# Midterm 2

General Chemistry 2

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When a 14.50 g of potassium chlorate is dissolved in 200.0 mL of water in a vacuum flask, the temperature falls from 25.0°C to 19.0°C. The heat capacity of the solution is 4.184 J.g-1.°C-1.

What is the enthalpy change ΔHdiss (in kJ.mol-1) for the dissolution of KClO3 in water?

cS = $\frac{{\mathrm{c}}_{\mathrm{P}}}{\mathrm{m}}$ =

cS = specific heat capacity of the solution (in J.g-1.°C-1)
qP = energy added as heat (in J) = nKClO3 ΔHdiss
n = number of moles (in mol)
ΔT = Tf – Ti (in °C)
m = mass (in g)

cS =      ⇒      ΔH =

nKClO3 = $\frac{{\mathrm{m}}_{\mathrm{KClO}3}}{{\mathrm{M}}_{\mathrm{KClO}3}}$ =  $\frac{14.50}{122.6}$ = 0.118 mol

ΔH =  = 45.6 x 104 J.mol-1 = 45.6 kJ.mol-1

0.522 g of solid Mg(OH)2 are dissolved in 25.0mL of a 4.0 M HNO3 solution. At the end, the solution remains acidic.

1) Write the chemical equation of this reaction

2) If you assume the reaction is complete, calculate the concentration of Mg2+, NO3- and H+.

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:

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

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

Mg(NO3)2 = Mg2+ + 2 NO3-

[Mg2+] = $\frac{{\mathrm{n}}_{\mathrm{Mg}2+}}{{\mathrm{V}}_{\mathrm{sol}}}$ = $\frac{{\mathrm{n}}_{\mathrm{Mg}\left(\mathrm{NO}3\right)2}}{{\mathrm{V}}_{\mathrm{sol}}}$ =  = 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:
nHNO3 = nHNO3 (t=0) – 2 x nMg(OH)2 (t=0)

nHNO3 = [HNO3] x Vsol – 2 x $\frac{{\mathrm{m}}_{\mathrm{Mg}\left(\mathrm{OH}\right)2}}{{\mathrm{M}}_{\mathrm{Mg}\left(\mathrm{OH}\right)2}}$

nHNO3 = 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+] = $\frac{{\mathrm{n}}_{\mathrm{H}+}}{{\mathrm{V}}_{\mathrm{sol}}}$ = $\frac{{\mathrm{n}}_{\mathrm{HNO}3}}{{\mathrm{V}}_{\mathrm{sol}}}$ =  = 3.28 M

The density of H2O(s) is 0.920 g.cm-3 at 0°C and is therefore lower than that of H2O(l). So water expands when it freezes.

How much work does 50.0 g of water do when it freezes at 0°C and bursts a water pipe with an opposing pressure of 1000 atm?

By definition, work W = -Pext ΔV

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

Pext = 1000 atm = 1.013 x 108 Pa

density d = $\frac{\mathrm{m}}{\mathrm{V}}$ ⇒ V = $\frac{\mathrm{m}}{\mathrm{d}}$
VH2O(l) = Vi = $\frac{50.0}{1.00}$ = 50.0 cm3 = 50.0 x 10-6 m3

VH2O(g) = Vf = $\frac{50.0}{0.920}$ =  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

Given:

4 NH3 (g) + 3 O2 (g) $⇌$ 2 N2 (g) + 6 H2O (l)              ΔH1 = -1534 kJ

N2O (g) + H2 (g) $⇌$ N2 (g) + H2O (l)                            ΔH2 = -367.4 kJ

Determine ΔH for the following reaction:

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

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

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

+ N2 (g) + H2O (l) $⇄$ N2O (g) + H2 (g)

According to Hess’s Law:

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

How many liters of carbon dioxide gas are produced by the complete combustion of 50.0 mol of glucose C6H12O6 at 25°C and 1 atm?

Combustion of glucose:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

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

⇒ nCO2 = 6 x nglucose = 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

Ethylene C2H4 reacts with diazene H2N2 to afford ethane C2H6 and nitrogen gas.

1) Draw the Lewis structure of each compound

2) Write a chemical equation for this reaction

3) Calculate the enthalpy change ΔHrxn for this reaction

Data:

Hbond(H-H) = 436 kJ.mol-1

Hbond(C-H) = 413 kJ.mol-1

Hbond(N-H) = 391 kJ.mol-1;

Hbond(C-C) = 348 kJ.mol-1

Hbond(C=C) = 614 kJ.mol-1

Hbond(N-N) = 163 kJ.mol-1;

Hbond(N=N) = 418 kJ.mol-1

Hbond(N$\equiv$N) = 941 kJ.mol-1

1) Lewis structures: 2) C2H4 + N2H2 → C2H6 + N2

3) ΔHrxn = ΣHbond broken  - ΣHbond formed

⇒ ΔHrxn = ΣHbond (reactants) - ΣHbond (products)

⇒ ΔHrxn = 4 Hbond (C-H) + Hbond (C=C) + 2 Hbond (N-H) + Hbond (N=N)
- 6 Hbond (C-H) - Hbond (C-C) - Hbond (N$\equiv$N)

⇒ ΔHrxn = -301 kJ.mol-1