html5-img
1 / 7

Standard Enthalpies of Formation

Standard Enthalpies of Formation. Chapter 8.5. Standard Enthalpies of Formation “Heat of Formation”.

finley
Télécharger la présentation

Standard Enthalpies of Formation

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. Standard Enthalpies of Formation Chapter 8.5

  2. Standard Enthalpies of Formation“Heat of Formation” • 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 its component elements in their standard state. • Note that the standard enthalpy of formation for a pure element in its standard state is zero.

  3. You have a similar table on page 209 of your text NH3 = -46.1

  4. Using DHfotoCalculate DHofor a Rxn • 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.

  5. Why use DHfotoCalculate DHowhen we can use Hess’s Law? • Simply put it is easier • Finding combinations of reactions to fit your particular reaction gets tedious • We have very accurate data tables that give us the DHfovalues • Pure elements are 0!

  6. Let’s do a Practice Problem!!! -183.6 + 0 361.2 + (-1450.8) -183.6 for reactants -1089.6 for products Prod. – React. = -1089.6 – (-183.6) = -906 kJ • *Hint: First record the values of DHfo under the formulas in the equation then multiplying them by the coefficients in the equation.You can then determine DHoby subtracting the values for the reactants from the values for the products.

  7. Homework for 8.5 Problems #38 (don’t forget DH in equations!), 39, 43, and 45 Chem Dollar problem #52 ($2)

More Related