October 11, 2011
CH-111
Chapter 5, Section 4
Thermochemistry
In previous section, we determined that ΔH = Hfinal - Hinitial. From here we can see that the enthalpy change for a chemical reaction would be given by ΔH = Hproducts - Hreactants. The enthalpy change accompanying this reaction is referred to as either "enthalpy of reaction" or "heat of reaction", and is written sometimes as ΔHrxn.
"rxn" is a common abbreviation for "reaction".
When giving the numerical value for ΔHrxn, the reaction involved must also be specified.
For example: if 2 mol H2(g) burn to form 2 mol of H2O (g), at a constant pressure, the system releases 483.6kJ of heat.
When giving the numerical value ΔHrxn= -483.6kJ (because the reaction is exothermic, system is losing energy in heat), we must also specify the accompanying reaction: 2H2(g) + O2(g) --> 2H2O, thus
2H2(g) + O2(g) = 2H2O(g) ΔH = -483.6kJ
Coefficients in the balanced equation represent number of moles of reactants and products accounting for enthalpy change.
These balanced chemical equations which show the associated change of enthalpy in this way are called "thermochemical equations". Slip that into conversation somehow and watch the girls swoon.
Guidelines for Using Thermochemical Equations and Enthalpy Diagrams:
1. Enthalpy is an extensive property.
The magnitude of ΔH is proportional to the amount of reactant consumed in the process.
2. The enthalpy change for a reaction is equal in magnitude, but opposite in sign to ΔH for the reverse reaction.
When a reaction is reversed, the ΔH is the same, but the sign is reversed, indicating a change in direction of the energy from products to reactants.
3. The enthalpy change for a reaction depends on the states of the reactants and products.
Let's imagine one chemical equation involving H2O(g) and another involving H2O(l), as in, one is gas and the other is liquid. The chemical equation where H2O is in gas form will produce less kJ than the reaction involving H2O in liquid form. This is because liquid water must be converted to water vapor, and that conversion process is endothermic.
CO2(g) + 2H2O(l) ---------> CH4(g) + 2O2(g) ΔH = +890kJ (+88kJ for reaction 2H2O(l)---->2H2O(g))
CO2(g) + 2H2O(g) ------------> CH4(g) + 2O2(g) ΔH = +802kJ (-88kJ)
Let's work an example to further distribute the information into our brain parentheses.
My book gives the example:
How much heat is released when 4.50g of methane gas is burned in a constant pressure system?
CH4(g) + 2O2(g) ----------------> CO2(g) +2H2O(l) ΔH = -890kJ
So let's take a look at this bad boy. We need to figure out how much energy 4.50g of methane (CH4) gas produces and we're given that equation and told that the whole reaction releases -890kJ out of the system and into the surroundings. So the first thing I can logically think of here would be to analyze the mole ratios and then convert our given value of 4.50g to moles of CH4. By the equation I see that 1 mole of CH4(g) burned in oxygen at constant pressure yields approx -890kJ (negative only informing that the energy is released unto the surroundings from the system). So with a little thought we come to the conclusion that our conversion path must be grams of CH4 ----> moles of CH4 ------ energy per mole kJ -----> result. So, let's do that!
To convert grams to moles, we divide by the molecular mass, which for CH4 is approx 16.0g, and therefore one mole of CH4 is equal to 16.0g of CH4:
4.50g CH4 = 1 mol CH4/16.0g CH4 (cancel out like terms and remember significant figures)
0.2813 moles CH4 in 4.50 g CH4
Now, we know that 1 mole of CH4 produces -890kj, so to find out how many kJ 0.2813 moles of CH4 produces use the conversion:
given unit X desired unit/given unit
In this case: 0.2813 mol CH4 X -890kJ/1 mol CH4 = -250kJ
Then we do our best happy dance and flail our arms around like loonies because, YES, that's correct!