6.00kJ + H2O(s) H2O(l)
H = +6.00kJ
! 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.
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
Problems 1: H Calculation
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
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.**
Heats of formation are usually negative quantities.
H = Hf products - Hf reactants To apply the above relation, use the following rules:
Heats of formation can be found in appendix A-4
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.
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
Which reaction yields more energy per gram of metal formed?
This page was made by Erik Epp.