Thermochemistry
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A. Enthalpy ( H) is the amount of heat content.
1. Heat content is accounted for by a change in "heat flow" or enthalpy of
the reaction system.
1. Endothermic reaction: H > 0
(i.e., H products >H reactants).
Heat absorbed goes to increase the enthalpy of the reaction system.
2. Exothermic reaction: H < 0
(i.e., H products < H reactants).
Heat is evolved at the expense of the reaction system.
2. Thermochemical Equation: specify H in kilojoules/mole.
a. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + 890.3 kJ
H = -890 kJ
6.00kJ + H2O(s)
H = +6.00kJ
H2O(l)
! In some textbooks H is written as a product or reactant !
The preceding is based upon the Law of Conservation of Energy
(James Joule, 1818-1889, Joule also developed the First Law of
Thermodynamics): energy is neither created nor destroyed in
ordinary chemical or physical changes.
b. Quantitative H
H = qreaction mixture (at constant temperature only)
q = (m)( t)(Cp)
q = heat absorbed by the water in joules (J)
m = mass of substance
t = tfinal - tinitial
Cp = specific heat of water = 4.184
When using moles, molar heat capacity is used. The units are
1 cal = 4.184 J
B. Calorimetry
1. Coffee-cup calorimeter (only used for reactions in solution, must be at
constant pressure)
qreaction=-qwater
2. Bomb calorimeter (reaction gases, and must have constant volume)
qreaction=-(qwater+qbomb)
qbomb=C t, where C is the calorimeter constant (Cv of bomb x mass of
bomb, really same equation)
3. H vs. E for chemical reactions
H=qp since E=qp-P V
substituting gives H= E+P V
where P will usually be in atmospheric pressure, and V is volume change
at that pressure.
C. Laws of Thermochemistry
1. The magnitude of H is directly proportional to the amount of reactant or
product.
-Thus H can be used as a conversion factor in a balanced equation to
obtain amounts of reactant/product or H itself. (mole to mole ratio's).
2. H for a reaction is equal in magnitude but opposite in sign to H for the
reverse reaction.
Problems 1: H Calculation
.
When 1 mol of methane is burned at constant pressure, 890.3kJ of
energy is released as heat. Calculate H for a process in which a
5.8 gram sample of methane is burned at constant pressure.
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + 890.3 kJ
= 320 kJ
H = -320 kJ
A. For the reaction of methane with oxygen given in the notes,
calculate the H in kJ if 5.8 grams of oxygen are consumed in the
process.
= 81 kJ
H= -81kJ
B. Ammonium nitrate, NH4NO3, is commonly used as an explosive. It
decomposes by the following reaction:
NH4NO3
N2O(g) + 2 H2O(g) + 37.0kJ
H = -37.0 kJ
If 72.0 grams of H2O are formed from the reaction, how much heat
was released?
= 73.9 kJ
3. Hess' Law: The value of H for a reaction is the same whether it occurs
directly or in a series of steps (state function).
Htotal = H1 + H2
often used to calculate H for one step, knowing H for all steps and for
the overall reaction.
**All of the laws of thermochemistry follow from the fact that the
enthalpy H of a substance is one of its properties.**
D. Heats of Formation
Molar heat of formation ( Hf) is equal to the enthalpy change, H when one mole
of the compound is formed from the elements in their stable forms at 25oC and 1
atm is Ho (pronounced 'delta h naught'). Ho of a solution is of a 1M solution, at
1 atm and 25 oC.
Heats of formation are usually negative quantities.
H=
rules:
Hf products -
Hf reactants To apply the above relation, use the following
a. The contribution for each compound is found by multiplying the heat of
formation in kJ per mole by the number of moles of compound, given by
its coefficient in the balanced equation.
Heats of formation can be found in appendix A-4
b. Any element in its stable form is omitted.
Can also apply to heats of formation to ions.
Arbitrarily assign H+ ion to be zero. Hf H+(aq) = 0
Having established the above, a scale can be established with Hydrogen ion as the
base.
3. Calculate H0rxn for the following reaction.
2 C3H6(g) + 9 O2(g) 6 CO2(g) + 6 H2O(l)
*Appendix 4*
Hrxn = Hf products - Hf reactants
Hrxn = [ 6 H2O(l) + 6 CO2(g) ] - [ 2 C3H6(g) + 9 O2(g) ]
Hrxn = [ 6(-286 kJ) + 6 (-393.2 kJ) ] - [ 2(20.9 kJ) + 9(0)]
Hrxn = [ -1716 kJ + -2361 kJ ] - [41.8 kJ ]
Hrxn =
= 2059
4.
. Calculate H0 for 2Al(s) + Cr2O3(s) Al2O3(s) + 2Cr(s).
A. Compare this reaction to sample exercise 6.10, "thermite" reaction.
Which reaction yields more energy per gram of metal formed?
B.
Hrxn = [ Al2O3(s) + 2 Cr(s) ] - [ 2 Al(s) + Cr2O3(s) ]
Hrxn = [ (-1676 kJ) + 2(0 kJ) ] - [ 2(0 kJ) + -1128 kJ ]
Hrxn = [ -1676 kJ ] - [ -1128 kJ ]
Hrxn =
C.
= -10.1
Hrxn =
= -15.75
"Thermite" reaction releases more energy per gram of metal
formed.
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