Here is an oxidation-reduction reaction: 2 N 2 H 4 + N 2 O 4 → 3 N 2 + 4 H 2 O Assign the oxidation numbers to the atoms and identify the oxidizing and reducing agents.

N 2 H 4 ⇒ H: +1 (Hydrogen atoms with a nonmetal: +1) and N: -2 (- 4x1/2) N 2 O 4 ⇒ O: -2 (Oxygen atoms in compounds: -2) and N: +4 (4x2/2) N 2 ⇒ N: 0 H 2 O ⇒ O: -2 and H: +1 N 2 H 4 is oxidized (oxidation state: -2 to 0) ⇒ reducing agent N 2 O 4 is reduced (oxidation state: +4 to 0) ⇒ oxidizing agent

Here is a reaction: Cl 2 (g) → 2 Cl (g) 1) Predict if this reaction is endothermic or exothermic 2) Predict if ΔS 0 is positive, negative or zero 3) Is this reaction spontaneous at all temperature? Explain

1) Chlorine is naturally a diatomic element ⇒ Cl 2 is most stable elemental state AND this reaction results in the breaking of a bond ⇒ heat taken ⇒ endothermic reaction (ΔH 0 rxn > 0) 2) The disorder is increasing: from 1 mole of gas to 2 moles of gas ⇒ ΔS 0 rxn > 0 3) The reaction is spontaneous when ΔG 0 rxn < 0 ⇒ ΔH 0 rxn – TΔS 0 rxn < 0 The reaction is only spontaneous at high temperature (when T > ΔH 0 rxn /ΔS 0 rxn )

For a first order reaction A → B: k 1 = 1.24 x 10 -5 s -1 at T 1 = 370 K and k 2 = 5.36 x 10 -3 s -1 at T 2 = 650 K Determine the activation energy of this reaction.

ln k 2 k 1 = E a R x 1 T 1 - 1 T 2 Ea = R ln k 2 k 1 x T 1 T 2 T 2 - T 1 Ea = 8.314 x ln 5 . 36 × 10 - 3 1 . 24 × 10 - 5 x 370 × 650 280 Ea = 4.33 x 10 4 J.mol -1

Final

General Chemistry 3

The test is over.

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

You can revisit questions later in your account.

Question 1

Balance the following redox reaction in basic solution: P₄ (s) + NO₃^- (aq) → PO₄^- (aq) + NO (g)

Solution

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) 6) Add the half-reactions together: 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)

Question 2

Calculate the values of ΔG⁰ and K_eq for a copper / zinc voltaic cell at 25°C. Data: E⁰ [Cu²⁺/Cu] = 0.340 V E⁰ [Zn²⁺/Zn] = - 0.763 V

Solution

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 = - ν_eFE_cell⁰

ν_e = stoichiometric coefficient of the electrons in the 2 half reaction equations = 2 F = Faraday’s constant = 96485 C.mol^-1 E_cell⁰ = electromotive force emf of the cell (in V)

E_cell⁰ = E⁰ (reduction) - E⁰ (oxydation) E_cell⁰ = 0.340 – (-0.763) = 1.10 V ΔG⁰ = - ν_eFE_cell⁰ = - 2 x 96485 x 1.10 = -212 kJ.mol^-1

ln K_eq =

\frac{ν_{e} F}{RT}

E⁰_cell

K_eq = equilibrium constant

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

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

E⁰_cell = 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 K_eq = $\frac{2 \times 96485 \times 1.10}{8.314 \times 298}$ = 85.7

K_eq = exp (85.7) = 1.61 x 10³⁷

Question 3

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.

Solution

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

Question 4

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

Solution

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 (ΔH⁰_rxn > 0)

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

⇒ ΔS⁰_rxn > 0

3) The reaction is spontaneous when ΔG⁰_rxn < 0 ⇒ ΔH⁰_rxn – TΔS⁰_rxn < 0

The reaction is only spontaneous at high temperature (when T > ΔH⁰_rxn/ΔS⁰_rxn)

Question 5

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

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

Determine the associated entropy change ΔS for this transition.

Solution

ΔS = $\frac{q_{rev}}{T}$

ΔS = $\frac{44.3 \times 10^{3}}{661 + 273}$

ΔS = 47.4 J.mol^-1.K^-1

Question 6

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

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

Solution

Henderson-Hasselbalch equation:

pH = pK_a + log $\frac{[base]}{[acid]}$

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

pH = 3.37

Question 7

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: K_sp [PbI₂] = 7.10 x 10^-9

Solution

Dissolution reaction:

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

PbI₂ (s) forms if Q_sp > K_sp

Let’s determine Q_sp:

Q_sp = [Pb²⁺]₀ [I^-]₀²

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

From the stoichiometry of this reaction: n_Pb2+,0 = n_Pb(NO3)2,0

⇒ n_Pb2+,0 = [Pb(NO₃)₂]₀ x V_Pb(NO3)2

⇒ n_Pb2+,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: n_I-,0 = n_NaI,0

⇒ n_I-,0 = [NaI]₀ x V_NaI= 1.72 x 10^-5 mol

Q_sp = [Pb²⁺]₀ [I^-]₀²

Q_sp = $\frac{n_{{Pb}^{2 +}, 0} \times {n_{I^{-}, 0}}^{2}}{V^{3}}$

Q_sp = 2.17 x 10^-13 < K_sp

PbI₂ (s) does not form

Question 8

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

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

Solution

LiCl (s) → Li⁺ (aq) + Cl^- (aq)

solution molality m = $\frac{n_{solute}}{m_{solution}}$

m = $\frac{m_{LiCl}}{M_{LiCl} \times m_{solution}}$

m = $\frac{3.25}{42.39 \times 10.0 \times 10^{- 3}}$

m = 7.67 mol.kg^-1

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

ΔT_f = - i x K_f x m

ΔT_f = T_f _solution – T_{f water}

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

m = solution molality = 7.67 mol.kg^-1

K_f = proportionality constant (freezing point depression constant)

T_{f solution} – 273 = - 2 x 1.86 x 7.67

T_{f solution} = 244.5 K

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

ΔT_b = i x K_b x m

ΔT_b = T_{b solution} – T_{b water}

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

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

K_b = proportionality constant (boiling point elevation constant)

T_{f solution} – 373 = 2 x 0.512 x 7.67

T_{f solution} = 380.9 K

Question 9

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.

Solution

At t = t_1/2:

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

⇒ $\frac{2}{{[A]}_{0}}$ = $\frac{1}{{[A]}_{0}}$ + kt_1/2

⇒ t_1/2 = $\frac{1}{k {[A]}_{0}}$

⇒ t_1/2 = $\frac{1}{4.72 \times 10^{- 4} \times 5.25 \times 10^{- 2}}$

⇒ t_1/2 = 4.04 x 10⁴ s

Question 10

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.

Solution

ln $\frac{k_{2}}{k_{1}}$ = $\frac{E_{a}}{R}$ x $(\frac{1}{T_{1}} - \frac{1}{T_{2}})$

Ea = R ln $\frac{k_{2}}{k_{1}}$ x $\frac{T_{1} T_{2}}{T_{2} - T_{1}}$

Ea = 8.314 x ln $(\frac{5.36 \times 10^{- 3}}{1.24 \times 10^{- 5}})$ x $\frac{370 \times 650}{280}$

Ea = 4.33 x 10⁴ J.mol^-1

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