Thermochemistry: Chapter 6 from Chem 1036 Textbook - Prof. Patricia G. Amateis, Study notes of Chemistry

Download Thermochemistry: Chapter 6 from Chem 1036 Textbook - Prof. Patricia G. Amateis and more Study notes Chemistry in PDF only on Docsity! Chem 1036 Chapter 6 1 of 30 Chapter 6 Thermochemistry Thermodynamics: Thermochemistry: System Surroundings Chem 1036 Chapter 6 2 of 30 First Law of Thermodynamics First Law: Potential Energy ⇔ Kinetic Energy Potential Energy: Kinetic Energy: Internal Energy: Chem 1036 Chapter 6 3 of 30 ∆E = Efinal - Einitial = Eproducts - Ereactants Units of energy: Chem 1036 Chapter 6 4 of 30 q = w = ∆E = q + w Printed with FinePrint - purchase at www.fineprint.com Chem 1036 Chapter 6 5 of 30 Work (w) • Pressure-volume work: ─work of expansion: ─work of contraction: • Net increase in moles of gas: • Net decrease in moles of gas: Chem 1036 Chapter 6 6 of 30 work = (─) work = (+) Units of work: w = ─P∆V Chem 1036 Chapter 6 7 of 30 Chem 1036 Chapter 6 8 of 30 Example Problem: Consider the following reactions. Does the system do work on the surroundings or does the surroundings do work on the system? a. C2H2(g) + 2 H2(g) → C2H6(g) b. 2 N2O5(g) → 4 NO2(g) + O2(g) Example Problem: A gas is heated, causing it to expand from a volume of 2.0 L to 4.5 L at 1 atm pressure. Calculate the work of the system in joules (watch the sign). 101.3 J = 1 L•atm Printed with FinePrint - purchase at www.fineprint.com Chem 1036 Chapter 6 17 of 30 Example: A sample of iron pellets is heated to 99.4 °C and dropped in 57.8 g of well insulated water at 19.3 °C. The final temperature is 28.4 °C. What was the mass of the iron sample? (Cs H2O = 4.184 J•g-1•°C-1, Cs Fe = 0.44 J•g-1•°C-1) Heat gained by H2O: Heat lost by Fe pellets: (Law of Conservation of Energy) Chem 1036 Chapter 6 18 of 30 Constant Volume (Bomb) Calorimetry • • • Chem 1036 Chapter 6 19 of 30 Chem 1036 Chapter 6 20 of 30 Problem: 0.100 mol of propane gas (C3H8) are injected into a bomb calorimeter and ignited with excess oxygen, according to the reaction below. The calorimeter has a heat capacity of 97.1 kJ / °C. C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O( ) (a) If the temperature rose from 25.000 °C to 27.282 °C, what is the heat of the reaction? (b) Calculate the heat of the reaction in units of kJ/mol Printed with FinePrint - purchase at www.fineprint.com Chem 1036 Chapter 6 21 of 30 Using Enthalpy ∆H Enthalpy: ∆H : ∆E = q + w = qP + (-P∆V) qP = q at constant pressure qP = ∆E + P∆V ∆H = ∆E + P∆V; ∆H = qP Both physical changes and chemical reactions have enthalpy values associated with them. ∆Hvap: ∆Hfus: Important facts to know about enthalpy change: Chem 1036 Chapter 6 22 of 30 • ∆Ho: • The magnitude of ∆Ho is proportional to the amount of substance reacting. Thermochemical equation: 3 Fe2O3(s) + CO(g) → 2 Fe3O4(aq) + CO2(g) ∆Ho = -59 kJ • The sign of ∆Ho is changed when an equation is reversed: H2O( ) → H2O(g) ∆Ho = H2O(g) → H2O( ) ∆Ho = • ∆Ho and numbers of moles are multiplied by the same factor: 2 H2(g) + O2(g) → 2 H2O( ) ∆Ho = 4 H2(g) + 2 O2(g) → 4 H2O( ) ∆Ho = Chem 1036 Chapter 6 23 of 30 Example Problem: Given the following thermochemical equation, calculate the amount of heat absorbed when 40.0 g of C reacts with excess sulfur. 4 C(s) + S8(s) → 4 CS2( ) ∆H = +358.8 kJ How do we know that heat is absorbed? Chem 1036 Chapter 6 24 of 30 Bonds: Source of Enthalpy of Reaction (Chapter 9 pages 356-361) A reaction is a process in which ∆Horxn = ∆Horeactant bonds broken + ∆Hoproduct bonds formed CH4 + 2O2 → CO2 + 2H2O 1652 kJ 996 kJ - 1598 kJ - 1868 kJ Bond energies 2648 kJ -3466 kJ Printed with FinePrint - purchase at www.fineprint.com Chem 1036 Chapter 6 25 of 30 Two Methods of Calculating ∆H 1. Hess's Law Hess's Law: overall enthalpy of a reaction = Used to determine enthalpy of reactions that can't be measured by calorimetry Reaction sequence: A B C D. A B ∆H1 = 100 kJ B C ∆H2 = 260 kJ C D ∆H3 = -500 kJ A D ∆Hrxn = ∆H1 + ∆H2 + ∆H3 = -140 kJ ∆ H (k J/ m ol ) ∆Η1 ∆Η2 ∆Η3 ∆Ηrxn A B C D 0 100 360 -140 Chem 1036 Chapter 6 26 of 30 Example: Calculate the enthalpy for the reaction 2F2(g) + 2H2O( ) 4HF(g) + O2(g) ∆H = ? given the following reaction enthalpies: 1. H2(g) + F2(g) 2HF(g) ∆H = -537 kJ 2. 2H2(g) + O2 2H2O( ) ∆H = -572 kJ Strategy: Reactions must be rearranged so they sum to give the overall reaction: Chem 1036 Chapter 6 27 of 30 2. Enthalpy (Heat) of Formation: ∆Hf° = Enthalpy of Formation: f implies ° implies Table of ∆Hf° values in Table 6.3 and Appendix B (p. A-5). by definition, elements in their standard state have ∆Hfo = e.g. ∆Hf° H2(g) = 0 Elements not in their standard state and all compounds have ∆Hf° ≠ 0 For stable compounds, ∆Hf° is negative. For unstable compounds (e.g. "self contained" explosives), ∆Hf° is positive Chem 1036 Chapter 6 28 of 30 Printed with FinePrint - purchase at www.fineprint.com