1 / 25

Thermodynamics 2

Thermodynamics 2. Energy in Reaction. Thermochemical Equations. Just like a regular chemical equation, with an energy term. CH 4 ( g ) + 2O 2 ( g )  CO 2 ( g ) + 2H 2 O( g )  r H o =  802 kJ CH 4 ( g ) + 2O 2 ( g )  CO 2 ( g ) + 2H 2 O( g ) + 802 kJ. energy out…

yehuda
Download Presentation

Thermodynamics 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Thermodynamics 2 Energy in Reaction

  2. Thermochemical Equations Just like a regular chemical equation, with an energy term. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) rHo = 802 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 802 kJ energy out… Exothermic Energy is a product just like CO2 or H2O! What does this imply? CONVERSION FACTORS!!! From the equation:

  3. Problem: How many kJ of energy are released when 5 moles of methane, CH4(g) is combusted? rHo = 802 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)  Reaction enthalpy If one mole of methane yield 802 kJ when burned , than 5 moles of methane yields………. 5 moles x 802 kJ/moles = 4010 kJ

  4. Problem: How many kJ of energy are released when 128.5 g of methane, CH4(g) is combusted? rHo = 802 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) g  mols  J  molar mass  Reaction enthalpy 6.43103 kJ

  5. Thermochemical Equations CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) rHo = 802 kJ When the reaction is reversed , the sign of DH reverses: Exothermic CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) rHo = +802 kJ Endothermic

  6. Thermochemical Equations CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) rHo =  802 kJ DH scales with the reaction: [ ] [ ] rHo =  401 kJ Yes you can write the reaction with fractions, so long as you are writing it on a mole basis…

  7. Thermochemical Equations Change in enthalpy depends on state: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) rH° = 802 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) rH° = 890 kJ =  88kJ This means that water’s liquid state lies 44 kJ/mol lower than the gas state From your text: H2O(g)  H2O(l) The difference in the rxn H is due to the change in state!! H = 44 kJ/mol  88kJ [ ]  2 =

  8. CH4(g) + 2O2(g) Enthalpy (H) CO2(g) + 2H2O(g) 2  44 kJ CO2(g) + 2H2O(l) Comparing the reaction with water as a gas or liquid: reactants either pathway gives the same results! rHo = 802 kJ products rHo = 890 kJ The difference is the 88 kJ released when 2 mols of water go from gas to liquid. products

  9. Constant Pressure Calorimetry, Measuring H • A constant pressure calorimeter can be used to measure the amount of energy transferred as heat under constant pressure conditions, that is, the enthalpy change for a chemical reaction.

  10. Constant Volume Calorimetry, Measuring U Heats of combustion (rUocombustion ) are measured using a device called a Bomb Calorimeter. A combustible sample is reacted with excess O2 The heat capacity of the bomb is constant. The heat of reaction is found by:

  11. Calculating Heat in an Exothermic Reaction Octane, the primary component of gasoline combusts by the reaction: C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9H2O(l) A 1.00 g sample of octane is burned in a bomb calorimeter that contains 1.20 kg of water surrounding the bomb. The temperature of the water rises to 33.20 °C from 25.00 °C when the octane is reacted. If the heat capacity of the bomb is 837 J/°C, calculate the heat of reaction per mole of octane. Since the temperature of the water rose, the reaction must have been exothermic: Therefore one can write: –qrxn = qwater + qbomb

  12. Calculating Heat in an Exothermic Reaction qbomb qwater –qrxn = mwaterCwaterTwater + qbombTwater qRXN = – 4803 J or – 48.0 kJ Heat transferred per mole qV:

  13. Hess’s Law The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Intermediate Reaction so we can write… rH1 known rH2 known rH1 + rH2 = rH Reactants Products What if the enthalpy changes through another path are know? rH = ? unknown! The sum of the H’s in one direction must equal the sum in the other direction. Why? Because enthalpy is a state function… Path independent!

  14. Hess’s law & Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆rHtotal is the same no matter which path is followed. You can look up the heat of formation from any substance in Appendix L or at http://www.update.uu.se/~jolkkonen/pdf/CRC_TD.pdf

  15. Hess’s Law Problem: Example:Determine the rH for the reaction: 3H2(g) + N2(g)  2NH3(g) rHo= ??? Given: (1) 2 H2(g) + N2(g)  N2H4(g) ∆rH°1 = +95.4 kJ (2) N2H4(g) + H2(g)  2NH3(g) ∆rH°2 = –187.6 kJ Notice that the path from reactants to products in the desired reaction goes through an “intermediate compound” in the given reactions. This means that the path for hydrogen and nitrogen to produce ammonia goes through hydrazine (N2H4). Therefore, the path to the enthalpy of the reaction must be a sum of the two given reactions!

  16. Hess’s Law Problem: Example:Determine the rH for the reaction: 3H2(g) + N2(g)  2NH3(g) rHo= ??? Given: (1) 2 H2(g) + N2(g)  N2H4(g) ∆rH°1 = +95.4 kJ (2) N2H4(g) + H2(g)  2NH3(g) ∆rH°2 = –187.6 kJ adding equations (1) & (2) yields:  N2H4(g) 2 H2(g) + N2(g) + N2H4(g) + H2(g) + 2NH3(g)

  17. Hess’s Law Problem: Example:Determine the rH for the reaction: 3H2(g) + N2(g)  2NH3(g) rHo= ??? Given: (1) 2 H2(g) + N2(g)  N2H4(g) ∆rH°1 = +95.4 kJ (2) N2H4(g) + H2(g)  2NH3(g) ∆rH°2 = –187.6 kJ adding equations (1) & (2) yields: Look what happens… / /  N2H4(g) 2 H2(g) + N2(g) + N2H4(g) + H2(g) + 2NH3(g) 3H2(g) + N2(g)  2NH3(g)

  18. Hess’s Law Problem: Example:Determine the rH for the reaction: 3H2(g) + N2(g)  2NH3(g) rHo= ??? Given: (1) 2 H2(g) + N2(g)  N2H4(g) ∆rH°1 = +95.4 kJ (2) N2H4(g) + H2(g)  2NH3(g) ∆rH°2 = –187.6 kJ adding equations (1) & (2) yields: Look what happens… / /  N2H4(g) 2 H2(g) + N2(g) + N2H4(g) + H2(g) + 2NH3(g) 3H2(g) + N2(g)  2NH3(g) therefore… rH= ∆rH°1 + ∆rH°2= +95.4 kJ + (–187.6 kJ) = –92.2 kJ

  19. intermediates H2(g) + N2H4(g) (Step 1) Ho = 95.4kJ (Step 2) Ho = 187.6kJ reactants 3H2(g) + N2(g) (Overall) Ho = 92.2kJ products 2N3 (g) • Reaction (1) is endothermic by 95.4 kJ • Reaction (2) is exothermic by 187.6 kJ • Since theexothermicityhas a greater magnitude than the endothermicity, the overall process is exothermic (rH < 0).

  20. Standard Enthalpies of Formation When 1 mole of compound is formed from its elements, the enthalpy change for the reaction is called the enthalpy of formation, fHo (kJ/mol). These enthalpies are always reported at Standard conditions: 1 atm and 25 °C (298 K). The standard enthalpies of formation of the most stable form of any element is zero: fH (element) = 0 fH° O2(g) = 0 fH°O(g) ≠ 0 NOT the elemental form elemental form

  21. Enthalpy Values Since Enthalpy is a state function, enthalpy values depend on the reaction conditions in terms of the phases of reactants and products. H2(g) + ½ O2(g) H2O(g) ∆rH˚ = -242 kJ H2(g) + ½ O2(g) H2O(liq) ∆rH˚ = -286 kJ Same reaction, different phases, different enthalpies. http://www.update.uu.se/~jolkkonen/pdf/CRC_TD.pdf

  22. Formation Reactions A chemical reaction that describes the formation of one mole of a compound from its elements at standard state conditions is known as a “formation reaction”. The formation of water is given by the reaction: H2(g) + ½ O2(g) H2O(l) Each element and the compound are represented by the physical state they take on at 25.0 °C and 1 atm pressure. (Standard State conditions)

  23. Question: What is the formation reaction for potassium permanganate? elements compound (g) K (s) + Mn (s) + O2 2 KMnO4 (s) • salts are solids at standard state conditions. • metals are solids at standard state conditions. • oxygen is a gas at standard state conditions. • balance for one mole of the product standard state conditions = 25oC and 1 atm http://www.update.uu.se/~jolkkonen/pdf/CRC_TD.pdf

  24. Enthalpy Changes for a Reaction:Using Standard Enthalpy Values • All components of a reaction can be related back to their original elements. • Each compound has an enthalpy of formation associated with it. • Reactants require energy to return to component elements. • Products release energy when formed form component elements. • Since enthalpy is a state function, the sum of the above must relate somehow to the overall enthalpy of a reaction.

  25. Consider the Combustion of Propane 3C(s) + 8H2(g) + 5O2(g) C3H8(g) + 5O2(g)  3CO2(g) + 4 H2O(l) • In order to make CO2(g) and H2O(l) one must break the propane up into its elements. • This takes energy. • The elements carbon, hydrogen and oxygen then combine to make the new compounds, CO2 and H2O. • This process releases energy.

More Related