# Final

General Chemistry 3

is over.

Check the solutions on the questions and answer that you have decided correctly

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Balance the following redox reaction in basic solution:

P4 (s) + NO3- (aq) → PO4- (aq) + NO (g)

1) Divide the reaction into half-reactions:

P4 (aq) → 4 PO4- (aq)

NO3- (aq) → NO (g)

2) Balance O atoms by adding H2O molecules:

P4 (aq) + 16 H2O (l) → 4 PO4- (aq)

NO3- (aq) → NO (g) + 2 H2O (l)

3) Balance H atoms by adding H+ ions:

P4 (aq) + 16 H2O (l) → 4 PO4- (aq) + 32 H+ (aq)

NO3- (aq) + 4 H+ (aq) → NO (g) + 2 H2O (l)

4) Balance charge by adding electrons:

P4 (aq) + 16 H2O (l) → 4 PO4- (aq) + 32 H+ (aq) + 28 e-

NO3- (aq) + 4 H+ (aq) + 3 e- → NO (g) + 2 H2O (l)

5) Multiply each half-reaction by an integer to have: number of electrons lost = number of electrons gained:

3 P4 (aq) + 48 H2O (l) → 12 PO4- (aq) + 96 H+ (aq) + 84 e-

28 NO3- (aq) + 112 H+ (aq) + 84 e- → 28 NO (g) + 56 H2O (l)

3 P4 (aq) + 28 NO3- (aq) + 16 H+ (aq) → 12 PO4- (aq) + 28 NO (g) + 8 H2O (l)

7) Add HO- ions to react with all H+ ions: HO- + H+ à H2O:

3 P4 (aq) + 28 NO3- (aq) + 16 H+ (aq) + 16 HO- (aq) → 12 PO4- (aq) + 28 NO (g) + 8 H2O (l) + 16 HO- (aq)

The balanced equation in basic solution is:

3 P4 (aq) + 28 NO3- (aq) + 8 H2O (l) → 12 PO4- (aq) + 28 NO (g) + 16 HO- (aq)

Calculate the values of ΔG0 and Keq for a copper / zinc voltaic cell at 25°C.

Data: E0 [Cu2+/Cu] = 0.340 V

E0 [Zn2+/Zn] = - 0.763 V

According to E0 values: copper preferentially reduced, zinc oxidized

Cu2+ (aq) + 2 e- → Cu (s) (reduction à cathode)

Zn (s) → Zn2+ (aq) + 2 e- (oxidation à anode)

ΔG0 = - νeFEcell0

νe = stoichiometric coefficient of the electrons in the 2 half reaction equations = 2

F = Faraday’s constant = 96485 C.mol-1

Ecell0 = electromotive force emf of the cell (in V)

Ecell0 = E0 (reduction) - E0 (oxydation)

Ecell0 = 0.340 – (-0.763) = 1.10 V

ΔG0 = - νeFEcell0 = - 2 x 96485 x 1.10 = -212 kJ.mol-1

ln Keq =  $\frac{{\mathrm{\nu }}_{\mathrm{e}}\mathrm{F}}{\mathrm{RT}}$ E0cell

Keq = equilibrium constant

νe = stoichiometric coefficient of the electrons in the 2 half reaction equations = 2

F = Faraday’s constant = 96485 C.mol-1

E0cell = standard cell voltage (in V) = 1.10 V

R = ideal gas constant = 8.314 J.mol-1.K-1

T = temperature (in K) = 298 K

ln Keq =  = 85.7

Keq = exp (85.7) = 1.61 x 1037

Here is an oxidation-reduction reaction:

2 N2H4 + N2O4 → 3 N2 + 4 H2O

Assign the oxidation numbers to the atoms and identify the oxidizing and reducing agents.

N2H4 ⇒ H: +1 (Hydrogen atoms with a nonmetal: +1) and N: -2 (- 4x1/2)

N2O4 ⇒ O: -2 (Oxygen atoms in compounds: -2) and N: +4 (4x2/2)

N2 ⇒ N: 0

H2O ⇒ O: -2 and H: +1

N2H4 is oxidized (oxidation state: -2 to 0) ⇒ reducing agent

N2O4 is reduced (oxidation state: +4 to 0) ⇒ oxidizing agent

Here is a reaction: Cl2 (g) → 2 Cl (g)

1) Predict if this reaction is endothermic or exothermic

2) Predict if ΔS0 is positive, negative or zero

3) Is this reaction spontaneous at all temperature? Explain

1) Chlorine is naturally a diatomic element ⇒ Cl2 is most stable elemental state

AND this reaction results in the breaking of a bond ⇒ heat taken

⇒ endothermic reaction (ΔH0rxn > 0)

2) The disorder is increasing: from 1 mole of gas to 2 moles of gas

⇒ ΔS0rxn > 0

3) The reaction is spontaneous when ΔG0rxn < 0 ⇒ ΔH0rxn – TΔS0rxn < 0

The reaction is only spontaneous at high temperature (when T > ΔH0rxn/ΔS0rxn)

NaI undergoes a solid → liquid phase transition at 661°C.

The heat energy qrev required for the phase transition at this temperature is 44.3 kJ.mol-1.

Determine the associated entropy change ΔS for this transition.

ΔS = $\frac{{\mathrm{q}}_{\mathrm{rev}}}{\mathrm{T}}$

ΔS =

ΔS = 47.4 J.mol-1.K-1

Determine the pH of a solution that contains 0.232 M formic acid HCOOH and 0.0980 M sodium formate NaCHO2.

Data: Ka [HCOOH/HCOO-] = 1.80 x 10-4

Henderson-Hasselbalch equation:

pH = pKa + log $\frac{\left[\mathrm{base}\right]}{\left[\mathrm{acid}\right]}$

pH = - log (1.80 x 10-4) + log $\left(\frac{0.0980}{0.232}\right)$

pH = 3.37

250 mL of 4.25 x 10-4 M  Pb(NO3)2 is added to 275 mL of 6.25 x 10-5 M  NaI.

Will a precipitate form?

Data: Ksp [PbI2] = 7.10 x 10-9

Dissolution reaction:

PbI2 (s) → Pb2+ (aq) + 2 I- (aq)    [Ksp]

PbI2 (s) forms if Qsp > Ksp

Let’s determine Qsp:

Qsp = [Pb2+]0 [I-]02

Pb(NO3)2 (s) → Pb2+ (aq) + 2 NO3- (aq)

From the stoichiometry of this reaction: nPb2+,0 = nPb(NO3)2,0

⇒ nPb2+,0 = [Pb(NO3)2]0 x VPb(NO3)2

⇒ nPb2+,0 = 0.250 x 4.25 x 10-4 = 1.06 x 10-4 mol

NaI (s) → Na+ (aq) + I- (aq)

From the stoichiometry of this reaction: nI-,0 = nNaI,0

⇒ nI-,0 = [NaI]0 x VNaI = 1.72 x 10-5 mol

Qsp = [Pb2+]0 [I-]02

Qsp =

Qsp = 2.17 x 10-13 < Ksp

PbI2 (s) does not form

Determine the freezing point and the boiling point of an aqueous solution of LiCl that contains 3.25 g of LiCl and 10.0 mL of water.

Data: Kf [H2O] = 1.86 K.kg.mol-1 [freezing point depression constant]

Kb [H2O] = 0.512 K.kg.mol-1 [boiling point elevation constant]

LiCl (s) → Li+ (aq) + Cl- (aq)

solution molality m = $\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{m}}_{\mathrm{solution}}}$

m =

m =

m = 7.67 mol.kg-1

The magnitude of the freezing point depression ΔTf (in K) is:

ΔTf = - i x Kf x m

ΔTf = Tf solution – Tf water

i = van’t Hoff i-factor = number of ions produced per formula unit = 2

m = solution molality = 7.67 mol.kg-1

Kf = proportionality constant (freezing point depression constant)

Tf solution – 273 = - 2 x 1.86 x 7.67

Tf solution = 244.5 K

The magnitude of the boiling point elevation ΔTb (in K) is:

ΔTb = i x Kb x m

ΔTb = Tb solution – Tb water

i = van’t Hoff i-factor = number of ions produced per formula unit = 2

m = solution molality (in mol.kg-1) = 7.67

Kb = proportionality constant (boiling point elevation constant)

Tf solution – 373 = 2 x 0.512 x 7.67

Tf solution = 380.9 K

A simple A → B reaction is determined to be a second order reaction with a rate constant k = 4.72 x 10-4 M-1.s-1.

Determine the half-life of the reaction if [A]0 = 5.25 x 10-2 M.

At t = t1/2:

[A] = $\frac{{\left[\mathrm{A}\right]}_{0}}{2}$

⇒ $\frac{2}{{\left[\mathrm{A}\right]}_{0}}$ = $\frac{1}{{\left[\mathrm{A}\right]}_{0}}$ + kt1/2

⇒ t1/2 = $\frac{1}{\mathrm{k}{\left[\mathrm{A}\right]}_{0}}$

⇒ t1/2 =

⇒ t1/2 = 4.04 x 104 s

For a first order reaction A → B:

k1 = 1.24 x 10-5 s-1 at T1 = 370 K

and k2 = 5.36 x 10-3 s-1 at T2 = 650 K

Determine the activation energy of this reaction.

ln $\frac{{\mathrm{k}}_{2}}{{\mathrm{k}}_{1}}$ = $\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}$ x

Ea = R ln $\frac{{\mathrm{k}}_{2}}{{\mathrm{k}}_{1}}$ x

Ea = 8.314 x ln x

Ea = 4.33 x 104 J.mol-1