Difference between revisions of "Ch6 lec 2"

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:Q: How much heat is given off when 2.50 moles of CH<sub>4</sub> is combusted?
 
:Q: How much heat is given off when 2.50 moles of CH<sub>4</sub> is combusted?
 
:A: 2.50 moles CH<sub>4</sub> * -890.4 kJ/mol of CH<sub>4</sub> = -2226 kJ.
 
:A: 2.50 moles CH<sub>4</sub> * -890.4 kJ/mol of CH<sub>4</sub> = -2226 kJ.
 +
 +
=====Example 3=====
 +
:Q: How much heat is given off when 1.25 moles of O<sub>2</sub> is combusted?
 +
:A: 1.25 moles O<sub>2</sub> * -890.4 kJ/ 2 mol of O<sub>2</sub> = -556.5 kJ.
  
 
To make this a bit more confusing, we don't measure out our reactants or products in terms of moles, we use grams. So a typical problem might look like this:
 
To make this a bit more confusing, we don't measure out our reactants or products in terms of moles, we use grams. So a typical problem might look like this:
  
=====Example 3=====
+
=====Example 4=====
 
:Q: How much heat is given off when 10.0 grams of CH<sub>4</sub> is combusted?
 
:Q: How much heat is given off when 10.0 grams of CH<sub>4</sub> is combusted?
 
:A: Knowing the balanced chemical equation above, the enthalpy of -890.4 kJ/mole of CH<sub>4</sub>, AND the molecular weight of CH<sub>4</sub>, then the first thing you need to do is convert the grams to moles:
 
:A: Knowing the balanced chemical equation above, the enthalpy of -890.4 kJ/mole of CH<sub>4</sub>, AND the molecular weight of CH<sub>4</sub>, then the first thing you need to do is convert the grams to moles:

Revision as of 16:48, 7 April 2020

(4/8/20, bes)

Sec 6.3 Introduction to Thermodynamics

Skim...

This book chapter is titled "thermochemistry". Thermochemistry is only a small part of a larger field of study called Thermodynamics (taught in Chem 312: Physical Chemistry 1). This section attempt to introduce you MANY additional topics and i personally find it a bit overwhelming...so just skim this section. The following are take home points:

- 1st Law of Thermodynamics, energy cannot be created or destroyed only converted from one form to another.
- the change in (internal) energy (abbreviated ΔU) is a result of either heat (q) being transferred or work (w) being done --> ΔU = q + w <--this is another way of writing the 1st Law.

Sec 6.4 Enthalpy of Chemical Reactions

The term enthalpy, abbreviated ΔH, was discussed in lab. Although this term is derived from the complexity of thermodynamics, it can be thought of as simply the amount of heat given off (or take up) during a chemical reaction...technically...a chemical reaction at constant pressure. All balanced chemical reactions have integer values for stoichiometric coefficients so when the amount of heat given off (or taken up) by a reaction is noted, it is usually given in units of J/mol. Here is an example:

CH4 (g) + 2 O2 (g) --> CO2 (g) + 2 H2O (l)

The enthalpy for this reaction could be presented in a variety of ways:

1) -890.4 kJ ...for the rxn as shown,
2) -890.4 kJ/mol of CH4 consumed,
3) -890.4 kJ/2 moles of O2 consumed, or -445.2 kJ/mol O2 consumed,
4) -890.4 kJ/mol of CO2 produced,
5) -890.4 kJ/2 moles of H2O produced or -445.2 kJ/mol H2 O produced...
this is why this stuff can be confusing!
Example 1
Q: How much heat is given off when 1.00 moles of CH4 is combusted?
A: Knowing the balanced chemical equation above and the enthalpy value given of -890.4 kJ/mole of CH4, then the answer is relatively straightforward:
1 mole CH4 * -890.4 kJ/mol of CH4 = -890.4 kJ.
Example 2
Q: How much heat is given off when 2.50 moles of CH4 is combusted?
A: 2.50 moles CH4 * -890.4 kJ/mol of CH4 = -2226 kJ.
Example 3
Q: How much heat is given off when 1.25 moles of O2 is combusted?
A: 1.25 moles O2 * -890.4 kJ/ 2 mol of O2 = -556.5 kJ.

To make this a bit more confusing, we don't measure out our reactants or products in terms of moles, we use grams. So a typical problem might look like this:

Example 4
Q: How much heat is given off when 10.0 grams of CH4 is combusted?
A: Knowing the balanced chemical equation above, the enthalpy of -890.4 kJ/mole of CH4, AND the molecular weight of CH4, then the first thing you need to do is convert the grams to moles:
10.0 g * (1 mol/16.04 g) = 0.623 moles CH4, then,
0.623 moles CH4 * -890.4 kJ/mol of CH4 = -554.7 kJ.

Sec 6.5 Calorimetry