Ch6 lec 2

(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

I wanted to follow up on some lab conversation before we leave for Easter break.

In lab we discussed how the temperature of water will increase when a certain amount of heat is added. The lab conversation was complicated by the fact that i had to introduce enthalpy, ΔH as well as the concept of heat capacity. If you recall, the heat capacity (aka specific heat) of water is 4.184 J/(g^oC)...and the equation used was:

q (heat) = (mass of water) * 4.184 J/(g^oC) * ΔT

Example 1

Q: If we added 1000 J of heat to 100 g of water, how much will the temperature rise?

A: 1000 J = (100 g) * 4.184 J/g^oC) * ΔT, solving for ΔT = 2.39 ^oC

Remember that given an equation that has 4 inputs (q, grams of H₂O, specific heat, ΔT), all you need is 3 of these and you can solve for the 4th...

Example 2

Q: If the temperature of 250 mL (= 250 g because the density is 1 g/mL) of water increased by 5.0 ^oC, how much heat was added?

A: q = (250 g) * 4.184 J/g^oC) * 5.0 ^oC, solving for q = 5,230 J or 5.23 kJ

One more point...ΔT = T_final - T_initial, so if the initial temperature of water is given, then you can calculate the final temperature of the water...

Example 3

Q: If we added 1234 J of heat to 321 g of water that was initially 25.0^oC, what will be the final temperature?

A: 1234 J = (321 g) * 4.184 J/g^oC) * ΔT, solve for ΔT = 0.919 ^oC

ΔT = T_final - T_initial; 0.919 ^oC = T_final - 25.0 ^oC, solving for T_final = = 25.9 ^oC

There is a WA waiting for you