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Laws of Thermochemistry - Understanding Enthalpy and ...

The laws of thermochemistry are used to describe the heat or energy flow of chemical reactions. Understanding the values of a reaction's enthalpy will help
   
 Enthalpy ( delta H) is the amount of heat content.
Heat content is accounted for by a change in "heat flow" or enthalpy of the reaction system.
Endothermic reaction: delta H > 0
(i.e., H products >H reactants).
Heat absorbed goes to increase the enthalpy of the reaction system.
Exothermic reaction: delta H < 0
(i.e., H products < H reactants).
Heat is evolved at the expense of the reaction system.
Thermochemical Equation: specify delta H in kilojoules/mole.
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l) + 890.3 kJ
delta H = -890 kJ

6.00kJ + H2O(s) --> H2O(l)
delta H = +6.00kJ

! In some textbooks delta 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.
Quantitative delta H
delta H = qreaction mixture (at constant temperature only)

q = (m)( delta t)(Cp)

q = heat absorbed by the water in joules (J)
m = mass of substance
delta t = tfinal - tinitial
Cp = specific heat of water = 4.184 J/g oC

When using moles, molar heat capacity is used. The units are kJ/mol K

1 cal = 4.184 J

Calorimetry
Coffee-cup calorimeter (only used for reactions in solution, must be at constant pressure)
qreaction=-qwater
Bomb calorimeter (reaction gases, and must have constant volume)
qreaction=-(qwater+qbomb)
qbomb=C delta t, where C is the calorimeter constant (Cv of bomb x mass of bomb, really same equation)
delta H vs. delta E for chemical reactions
delta H=qp since delta E=qp-P delta V
substituting gives delta H= delta E+P delta V
where P will usually be in atmospheric pressure, and delta V is volume change at that pressure.
Laws of Thermochemistry
The magnitude of delta H is directly proportional to the amount of reactant or product.
-Thus delta H can be used as a conversion factor in a balanced equation to obtain amounts of reactant/product or delta H itself. (mole to mole ratio's).
delta H for a reaction is equal in magnitude but opposite in sign to delta H for the reverse reaction.

Problems 1: delta H Calculation

When 1 mol of methane is burned at constant pressure, 890.3kJ of energy is released as heat. Calculate delta 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
delta H = -320 kJ

For the reaction of methane with oxygen given in the notes, calculate the delta H in kJ if 5.8 grams of oxygen are consumed in the process.
= 81 kJ
delta H= -81kJ

Ammonium nitrate, NH4NO3, is commonly used as an explosive. It decomposes by the following reaction:

NH4NO3 --> N2O(g) + 2 H2O(g) + 37.0kJ

delta H = -37.0 kJ

If 72.0 grams of H2O are formed from the reaction, how much heat was released?
= 73.9 kJ
Hess' Law: The value of delta H for a reaction is the same whether it occurs directly or in a series of steps (state function).

delta Htotal = delta H1 + delta H2

often used to calculate delta H for one step, knowing delta 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.**
Heats of Formation
Molar heat of formation ( delta Hf) is equal to the enthalpy change, delta H when one mole of the compound is formed from the elements in their stable forms at 25oC and 1 atm is delta Ho (pronounced 'delta h naught'). delta Ho of a solution is of a 1M solution, at 1 atm and 25 oC.

Heats of formation are usually negative quantities.

delta H = sigma delta Hf products - sigma delta Hf reactants To apply the above relation, use the following rules:

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

Any element in its stable form is omitted.
Can also apply to heats of formation to ions.

Arbitrarily assign H+ ion to be zero. delta Hf H+(aq) = 0

Having established the above, a scale can be established with Hydrogen ion as the base.
Calculate delta H0rxn for the following reaction.

2 C3H6(g) + 9 O2(g) --> 6 CO2(g) + 6 H2O(l)
*Appendix 4*
delta Hrxn = sigma delta Hf products - sigma delta Hf reactants
delta Hrxn = [ 6 H2O(l) + 6 CO2(g) ] - [ 2 C3H6(g) + 9 O2(g) ]
delta Hrxn = [ 6(-286 kJ) + 6 (-393.2 kJ) ] - [ 2(20.9 kJ) + 9(0)]
delta Hrxn = [ -1716 kJ + -2361 kJ ] - [41.8 kJ ]
delta Hrxn = -4118.8 kJ / 2 mol = 2059 kJ / mol

Calculate delta H0 for 2Al(s) + Cr2O3(s) --> Al2O3(s) + 2Cr(s).
Compare this reaction to sample exercise 6.10, "thermite" reaction.

Which reaction yields more energy per gram of metal formed?

delta Hrxn = [ Al2O3(s) + 2 Cr(s) ] - [ 2 Al(s) + Cr2O3(s) ]
delta Hrxn = [ (-1676 kJ) + 2(0 kJ) ] - [ 2(0 kJ) + -1128 kJ ]
delta Hrxn = [ -1676 kJ ] - [ -1128 kJ ]
delta Hrxn = = -10.1 kJ / g Al
delta Hrxn = = -15.75 kJ / g Al
"Thermite" reaction releases more energy per gram of metal formed.


 

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