# Midterm 2

General Chemistry 3

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

CO (g) + ½ O_{2} (g) → CO_{2} (g)

1) Determine the standard enthalpy change ΔH^{0}_{rxn} for the combustion of CO at 298 K

2) Determine the standard entropy change ΔS^{0}_{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 O_{2} (g) → CO (g) [ΔH^{0} = - 110.5 kJ.mol^{-1}]

C (s) + O_{2} (g) → CO_{2} (g) [ΔH^{0} = - 393.5 kJ.mol^{-1}]

S^{0} [CO (g)] = 197.7 J.mol^{-1}.K^{-1}

S^{0} [CO_{2} (g)] = 213.7 J.mol^{-1}.K^{-1}

S^{0} [O_{2} (g)] = 205.1 J.mol^{-1}.K^{-1}

1) CO (g) → C (s) + ½ O_{2} (g) [ΔH^{0} = 110.5 kJ.mol^{-1}]

+ C (s) + O_{2} (g) → CO_{2} (g) [ΔH^{0} = - 393.5 kJ.mol^{-1}]

= CO (g) + ½ O_{2} (g) → CO_{2} (g) [ΔH^{0}_{rxn}]

ΔH^{0}_{rxn} = 110.5 – 393.5 = - 283.0 kJ.mol^{-1}

2) ΔS^{0}_{rxn} = S^{0} [CO_{2} (g)] - S^{0} [CO (g)] – $\frac{1}{2}$ x S^{0} [O_{2} (g)]

ΔS^{0}_{rxn} = 213.7 – 197.7 – $\frac{1}{2}$ x 205.1

ΔS^{0}_{rxn} = - 86.5 J.mol^{-1}.K^{-1}

3) ΔG^{0}_{rxn} = ΔH^{0}_{rxn} - TΔS^{0}_{rxn}

ΔG^{0}_{rxn} = - 283.0 x 10^{3} – 298 x (- 86.5)

ΔG^{0}_{rxn} = - 257.2 x 10^{3} J.mol^{-1} = - 257.2 kJ.mol^{-1}

ΔG^{0}_{rxn} < 0 ⇒ The reaction is spontaneous at 298 K

4) ΔG^{0}_{rxn} = - RT ln K_{eq}

K_{eq} = exp (-ΔG^{0}_{rxn}/RT)

K_{eq} = 1.219 x 10^{45}

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

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

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

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 H_{2} form 2 moles of NH_{3}

n_{NH3} / 2 = n_{H2} / 3

n_{NH3 }= $\frac{2}{3}$ 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) $\rightleftharpoons $ 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} = $\frac{{\mathrm{n}}_{\mathrm{Ca}2+,0}}{\mathrm{V}}$ = $\frac{{\mathrm{n}}_{\mathrm{CaCl}2,0}}{\mathrm{V}}$ = $\frac{{\mathrm{m}}_{\mathrm{CaCl}2,0}}{{\mathrm{M}}_{\mathrm{CaCl}2}\times \mathrm{V}}$

[Ca^{2+}]_{0} = $\frac{1.12}{4.80\times 110.98}$

[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

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

Calculate the pH of this 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} = $\frac{{\mathrm{n}}_{{\mathrm{HO}}^{-},\mathrm{f}}}{\mathrm{V}}$

[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