# Midterm 2

General Chemistry 3

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The combustion of carbon monoxide is represented by the following equation:

CO (g) + ½ O2 (g) → CO2 (g)

1) Determine the standard enthalpy change ΔH0rxn for the combustion of CO at 298 K

2) Determine the standard entropy change ΔS0rxn for this reaction at 298 K

3) Is the reaction spontaneous under standard conditions at 298 K? Justify.

4) Calculate the value of the equilibrium constant Keq at 298 K

Data: At 298 K:

C (s) + ½ O2 (g) → CO (g)     [ΔH0 = - 110.5 kJ.mol-1]

C (s) + O2 (g) → CO2 (g)     [ΔH0 = - 393.5 kJ.mol-1]

S0 [CO (g)] = 197.7 J.mol-1.K-1

S0 [CO2 (g)] = 213.7 J.mol-1.K-1

S0 [O2 (g)] = 205.1 J.mol-1.K-1

1) CO (g) → C (s) + ½ O2 (g)        [ΔH0 = 110.5 kJ.mol-1]

+ C (s) + O2 (g) → CO2 (g)          [ΔH0 = - 393.5 kJ.mol-1]

= CO (g) + ½ O2 (g) → CO2 (g)  [ΔH0rxn]

ΔH0rxn = 110.5 – 393.5 = - 283.0 kJ.mol-1

2) ΔS0rxn = S0 [CO2 (g)] - S0 [CO (g)] – $\frac{1}{2}$ x S0 [O2 (g)]

ΔS0rxn = 213.7 – 197.7 – $\frac{1}{2}$ x 205.1

ΔS0rxn = - 86.5 J.mol-1.K-1

3) ΔG0rxn = ΔH0rxn - TΔS0rxn

ΔG0rxn = - 283.0 x 103 – 298 x (- 86.5)

ΔG0rxn = - 257.2 x 103 J.mol-1 = - 257.2 kJ.mol-1

ΔG0rxn < 0 ⇒ The reaction is spontaneous at 298 K

4) ΔG0rxn = - RT ln Keq

Keq = exp (-ΔG0rxn/RT)

Keq = 1.219 x 1045

1) What is the sign of ΔHrxn for a reaction in which the bonds are stronger in the products than in the reactants. Justify

2) A reaction is only spontaneous at high temperature. Predict and justify the sign of ΔHrxn for this reaction

1) ΔHrxn = Σ Hbonds (reactants) - Σ Hbonds (products)

If Σ Hbonds (products) > Σ Hbonds (reactants), ΔHrxn < 0

2) The reaction is spontaneous when ΔGrxn < 0

ΔGrxn = ΔHrxn - TΔSrxn

If a reaction is only spontaneous at high temperature ⇒ at low temperature ΔGrxn > 0

At low temperature: ΔHrxn >> TΔSrxn ⇒ ΔGrxn ≈ ΔHrxn > 0

Ammonia can be produced by the following reaction:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

1) Calculate the value for ΔH0 for this reaction

2) Can the yield of ammonia be increased by raising the temperature? Justify

3) What is the equilibrium constant for the reaction?

4) If 235 mL of H2 gas (25°C and 7.60 x 104 Pa) were completely converted to ammonia and the ammonia were dissolved in water to make 0.500 L of solution, what would be the molarity of the resulting solution?

Data: At 298K:

ΔG0f [NH3 (g)] = - 33.0 kJ.mol-1

S0 [N2 (g)] = 191.5 J.mol-1.K-1

S0 [H2 (g)] = 198.6 J.mol-1.K-1

S0 [NH3 (g)] = 192.3 J.mol-1.K-1

1) ΔS0f = 2 x S0 [NH3 (g)] - S0 [N2 (g)] – 3 x S0 [H2 (g)]

ΔS0f = 2 x 192.3 – 191.5 – 3 x 198.6

ΔS0f = - 402.7 J.mol-1.K-1

ΔG0f = ΔH0f - TΔS0f

ΔH0f = ΔG0f + TΔS0f

ΔH0f = - 33.0 x 103 - 298 x 402.7

ΔH0f = - 153.0 kJ.mol-1

2) ΔH0f < 0 ⇒ exothermic reaction

According to Le Chatelier’s principle, an increase in T shifts equilibrium to the left

⇒ this decreases equilibrium yield of NH3

3) ΔG0f = RT ln Keq

Keq = exp (-ΔG0rxn/RT)

Keq = 6.09 x 105

4)

PV = nRT

P = pressure in Pa

V = volume in m3 ⇒ 235 mL = 0.235 dm3 = 0.235 x 10-3 m3

T = temperature in K ⇒ 25°C = 298 K

nH2 = $\frac{\mathrm{PV}}{\mathrm{RT}}$

nH2 =

nH2 = 7.21 x 10-3 mol

According to the stoichiometry of the reaction:

3 moles of H2 form 2 moles of NH3

nNH3 / 2 = nH2 / 3
nNH3 = $\frac{2}{3}$ nH2 = 4.81 x 10-3 mol

nNH3 = [NH3] x V

[NH3] = 9.61 x 10-3 M

1.12 g of CaCl2 is added to 4.80 L of a 5.00 x 10-6 M solution of Na2CO3.

Will a precipitate of CaCO3 (s) form?

Data: Ksp [CaCO3 (s)] = 2.8 x 10-9

Dissolution reaction: CaCO3 (s) $⇌$ Ca2+ (aq) + CO32- (aq)     [Ksp = 2.8 x 10-9]

CaCO3 (s) forms if Qsp > Ksp

Let’s determine Qsp:

Qsp = [Ca2+]0 [CO32-]0

CaCl2 (s) → Ca2+ (aq) + 2 Cl- (aq)

From the stoichiometry of this reaction: nCa2+,0 = nCaCl2,0

[Ca2+]0$\frac{{\mathrm{n}}_{\mathrm{Ca}2+,0}}{\mathrm{V}}$ = $\frac{{\mathrm{n}}_{\mathrm{CaCl}2,0}}{\mathrm{V}}$ =

[Ca2+]0 =

[Ca2+]0 = 2.10 x 10-3 M

Na2CO3 (s) → 2 Na+ (aq) + CO32- (aq)

From the stoichiometry of this reaction: nCO32-,0 = nNa2CO3,0

⇒ [CO32-]0 = [Na2CO3]0 = 5.00 x 10-6 M

Qsp = [Ca2+]0 [CO32-]0

Qsp = 2.10 x 10-3 x 5.00 x 10-6

Qsp = 1.05 x 10-8 > Ksp

⇒ CaCO3 (s) forms

140 mL of 0.175 M HCl and 350 mL of 0.0750 M NaOH are combined.

Calculate the pH of this solution.

nH+,0 = [H+]0 x VHCl = 140 x 10-3 x 0.175

nH+,0 = 2.45 x 10-2 mol

nHO-,0 = [HO-]0 x VNaOH = 350 x 10-3 x 0.0750

nHO-,0 = 2.63 x 10-2 mol

There is an excess of HO- which will neutralize all the [H+].

nHO-,f = nHO-,0 - nH+,0

nHO-,f = 2.63 x 10-2 - 2.45 x 10-2 = 1.75 x 10-3 mol

[HO-]f = $\frac{{\mathrm{n}}_{{\mathrm{HO}}^{-},\mathrm{f}}}{\mathrm{V}}$

[HO-]f =

[HO-]f = 3.57 x 10-3 M

pOH = -log [HO-]f = 2.45

pH = 14 – pOH = 11.6