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CHEMISTRY THE CENTRAL SCIENCE

CHAPTER 5: THERMOCHEMISTRY


#33a
The reaction:
2Mg(s) + O2(g) --> 2MgO(s) ΔH = -1204 kJ
The negative sign means that energy is lost by the reacting particles so the reaction is exothermic.

#33b
The number by the side of the equation refers to the amount of energy per mol.
Mol of Mg = [2.40 g] / [24.305 g/mol] = 0.0987 mol
There are two mol of Mg per every 1204 kJ of energy (2:1 ratio).
Heat transferred = [-1204 kJ/mol] X [(1/2) X 0.0987 mol] = 59.4 kJ

#33c
Notice that the ratio MgO:energy is 2:1
Mol of MgO = [2 X 96.0 kJ] / [1204 kJ/mol] = 0.159 mol
grams of MgO = [0.159 mol] X [40.3 g/mol] = 6.43 g

#33d
Notice that the ratio MgO:energy is 2:1
Mol of MgO = [7.50 g] / [40.3 g/mol] = 0.186 mol
energy absorbed = [(1/2) X 0.186 mol] X [1204 kJ/mol] = 112 kJ

#49
mcΔT = Q
m = 100.0 g of water + 9.55 g of sodium hydroxide = 109.55 g
ΔT = T2 T1 = [47.4 23.6] 0C = 23.80C
[109.55 g] X [4.184 J/g x 0C] X [23.80C] = 10900 J
Mol of NaOH = [9.55 g] / [40.0] = 0.23875 mol
ΔH = [-Q] / [n] = [-10900 J] / [0.23875 mol] = -45654 J/mol --> -45.65 kJ/mol

#77a
C8H18(liquid) + (25/2)O2(g) --> 8CO2(g) + 9H2O(g) ΔH = -5069 kJ

#77b
The reaction for the formation of the octane from its elements is:
8C(s) + 9H2(g) --> C8H18(liquid)
Carbon is graphite because diamond is too expensive.
ΔH0 = Σ ΔHf0 Products - ΣΔHf0Reactants
Σ ΔHf0 Products = ΣΔHf0Reactants + ΔH0
Σ ΔHf0 C8H18(liquid) = [8 X (-393.5 kJ)] + [9 X (-241.82 kJ)] + 5069 kJ = -255 kJ

#97a
ΔT for the copper block = [30.1 100.4] 0C = 70.3 0C
Q = mcΔT = [121.0 g] X [0.385 J/g x 0C] X [70.3 0C] = -3.27 X 103 J
Heat is lost because Q is negative.

#97b
ΔT for water = [30.1 25.1] 0C = 5.00 0C
Q = mcΔT = [150.0 g] X [4.184 J/g x 0C] X [5.00 0C] = 3.14 X 103 J
Heat is gained because Q is positive.

#97c
Energy difference = [3.27 X 103 J 3.14 X 103 J] = 130 J
Heat Capacity = Q / [ΔT] = [130 J] / [5.00 0C] = 26.0 J/deg
deg applies to both Kelvin(K) and Celcius(0C) since they have the same magnitude and ΔT only measures the difference between two numbers. 0C = K + 273

#97d
If the cup is made of a perfectly insulating material, all the heat from copper gets transferred only to the water.
Q / [mc] = ΔT = [3.27 X 103 J] / [150 g X 4.184 J/g x 0C] = 5.21 0C
T2 = T1 + 5.21 0C = 30.3 0C

#107a
You can find this equation in any general chemistry textbook:
CH4(g) + 2O2(g) --> CO2(g) + H2(liquid) ΔH = -890.36 kJ/mol CH4
But there is a more difficult way of doing it. First get the enthalpies of formation of reactants and products:
ΔHf0 CO2(g) = -393.5 kJ/mol
ΔHf0 H2(liquid) (g) = -285.83 kJ/mol
ΔHf0 O2(g) = 0.0 kJ/mol
ΔHf0 CH4(g) = -74.8 kJ/mol
Then solve the equation:
ΔH0 = Σ ΔHf0 Products - ΣΔHf0Reactants
ΔH0 = -393.5 kJ + [2 X (-285.83 kJ)] [-74.8 kJ - 2(0) kJ] = 890.36 kJ
That is the energy per 6.022 X 1023 molecules of CH4 reacted.
Energy per molecule = 890.36 kJ / 6.022 X 1023 = 1.479 X 10-21 kJ = 1.479 X 10-18 J
(There are 1000 Joules in one kiloJoule)

#107b
From the back of the book:
1.0 keV = 96.485 X 106 J/mol, and since each mol of photon carries 8 keV,
Energy per photon = [8 X 96.485 X 106 J] / 6.022 X 1023 = 1.479 X 10-21 kJ = 1.282 X 10-15 J
[1.282 X 10-15 J from X-ray] / [1.479 X 10-18 J from CH4(g)] = 866.6
X-rays have more energy per unit than CH4(g)

#111a
Mol of CuSO4 = Molarity X Volume
Mol of CuSO4 = [1.00 mol/L] X [50.0 X 10-3 L] = 50.0 X 10-3 mol
Since there is one Cu per every CuSO4
Grams of Cu = [50.0 X 10-3 mol] X [63.55 g/mol] = 3.177 g of Cu

#111b
From the equation:
Cu2+ + SO42- + 2K+(aq) + 2OH-(aq) --> 2K+(aq) + SO42-(aq) + Cu(OH)2(s)
The precipitate is Cu(OH)2

#111c
Complete equation from above, net ionic:
Cu2+ + 2OH-(aq) --> Cu(OH)2(s)

#111d
ΔT = T2 - T1 = [27.7 21.5] 0 C = 6.2 0 C
m = 100.0 mL X [1.000 g/mL] = 100.0 g of solution
Q = mcΔT = 100.0 g X [4.184 J/g x 0 C] X 6.2 0 C = 2584.08 J
Mol of KOH = [2.00 mol/L] X [50.0 X 10-3 L] = 100.0 X 10-3 mol
From the stoichiometry get 1:2 ratio for Cu:OH- so there is just enough of both reactants. There is no limiting reactant to worry about in this problem; but this is not always the case.
Number of moles of Product = 50.0 X 10-3 mol
ΔH = -Q / n = [2584.08 J] / [50.0 X 10-3 mol] = -51881.6 J/mol

#112a Complete equation:
Ag+ + NO3- + Na+(aq) + Cl-(aq) --> Na+(aq) + NO3-(aq) + AgCl(s)
Net ionic:
Ag+ + Cl-(aq) --> AgCl(s)
ΔHf0 AgCl(s) = -127.0 kJ/mol
ΔHf0 Ag+(aq) = 105.90 kJ/mol
ΔHf0 Cl-(aq) = -167.2 kJ/mol
ΔH0 = Σ ΔHf0 Products - ΣΔHf0Reactants
ΔH0 = -127.0 kJ -105.90 kJ [-167.2 kJ] = -65.7 kJ

#112b
The net ionic equation defines the reaction outcome. The other ions appear on both sides of the equation (spectator ions) so the value obtained above applies to both the complete and the net ionic equation.

#112c
Now AgNO3(aq) is the product so the equation:
ΔH0 = Σ ΔHf0 Products - ΣΔHf0Reactants
becomes:
Σ ΔHf0 + Σ ΔHf0 Products = ΣΔHf0Reactants + ΣΔH0
ΔHf0 AgNO3(aq) = -446.2 kJ + (-127.0 kJ) (-407.1 kJ) (-65.7 kJ)
ΔHf0 AgNO3(aq) = -100.4 kJ/mol

#113a
Mol of CO2 = [21.83 g] / [44.01 g/mol] = 0.4960 mol
From mole ratio CO2:C is 1:1
grams of C = [0.4960 mol] X [12.0 g/mol] = 5.95 g
Mol of H2O = [4.47 g] / [18.02 g/mol] = 0.2481 mol
From mole ratio H2O:H is 2:1
grams of H = [0.2481 mol] X [1.000 g/mol] = 0.4961 g
Mass of compound = [5.95 + 0.4961] g = 6.45 g

#113b
Carbon --> 5.95 / 12.01 = 0.4960
Hydrogen --> 0.4961 / 1.000 = 0.4960
Carbon --> 0.4960 / 0.4960 = 1
Hydrogen --> 0.4960 / 0.4960 = 1
From 1:1 ratio the empirical formula is CH

#113c
6.45 g CH + O2(g) --> 21.83 g CO2(g) + 4.47 g H2O(g) + 311 kJ
ΔH0 = Σ ΔHf0 Products - ΣΔHf0Reactants
ΔHf0 CH Sample = ΔHf0CO2(g) + ΔHf0H2O(g) + ΣΔH0
Use number of mol from part a
ΔHf0CO2(g) = [-393.5 kJ/mol] X [0.4960 mol] = 195.2 kJ
ΔHf0H2O(g) = [-241.82 kJ/mol] X [0.2481 mol] = 60.00 kJ
ΔHf0 CH Sample = -195.2 kJ 60.00 kJ - [-311 kJ] = 55.8 kJ
Mol of CH units = [6.45 g] / [13.0 g/mol] = 0.496 mol CH
ΔHf0 CH Sample = [55.8 kJ] / [0.496 mol] = 112.5 kJ/mol





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