Ch6 lec 2 - Revision history

Bes at 11:22, 8 April 2020

2020-04-08T11:22:40Z

Bes: /* Example 3 */

2020-04-07T18:07:21Z

Example 3

Bes: /* Example 2 */

2020-04-07T17:50:17Z

Example 2

Bes: /* Sec 6.5 Calorimetry */

2020-04-07T17:43:50Z

Sec 6.5 Calorimetry

Bes: /* Sec 6.4 Enthalpy of Chemical Reactions */

2020-04-07T16:48:43Z

Sec 6.4 Enthalpy of Chemical Reactions

Bes: /* Example 3 */

2020-04-07T16:35:32Z

Example 3

Bes: /* Example 3 */

2020-04-07T16:34:54Z

Example 3

Bes: /* Sec 6.4 Enthalpy of Chemical Reactions */

2020-04-07T16:33:13Z

Sec 6.4 Enthalpy of Chemical Reactions

Bes: /* Sec 6.4 Enthalpy of Chemical Reactions */

2020-04-07T16:21:05Z

Sec 6.4 Enthalpy of Chemical Reactions

Bes at 16:19, 7 April 2020

2020-04-07T16:19:24Z

@@ Line 1: / Line 1: @@
 (4/8/20, bes)
-====Sec 6.3 Introduction to Thermodynamics====
+===Sec 6.3 Introduction to Thermodynamics===
 Skim...
@@ Line 8: / Line 8: @@
 :- 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====
+===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:
@@ Line 43: / Line 43: @@
 ::: 0.623 moles CH<sub>4</sub> * -890.4 kJ/mol of CH<sub>4</sub> = -554.7 kJ.
-====Sec 6.5 Calorimetry====
+===Sec 6.5 Calorimetry===
 I wanted to follow up on some lab conversation before we leave for Easter break.

← Older revision		Revision as of 18:07, 7 April 2020
Line 66:		Line 66:

	ΔT = T<sub>final</sub> - T<sub>initial</sub>; 0.919 <sup>o</sup>C = T<sub>final</sub> - 25.0 <sup>o</sup>C, solving for T<sub>final</sub> = = 25.9 <sup>o</sup>C		ΔT = T<sub>final</sub> - T<sub>initial</sub>; 0.919 <sup>o</sup>C = T<sub>final</sub> - 25.0 <sup>o</sup>C, solving for T<sub>final</sub> = = 25.9 <sup>o</sup>C
		+
		+	'''''There is a WA waiting for you'''''

@@ Line 56: / Line 56: @@
 =====Example 2=====
-:Q: If the temperature of 250 mL (=250 mL because the density is 1 g/mL) of water increased by 5.0 <sup>o</sup>C, how much heat was added?
+:Q: If the temperature of 250 mL (= 250 g because the density is 1 g/mL) of water increased by 5.0 <sup>o</sup>C, how much heat was added?
 :A: q = (250 g) * 4.184 J/g<sup>o</sup>C) * 5.0 <sup>o</sup>C, solving for q = 5,230 J or 5.23 kJ

← Older revision		Revision as of 17:43, 7 April 2020
Line 44:		Line 44:

	====Sec 6.5 Calorimetry====		====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<sup>o</sup>C)...and the equation used was:
		+	:::q (heat) = (mass of water) * 4.184 J/(g<sup>o</sup>C) * 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<sup>o</sup>C) * ΔT, solving for ΔT = 2.39 <sup>o</sup>C
		+
		+	Remember that given an equation that has 4 inputs (q, grams of H<sub>2</sub>O, 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 mL because the density is 1 g/mL) of water increased by 5.0 <sup>o</sup>C, how much heat was added?
		+	:A: q = (250 g) * 4.184 J/g<sup>o</sup>C) * 5.0 <sup>o</sup>C, solving for q = 5,230 J or 5.23 kJ
		+
		+	One more point...ΔT = T<sub>final</sub> - T<sub>initial</sub>, 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<sup>o</sup>C, what will be the final temperature?
		+	:A: 1234 J = (321 g) * 4.184 J/g<sup>o</sup>C) * ΔT, solve for ΔT = 0.919 <sup>o</sup>C
		+
		+	ΔT = T<sub>final</sub> - T<sub>initial</sub>; 0.919 <sup>o</sup>C = T<sub>final</sub> - 25.0 <sup>o</sup>C, solving for T<sub>final</sub> = = 25.9 <sup>o</sup>C

@@ Line 30: / Line 30: @@
 :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.
 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?
 :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:

← Older revision		Revision as of 16:33, 7 April 2020
Line 21:		Line 21:

	::::'''''this is why this stuff can be confusing!'''''		::::'''''this is why this stuff can be confusing!'''''
		+
		+	=====Example 1=====
		+	:Q: How much heat is given off when 1.00 moles of CH<sub>4</sub> is combusted?
		+	:A: Knowing the balanced chemical equation above and the enthalpy value given of -890.4 kJ/mole of CH<sub>4</sub>, then the answer is relatively straightforward:
		+	:: 1 mole CH<sub>4</sub> * -890.4 kJ/mol of CH<sub>4</sub> = -890.4 kJ.
		+
		+	=====Example 2=====
		+	: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.
		+
		+	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=====
		+	: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 and the enthalpy value given of -890.4 kJ/mole of CH<sub>4</sub>, 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 CH<sub>4</sub>, then,
		+	::: 0.623 moles CH<sub>4</sub> * -890.4 kJ/mol of CH<sub>4</sub> = -554.7 kJ.

	====Sec 6.5 Calorimetry====		====Sec 6.5 Calorimetry====

@@ Line 13: / Line 13: @@
 ::CH<sub>4</sub> (g) + 2 O<sub>2</sub> (g) --> CO<sub>2</sub> (g) + 2 H<sub>2</sub>O (l)
-The enthalpy for this reaction could be presented in a varity of ways:
+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 CH<sub>4</sub> consumed,
@@ Line 21: / Line 21: @@
 ::::'''''this is why this stuff can be confusing!'''''
 ====Sec 6.5 Calorimetry====

← Older revision		Revision as of 16:19, 7 April 2020
Line 1:		Line 1:
		+	(4/8/20, bes)
		+
	====Sec 6.3 Introduction to Thermodynamics====		====Sec 6.3 Introduction to Thermodynamics====
	Skim...		Skim...