Molar enthalpy is the energy change per mole of substance undergoing a change.
· DH represents the difference between the enthalpy of the system at the beginning of the reaction compared to what it is at the end of the reaction:
DH = Hproducts – Hreactants
· We are considering the heat content of the system. Thus:
1. if the system has higher enthalpy at the end of the reaction, then it absorbed heat from the surroundings (endothermic reaction)
The heat content of products is greater then the heat content of the reactants.
For endothermic reactions Hproducts > Hreactants and DH is positive (+DH)
if the system has a lower enthalpy at the end of the reaction, then it gave off heat during the reaction (exothermic reaction)
The heat content of the reactants is greater than the heat content of the products.
For exothermic reactions Hproducts < Hreactants and DH is negative (-DH)
· We can also describe DH for a reaction by comparing the enthalpies of the products and the reactants:
DH = H(products) - H(reactants)
· The enthalpy change that accompanies a reaction is called the enthalpy of reaction (DHr).
· It is sometimes convenient to provide the value for DHr along with the balanced chemical equation for a reaction (also known as a thermochemical equation):
2H2(g) + O2(g) → 2H2O(g) DH = -483.6 kJ
Note the following:
DH is negative, indicating that this reaction results in the release of heat (exothermic)
The reaction gives of 483.6 kJ of energy when 2 moles of H2 combine with 1 mole of O2 to produce 2 moles of H2O.
The equation can be rewritten to represent the formation of 1 mole H2O.
H2(g) + ½ O2(g) → H2O(g) DH = -241.8 kJ/mol
The relative enthalpies of the reactants and products can also be shown on an energy diagram:
Stating the molar enthalpy is a convenient way of describing the energy changes involved in a variety of physical and chemical changes.
The molar enthalpy of a physical change can be expressed as follows:
H2O(l) + 40.8 kJ → H2O(g)
The heat of vapourization is expressed as ΔHvap = 40.8 kJ/mol
The amount of energy involved in a change, ΔH, for either physical or chemical changes, depends on the quantity of matter undergoing the change. To calculate the enthalpy change, ΔH, for some amount other than one mole use the followingformula:
ΔH = nΔHx, where x represents the type of reaction
The refrigerant Freon -12, molar mass 120.91 g/mol, is used in a common refrigerator to cool your food to prevent it from decaying rapidly. The molar enthalpy of vapourization for Freon-12 is 34.99 kJ/mol. If 1 kg of the refrigerant is vapourized, what is the expected enthalpy change.
ΔHvap = 34.99 kJ/mol
ΔH = ?
MFreon-12 = 120.91 g
Find # moles of Freon-12
Since n = m/M
Then n = 1000g/120.91
= 8.27 mol
Calculate the enthalpy change, ΔH
ΔH = n ΔHvap
= 8.27 mol x 34.99 kJ/mol
= 289.4 kJ
Calorimetry of Physical & Chemical Changes
When investigating energy changes we base our analysis on the law of conservation of energy.
There are three assumptions often used in calorimetry
a. No heat transfer between the calorimeter and the outside environmet;
b. Any heat absorbed or released by the calorimeter materials, such as the container, is negligible;
c. A dilute aqueous solution is assumed to have a density and specific heat capacity equal to that of pure water (1.00g/ml and 4.18J/g•oC)
Pg. 308 do 1,2,3
Pg. 310 do 4,5
Pg. 311 do 7, 9, & 10
Pg. 312 do 1 to 5
Representing Enthalpy Changes
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