Midterm 2

c_S = $\frac{{\mathrm{c}}_{\mathrm{P}}}{\mathrm{m}}$ = $\frac{{\mathrm{q}}_{\mathrm{P}}}{∆\mathrm{T} \times \mathrm{m}}$

c_S = specific heat capacity of the solution (in J.g^-1.°C^-1)
q_P = energy added as heat (in J) = n_KClO3 ΔH_diss
n = number of moles (in mol)
ΔT = Tf – Ti (in °C)
m = mass (in g)

c_S = $\frac{{\mathrm{n}}_{\mathrm{KClO}3} \times ∆\mathrm{H}}{∆\mathrm{T} \times {\mathrm{m}}_{\mathrm{solution}}}$     ⇒      ΔH = $\frac{{\mathrm{c}}_{\mathrm{S}} \times ∆\mathrm{T} \times {\mathrm{m}}_{\mathrm{solution}}}{{\mathrm{n}}_{\mathrm{KClO}3}}$

n_KClO3 = $\frac{{\mathrm{m}}_{\mathrm{KClO}3}}{{\mathrm{M}}_{\mathrm{KClO}3}}$ =  $\frac{14.50}{122.6}$ = 0.118 mol

ΔH = $\frac{4.184 \times 6.0 \times 214.50}{0.118}$ = 45.6 x 10⁴ J.mol^-1 = 45.6 kJ.mol^-1

1) Mg(OH)₂ + 2 HNO₃ → 2 H₂O + Mg(NO₃)₂

2) The solution remains acidic: Mg(OH)₂ is the limiting reactant

Concentration of Mg²⁺:

According to the equation, if the reaction is complete:

n_Mg(NO3)2 = n_Mg(OH)2 = $\frac{{\mathrm{m}}_{\mathrm{Mg}\left(\mathrm{OH}\right)2}}{{\mathrm{M}}_{\mathrm{Mg}\left(\mathrm{OH}\right)2}}$

n_Mg(NO3)2 = $\frac{0.612}{58.309}$ = 8.95 x 10^-3 mol

Mg(NO₃)₂ = Mg²⁺ + 2 NO₃^-

[Mg²⁺] = $\frac{{\mathrm{n}}_{\mathrm{Mg}2+}}{{\mathrm{V}}_{\mathrm{sol}}}$ = $\frac{{\mathrm{n}}_{\mathrm{Mg}\left(\mathrm{NO}3\right)2}}{{\mathrm{V}}_{\mathrm{sol}}}$ = $\frac{8.95 \times {10}^{-3}}{25.0 \times {10}^{-3}}$ = 3.58 x 10^-1 M

Concentration of NO₃^-:

According to the equation 2 moles of NO₃^- are consumed to form 2 moles of NO₃^-
⇒ [NO₃^-] is unchanged ⇒ [NO₃^-] = [HNO₃]₀ = 4.0 M

Concentration of H⁺:

According to the equation: 2 moles of HNO₃ react with 1 mole of Mg(OH)₂.

Mg(OH)₂ is the limiting reactant:
n_HNO3 = n_HNO3 (t=0) – 2 x n_Mg(OH)2 (t=0)

n_HNO3 = [HNO₃] x V_sol – 2 x $\frac{{\mathrm{m}}_{\mathrm{Mg}\left(\mathrm{OH}\right)2}}{{\mathrm{M}}_{\mathrm{Mg}\left(\mathrm{OH}\right)2}}$

n_HNO3 = 4.0 x 25.0 x 10^-3 – 2 x $\frac{0.522}{58.309}$ = 8.21 x 10^-2 mol

HNO₃ = H⁺ + NO₃^-

[H⁺] = $\frac{{\mathrm{n}}_{\mathrm{H}+}}{{\mathrm{V}}_{\mathrm{sol}}}$ = $\frac{{\mathrm{n}}_{\mathrm{HNO}3}}{{\mathrm{V}}_{\mathrm{sol}}}$ = $\frac{5.83 \times {10}^{-2}}{25.0 \times {10}^{-3}}$ = 3.28 M

By definition, work W = -Pext ΔV

with Pext = constant external pressure (in Pa) and ΔV = Vf – V_i (in m³)

P_ext = 1000 atm = 1.013 x 10⁸ Pa

density d = $\frac{\mathrm{m}}{\mathrm{V}}$ ⇒ V = $\frac{\mathrm{m}}{\mathrm{d}}$
V_H2O(l) = V_i = $\frac{50.0}{1.00}$ = 50.0 cm³ = 50.0 x 10^-6 m³

V_H2O(g) = V_f = $\frac{50.0}{0.920}$ =  54.3 cm³ = 54.3 x 10^-6 m³

W = -Pext ΔV

⇒ W = -(1.013 x 10⁸) x (54.3-50.0) x 10^-6 = - 4.36 x 10² J

   2 NH₃ + 3/2 O₂ (g) $\rightleftharpoons$ N₂O (g) + 2 H₂O (l) + H₂ (g)

= 2 NH₃ + 3/2 O₂ (g) $\rightleftharpoons$ N₂ (g) + 3 H₂O (l)

+ N₂ (g) + H₂O (l) $\rightleftarrows$ N₂O (g) + H₂ (g)

According to Hess’s Law:

ΔH = ΔH₁ / 2 - ΔH₂ = -767 + 367.4 = -399.6 kJ

Combustion of glucose:

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

According to this equation: 1 mole of glucose affords 6 moles of CO₂

⇒ n_CO2 = 6 x n_glucose = 300 mol

Ideal gas law: PV = nRT

with P = pressure (in Pa) à 1 atm = 1.013 x 10⁵ Pa
V = volume (in m³)
n = number of moles (in mol)
R = ideal gas constant = 8.314 (in J.K^-1.mol^-1)
T = temperature (in K) à 0°C = 298 K

V = nRT / P = (300 x 8.314 x 298) / (1.013 x 10⁵)

⇒ V = 7.34 m³ = 7.34 x 10³ dm³ = 7.34 x 10³ L

1) Lewis structures:

2) C₂H₄ + N₂H₂ → C₂H₆ + N₂

3) ΔH_rxn = ΣH_{bond broken}  - ΣH_{bond formed}

⇒ ΔH_rxn = ΣH_bond (reactants) - ΣH_bond (products)

⇒ ΔH_rxn = 4 H_bond (C-H) + H_bond (C=C) + 2 H_bond (N-H) + H_bond (N=N)
                - 6 H_bond (C-H) - H_bond (C-C) - H_bond (N$\equiv$N)

⇒ ΔH_rxn = -301 kJ.mol^-1