = 0.0813

(c) Calculate the equilibrium vapor pressure (in torr) at 25.0°C.

For evaporation and sublimation the equilibrium constant K is equal to the equilibrium vapor pressure                 divided by the standard pressure P°. The standard pressure by convention is 1 bar.

K = P_eq/P° so P_eq = K x P°

P_eq = 0.0813 x 1bar = 0.0813 bar

P_eq = (0.0813 bar) x (1 atm/1.01325 bar) x (760 torr/1 atm)

P_eq = 61.0 torr

(d) Calculate DG_rxn at 25.0°C if the partial pressure of ethanol is 30.0 torr. Is the evaporation of ethanol

spontaneous under these conditions? Explain your reasoning.

DG_rxn = DG_rxn° +  RTlnQ where Q is the reaction quotient.

For evaporation and sublimation Q = P/P° where P is the vapor pressure of   the liquid or solid and

P° = 1 bar.

We first need to convert the given pressure (30.0 torr) to bar.

(30.0 torr) x (1 atm/760 torr) x (1.01325 bar/1 atm) = 0.0400 bar

Q = 0.0400 bar/1 bar = 0.0400

DG_rxn = 6.22 kJ/mol + [(8.314 x 10^-3 kJ/K·mol)(298.15 K)ln(0.0400)]

DG_rxn = 6.22 kJ/mol + (-7.979 kJ/mol) = – 1.76 kJ/mol

Since DG_rxn is < 0, the evaporation of ethanol is spontaneous under these conditions (P = 30.0 torr and

T = 25°C).

(e) Calculate DG_rxn at 25.0°C if the partial pressure of ethanol is 75.0 torr. Is the evaporation of ethanol

spontaneous under these conditions? Explain your reasoning.

DG_rxn = DG_rxn° +  RTlnQ where Q is the reaction quotient.

We first need to convert the given pressure (75.0 torr) to bar.

(75.0 torr) x (1 atm/760 torr) x (1.01325 bar/1 atm) = 0.100 bar

Q = 0.100 bar/1 bar = 0.100

DG_rxn = 6.22 kJ/mol + [(8.314 x 10^-3 kJ/K·mol)(298.15 K)ln(0.100)]

DG_rxn = 6.22 kJ/mol + (- 5.708 kJ/mol) = 0.51 kJ/mol

Since DG_rxn is > 0, the evaporation of ethanol is not spontaneous under these conditions (P = 75.0 torr and         T = 25°C).

3.   Using the values of DH_f° and S° provided, calculate the quantities listed below for the following phase change in

naphthalene, C₁₀H₈.

naphthalene

C₁₀H₈(s)     ⇄     C₁₀H₈(g)

DH_f° [C₁₀H₈(s)] = 77.9 kJ/mol                                            DH_f° [C₁₀H₈(g)] = 150.6 kJ/mol

S° [C₁₀H₈(s)] = 167.5 J/K·mol                                           S° [C₁₀H₈(g)] = 333.2 J/K·mol

(a) What intermolecular attractive forces are being overcome in this phase change?

Since naphthalene is nonpolar, dispersion forces must be overcome in order to sublime naphthalene.

(b) Calculate DG_rxn° at 25.0°C. Is the sublimation of naphthalene spontaneous under standard conditions? Explain       your reasoning.

DG_rxn° = DH_rxn° – TDS_rxn°

DH_rxn° = (1)(150.6 kJ/mol) – (1)(77.9 kJ/mol) = 72.7 kJ/mol

DS_rxn° = (1)(333.2 J/mol·K) – (1)(167.5 J/mol·K) = 165.7 J/mol·K

DS_rxn° = 0.1657 kJ/mol·K

DG_rxn° = 72.7 kJ/mol – [(298.15 K)(0.1657 kJ/mol·K)]

DG_rxn° = 72.7 kJ/mol – 49.40 kJ/mol = 23.3 kJ/mol

Since DG_rxn° is > 0, the sublimation of naphthalene is not spontaneous under standard conditions.

(c) Calculate the value of the equilibrium constant at 25.0°C.

K = e^{–DG°/ RT} = e ^{-23.3 kJ/mol / [(0.008314 kJ/K·mol)(298.15 K)]}

K = e^-9.40 = 8.28 x 10^-5

(d) Calculate the equilibrium vapor pressure (in torr) at 25.0°C.

For evaporation and sublimation the equilibrium constant K is equal to the equilibrium vapor pressure                 divided by the standard pressure P°. The standard pressure by convention is 1 bar.

K = P_eq/P° so P_eq = K x P°

P_eq = (8.28 x 10^-5) x 1bar = 8.28 x 10^-5 bar

P_eq = (8.28 x 10^-5 bar) x (1 atm/1.01325 bar) x (760 torr/1 atm)

P_eq = 0.0621 torr

(e) Calculate DG_rxn at 25.0°C if the partial pressure of naphthalene is 1.00 torr. Is the sublimation of naphthalene            spontaneous under these conditions? Explain your reasoning.

DG_rxn = DG_rxn° +  RTlnQ where Q is the reaction quotient.

For evaporation and sublimation Q = P/P° where P is the vapor pressure of   the liquid or solid and

P° = 1 bar.

We first need to convert the given pressure (1.00 torr) to bar.

(1.00 torr) x (1 atm/760 torr) x (1.01325 bar/1 atm) = 0.00133 bar

Q = 0.00133 bar/1 bar = 0.00133

DG_rxn = 23.3 kJ/mol + [(8.314 x 10^-3 kJ/mol·K)(298.15 K)ln(0.00133)]

DG_rxn = 23.3 kJ/mol + (-16.42 kJ/mol) = 6.9 kJ/mol

Since DG_rxn is > 0, the sublimation of naphthalene is not spontaneous under these conditions (P = 1.00 torr        and T = 25°C).

4.  Use the following data to calculate the value of K_p at 298 K for the reaction

N₂(g)   +   2 O₂(g)     ⇄     2 NO₂(g)

N₂(g)   +   O₂(g)     ⇄     2 NO(g)                        DG^° =  173.2 kJ/mol

2 NO(g)   +   O₂(g)    ⇄     2 NO₂(g)                   DG^° =  – 69.7 kJ/mol

Remember that K_p and DG^° are related to each other by the following equation

K_p = e^{–DG° / RT}

You need to recognize that the desired chemical equation is the sum of the two chemical equations given.         Hence, the standard Gibbs free-energy change for the desired reaction is obtained by adding the DG^° values of the given reactions.

DG^° = 173.2 kJ/mol  +  (- 69.7 kJ/mol) = 103.5 kJ/mol

K_p = e^{–DG° / RT} = e ^{– (103.5 kJ/mol) / [(0.008314 kJ/K·mol)(298.15 K)]}

K_p = e^-41.75 = 7.4 x 10^-19

5.  In the gas phase, formic acid exits as an equilibrium of monomer and dimer molecules. (The dimer consists of two

molecules linked through hydrogen “bonds”.)

(a)  Is the formation of a dimer molecule from monomer molecules endothermic or exothermic? Justify your reasoning.

A formic acid dimer results from the formation of two hydrogen bonds between two formic acid molecules.       Thus, this process is exothermic.

(b) Calculate the equilibrium constant, K_p, at 25.0°C for the monomer-dimer equilibrium

2  HCOOH(g)    ⇄   (HCOOH)₂(g)

using the data given below. Is this process spontaneous under standard conditions? Explain your reasoning.

DH_f° (kJ/mol)                       S° (J/K·mol)

HCOOH (g)                               – 362.63                                   251.0

(HCOOH)₂ (g)                           – 785.34                                   347.7

DG° = DH°– TDS°

DH° = [(1)(- 785.34 kJ/mol)] – [(2)(- 362.63 kJ/mol)] = – 60.08 kJ/mol

DS° = [(1)(347.7 J/mol·K)] – [(2)(251.0 J/mol·K)] = – 154.3 J/mol·K = – 0.1543 kJ/mol·K

DG° = – 60.08 kJ/mol – [(298.15 K)(- 0.1543 kJ/mol·K)]

DG° = – 60.08 kJmol – (- 46.00 kJ/mol) = – 14.08 kJ/mol

Since DG^° is < 0, this process is spontaneous under standard conditions.

K_p = e^{–DG° / RT} = e ^{– (-14.08 kJ/mol) / [(0.0083145 kJ/KCmol)(298.15 K)]}

K_p = e^+5.680 = 292.9

(c) Calculate the enthalpy change per hydrogen bond formed in the gas phase. Show your reasoning clearly.

The enthalpy change (DH) for the formation of two moles of hydrogen bonds is – 60.08 kJ. Thus, the       enthalpy change per mole of hydrogen bond formed is – 30.04 kJ.

(d) As the temperature increases, in which direction does the equilibrium shift? How does an increase in temperature

affect the value of K_p? Justify your reasoning.

An increase in temperature will shift an equilibrium mixture in the endothermic direction. Since this is an            exothermic process, the equilibrium mixture will shift to the left (to the reactants side). The value of K_p decreases with an increase in temperature.

6.  (a) Calculate the Gibbs free energy of the following reaction at 1200 K (K = 6.8)

I₂(g)   ⇄ 2 I(g)

when the partial pressures of I₂ and I are 0.13 bar and 0.98 bar, respectively.

Since the partial pressures of the two gases are not equal to 1 bar we need to calculate ΔG not ΔG°.

DG = DG° +  RTlnQ where Q is the reaction quotient

ΔG° = – RTlnK = – [(8.314 x 10^-3 kJ/mol·K)(1200 K)ln(6.8)] = – 19.12 kJ/mol

Q = P_I²/P_I2 = (0.98)²/0.13 = 7.4

RTlnQ = [(8.314 x 10^-3 kJ/mol·K)(1200 K)ln(7.4)] = 19.97 kJ/mol

ΔG = – 19.12 kJ/mol  +  19.97 kJ/mol = 0.85 kJ/mol

(b) What is the spontaneous direction of the reaction at these conditions? Explain your reasoning.

Since ΔG > 0 this reaction is spontaneous to the left and will produce more I₂. Also note that since

Q > K (7.4 > 6.8) this reaction will proceed to the left.

7.   Use the best-fit line obtained for the data plotted in the graph below to determine DH_vap and DS_vap for the evaporation of

1-propanol.

CH₃CH₂CH₂OH(l)     ⇄    CH₃CH₂CH₂OH(g)

slope = – DH_vap/R Þ DH_vap = – slope x R

y-intercept = DS_vap/R Þ DS_vap = y-intercept x R

DH_vap = – (- 5.70×10³ K) x (8.314 J/mol·K) = 4.74×10⁴ J/mol = 47.4 kJ/mol

DS_vap = 15.4 x 8.314 J/mol·K = 128 J/mol·K

How does the value of DH_vap for the evaporation of 1-propanol compare (greater, the same, or lesser) to the value of

DH_vap for the evaporation of methanol (CH₃OH)? Explain your reasoning.

The value of DH_vap for 1-propanol is greater than the value of DH_vap for methanol because the intermolecular      attractive forces are stronger between 1-propanol    molecules due to the longer hydrocarbon chain present        in 1-propanol which results in stronger dispersion forces.