Ammonia can be produced by the following reaction: N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) 1) Calculate the value for ΔH 0 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 H 2 gas (25°C and 7.60 x 10 4 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: ΔG 0 f [NH 3 (g)] = - 33.0 kJ.mol -1 S 0 [N 2 (g)] = 191.5 J.mol -1 .K -1 S 0 [H 2 (g)] = 198.6 J.mol -1 .K -1 S 0 [NH 3 (g)] = 192.3 J.mol -1 .K -1

1) ΔS 0 f = 2 x S 0 [NH 3 (g)] - S 0 [N 2 (g)] – 3 x S 0 [H 2 (g)] ΔS 0 f = 2 x 192.3 – 191.5 – 3 x 198.6 ΔS 0 f = - 402.7 J.mol -1 .K -1 ΔG 0 f = ΔH 0 f - TΔS 0 f ΔH 0 f = ΔG 0 f + TΔS 0 f ΔH 0 f = - 33.0 x 10 3 - 298 x 402.7 ΔH 0 f = - 153.0 kJ.mol -1 2) ΔH 0 f < 0 ⇒ exothermic reaction According to Le Chatelier’s principle, an increase in T shifts equilibrium to the left ⇒ this decreases equilibrium yield of NH 3 3) ΔG 0 f = RT ln K eq K eq = exp (-ΔG 0 rxn /RT) K eq = 6.09 x 10 5 4) PV = nRT P = pressure in Pa V = volume in m 3 ⇒ 235 mL = 0.235 dm 3 = 0.235 x 10 -3 m 3 T = temperature in K ⇒ 25°C = 298 K n H2 = PV RT n H2 = 7 . 60 × 10 4 × 0 . 235 × 10 - 3 8 . 314 × 298 n H2 = 7.21 x 10 -3 mol According to the stoichiometry of the reaction: 3 moles of H 2 form 2 moles of NH 3 n NH3 / 2 = n H2 / 3 n NH3 = 2 3 n H2 = 4.81 x 10 -3 mol n NH3 = [NH 3 ] x V [NH 3 ] = 9.61 x 10 -3 M

1.12 g of CaCl 2 is added to 4.80 L of a 5.00 x 10 -6 M solution of Na 2 CO 3 . Will a precipitate of CaCO 3 (s) form? Data: K sp [CaCO 3 (s)] = 2.8 x 10 -9

Dissolution reaction: CaCO 3 (s) ⇌ Ca 2+ (aq) + CO 3 2- (aq) [K sp = 2.8 x 10 -9 ] CaCO 3 (s) forms if Q sp > K sp Let’s determine Q sp : Q sp = [Ca 2+ ] 0 [CO 3 2- ] 0 CaCl 2 (s) → Ca 2+ (aq) + 2 Cl - (aq) From the stoichiometry of this reaction: n Ca2+,0 = n CaCl2,0 [Ca 2+ ] 0 = n Ca 2 + , 0 V = n CaCl 2 , 0 V = m CaCl 2 , 0 M CaCl 2 × V [Ca 2+ ] 0 = 1 . 12 4 . 80 × 110 . 98 [Ca 2+ ] 0 = 2.10 x 10 -3 M Na 2 CO 3 (s) → 2 Na + (aq) + CO 3 2- (aq) From the stoichiometry of this reaction: n CO32-,0 = n Na2CO3,0 ⇒ [CO 3 2- ] 0 = [Na 2 CO 3 ] 0 = 5.00 x 10 -6 M Q sp = [Ca 2+ ] 0 [CO 3 2- ] 0 Q sp = 2.10 x 10 -3 x 5.00 x 10 -6 Q sp = 1.05 x 10 -8 > K sp ⇒ CaCO 3 (s) forms

Midterm 2

General Chemistry 3

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Question 1

The combustion of carbon monoxide is represented by the following equation: CO (g) + ½ O₂ (g) → CO₂ (g) 1) Determine the standard enthalpy change ΔH⁰_rxn for the combustion of CO at 298 K 2) Determine the standard entropy change ΔS⁰_rxn 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 K_eq 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

Solution

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) [ΔH⁰_rxn] ΔH⁰_rxn = 110.5 – 393.5 = - 283.0 kJ.mol^-1 2) ΔS⁰_rxn = S⁰ [CO₂ (g)] - S⁰ [CO (g)] –

\frac{1}{2}

x S⁰ [O₂ (g)]

ΔS⁰_rxn = 213.7 – 197.7 – $\frac{1}{2}$ x 205.1

ΔS⁰_rxn = - 86.5 J.mol^-1.K^-1

3) ΔG⁰_rxn = ΔH⁰_rxn - TΔS⁰_rxn

ΔG⁰_rxn = - 283.0 x 10³ – 298 x (- 86.5)

ΔG⁰_rxn = - 257.2 x 10³ J.mol^-1 = - 257.2 kJ.mol^-1

ΔG⁰_rxn < 0 ⇒ The reaction is spontaneous at 298 K

4) ΔG⁰_rxn = - RT ln K_eq

K_eq = exp (-ΔG⁰_rxn/RT)

K_eq = 1.219 x 10⁴⁵

Question 2

1) What is the sign of ΔH_rxn 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 ΔH_rxn for this reaction

Solution

1) ΔH_rxn = Σ H_bonds(reactants) - Σ H_bonds(products)

If Σ H_bonds(products) > Σ H_bonds(reactants), ΔH_rxn < 0

2) The reaction is spontaneous when ΔG_rxn < 0

ΔG_rxn = ΔH_rxn - TΔS_rxn

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

At low temperature: ΔH_rxn >> TΔS_rxn ⇒ ΔG_rxn ≈ ΔH_rxn> 0

Question 3

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:

ΔG⁰_f [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

Solution

1) ΔS⁰_f = 2 x S⁰[NH₃ (g)] - S⁰[N₂ (g)] – 3 x S⁰[H₂ (g)]

ΔS⁰_f = 2 x 192.3 – 191.5 – 3 x 198.6

ΔS⁰_f = - 402.7 J.mol^-1.K^-1

ΔG⁰_f = ΔH⁰_f - TΔS⁰_f

ΔH⁰_f = ΔG⁰_f + TΔS⁰_f

ΔH⁰_f = - 33.0 x 10³ - 298 x 402.7

ΔH⁰_f = - 153.0 kJ.mol^-1

2) ΔH⁰_f < 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) ΔG⁰_f = RT ln K_eq

K_eq = exp (-ΔG⁰_rxn/RT)

K_eq = 6.09 x 10⁵

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

n_H2 = $\frac{PV}{RT}$

n_H2 = $\frac{7.60 \times 10^{4} \times 0.235 \times 10^{- 3}}{8.314 \times 298}$

n_H2 = 7.21 x 10^-3 mol

According to the stoichiometry of the reaction:

3 moles of H₂ form 2 moles of NH₃

n_NH3 / 2 = n_H2 / 3
n_NH3= $\frac{2}{3}$ n_H2 = 4.81 x 10^-3 mol

n_NH3 = [NH₃] x V

[NH₃] = 9.61 x 10^-3 M

Question 4

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: K_sp [CaCO₃ (s)] = 2.8 x 10^-9

Solution

Dissolution reaction: CaCO₃ (s) $⇌$ Ca²⁺ (aq) + CO₃^2- (aq) [K_sp = 2.8 x 10^-9]

CaCO₃ (s) forms if Q_sp > K_sp

Let’s determine Q_sp:

Q_sp = [Ca²⁺]₀ [CO₃^2-]₀

CaCl₂ (s) → Ca²⁺ (aq) + 2 Cl^- (aq)

From the stoichiometry of this reaction: n_Ca2+,0 = n_CaCl2,0

[Ca²⁺]₀ = $\frac{n_{Ca 2 +, 0}}{V}$ = $\frac{n_{CaCl 2, 0}}{V}$ = $\frac{m_{CaCl 2, 0}}{M_{CaCl 2} \times V}$

[Ca²⁺]₀ = $\frac{1.12}{4.80 \times 110.98}$

[Ca²⁺]₀ = 2.10 x 10^-3 M

Na₂CO₃ (s) → 2 Na⁺ (aq) + CO₃^2- (aq)

From the stoichiometry of this reaction: n_CO32-,0 = n_Na2CO3,0

⇒ [CO₃^2-]₀ = [Na₂CO₃]₀ = 5.00 x 10^-6 M

Q_sp = [Ca²⁺]₀ [CO₃^2-]₀

Q_sp = 2.10 x 10^-3 x 5.00 x 10^-6

Q_sp = 1.05 x 10^-8 > K_sp

⇒ CaCO₃ (s) forms

Question 5

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

Calculate the pH of this solution.

Solution

n_H+,0 = [H⁺]₀ x V_HCl = 140 x 10^-3 x 0.175

n_H+,0 = 2.45 x 10^-2 mol

n_HO-,0 = [HO^-]₀ x V_NaOH = 350 x 10^-3 x 0.0750

n_HO-,0 = 2.63 x 10^-2 mol

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

n_HO-,f = n_HO-,0 - n_H+,0

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

[HO^-]_f = $\frac{n_{{HO}^{-}, f}}{V}$

[HO^-]_f = $\frac{1.75 \times 10^{- 3}}{490 \times 10^{- 3}}$

[HO^-]_f = 3.57 x 10^-3 M

pOH = -log [HO^-]_f = 2.45

pH = 14 – pOH = 11.6

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