1 / 26

Thermochemistry

Thermochemistry. Kinetics (∆H) Potential Energy graphs. Thermochemistry. Thermodynamics is the science of the relationship between heat and other forms of energy. Thermochemistry is the study of the quantity of heat absorbed or evolved by chemical reactions.

badru
Download Presentation

Thermochemistry

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. Thermochemistry Kinetics (∆H) Potential Energy graphs

  2. Thermochemistry • Thermodynamics is the science of the relationship between heat and other forms of energy. • Thermochemistry is the study of the quantity of heat absorbed or evolved by chemical reactions.

  3. Figure 6.11: Coffee-cup calorimeter.

  4. Figure 6.12: A bomb calorimeter.

  5. Measuring Heats of Reaction • To see how heats of reactions are measured, we must look at the heat required to raise or lower the temperature of a substance, because a thermochemical measurement is based on the relationship between heat and temperature change. • Heat absorbed = Endothermic (+∆H) • Heat released = Exothermic (-∆H)

  6. Hess’s Law • Hess’s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps or chemical equations and their energies can be added as if they were math equations. • Similar to: 4A + 3B = 5C + 2e • e + 3C + 2B = 4A • 5B = 2C + e

  7. Applying Stoichiometry and Heats of Reactions: ∆H • Consider the reaction of methane, CH4, burning in the presence of oxygen at constant pressure. Given the following equation, how much heat could be obtained by the combustion of 10.0 grams CH4? • = • 556 kJ released

  8. Enthalpy diagram illustrating Hess’s law.

  9. Potential Energy GraphNO  ½ N2 + ½ O2

  10. EXOTHERMIC

  11. Endothermic

  12. Parts of the Graph • a) EF: Activation Energy of the forward reaction, to get from the reactants to the Activated Complex • b) ER: Activatiton Energy of the reverse reaction, to get from the products to the Activated Complex • c) ∆H: Energy level difference between reactants and products. Negative for Exothermic. Positive for Endothermic. • Reactant level: energy content of the reactant bonds • Activated Complex: energy needed to break reactant bonds and prepare for product bonds • Product level: energy content of the product bonds

  13. Could you use these data to obtain the enthalpy change (∆H) for the following reaction? Hess’s Law • For example, suppose you are given the following data:

  14. Standard Enthalpy of Formation from the elements • The Standard Enthalpy of formation of a substance, denotedDHfo, is the enthalpy change for the formation of one mole of a substance in its standard state FROM THE COMPONENT ELEMENTS IN THEIR STATES AT 25°C. • You will have to write the equation. • Note that the standard enthalpy of formation for a pure element in its standard state is zero.

  15. Chart of Standard Enthalpy of Formation • See page R66 table B-14 from your textbook • YOU WILL HAVE TO WRITE THE EQUATION FROM ELEMENTS

  16. Heats of Combustion • See page R60 • in your book • Table B-5 • You will have to write the equations of combusiton

  17. Thermochemical Equations • The following are two important rules for manipulating thermochemical equations: • When a thermochemical equation is multiplied by any factor, the value of DH for the new equation is obtained by multiplying the DH in the original equation by that same factor. • When a chemical equation is reversed, the value of DH is reversed in sign.

  18. Hess’s Law • sub equations: • S(s) + O2(g)  SO2(g) -296.8kj/mole • SO3(g)  S(s) + 3/2 O2(g) 395.7kj/mole • ------------------------------------------------------------- • (Net) SO3(g)  SO2(g) + ½ O2(g) 98.kj/mole

  19. Standard Enthalpies of Formation • The law of summation of heats of formation states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants. • S is the mathematical symbol meaning “the sum of”, and m and n are the coefficients of the substances in the chemical equation.

  20. A Problem to Consider • Large quantities of ammonia are used to prepare Nitric Oxide according to the following equation: Net(BOSS) ∆H = ? • What is the standard enthalpy change (∆H) for this reaction? Use Table B-14 for data. • If 500 pounds of ammonia are to be consumed, find the kJ that will be involved. (Endo vs. Exo) • Draw the Potential Energy Diagram of the reaction if the reactants are at 1100kJ and the AC is 200kJ.

  21. Net:(BOSS) 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) ∆H=? • Using Table • ½ N2 +3/2H2 NH3(g) ∆H=-45.9kJ/m • ½ N2 + ½ O2  NO(g) ∆H=90.3kJ/m • H2 + ½ O2  H2O(g) ∆H=-241.8kJ/m • Reversing first equation, multiplying 1st and 2nd by 4 and 3rd by 6 because of the Net equation • 4(NH3(g)  ½ N2 +3/2H2) 4(∆H= 45.9kJ/m) • 4(½ N2 + ½ O2  NO(g))4(∆H=90.3kJ/m) • 6(H2 + ½ O2  H2O(g)) 6(∆H=-241.8kJ/m) • Net:4NH3(g)+5O2(g)4NO(g) + 6H2O(g) ∆H=-906kJ

  22. Net:4NH3(g)+5O2(g)4NO(g) + 6H2O(g) ∆H=-906kJ • 1.21E7kJ • 1.21E7 kJ released (Exothermic reaction)

  23. Potential Energy Graphing • Reactants=1100kJ ∆H= -906kJ Products=194kJ AC=1300kJ EF=200kJ ER=1106kJ

  24. A Problem to Consider • You record the values of DHfo under the formulas in the equation, multiplying them by the coefficients in the equation. • You can calculate DHo by subtracting the values for the reactants from the values for the products. • ∆H=[6(-241.8)+4(90.3)] – [ 4(-45.9)+5(0)]= -906 kj

More Related