1 / 9

Calculating Enthalpy Change

Calculating Enthalpy Change. 15.4: Pgs. 534 - 541. Main Idea…. The enthalpy change for a reaction is the enthalpy of the products minus the enthalpy of the reactants. Hess’s Law. Most reactions occur in steps What if you need to know the heat change of a step, but you can’t measure it?

lamont
Download Presentation

Calculating Enthalpy Change

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. Calculating Enthalpy Change 15.4: Pgs. 534 - 541

  2. Main Idea… • The enthalpy change for a reaction is the enthalpy of the products minus the enthalpy of the reactants

  3. Hess’s Law • Most reactions occur in steps • What if you need to know the heat change of a step, but you can’t measure it? • Hess’s law allows you to gather this information indirectly • Elemental carbon can be found as graphite and diamond at 25oC C(diamond)  C(graphite) • Takes millions of years to go from diamond to graphite • Reaction is too slow to measure the heat change

  4. Hess’s Law • Hess’s Law of Heat Summation: • If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction • Example: Conversion of diamond to graphite C(diamond)  C(graphite) C(s, diamond) + O2 (g)  CO2 (g) ΔH = -395.4kJ C(s, graphite) + O2 (g)  CO2 (g) ΔH = -393.5kJ • REVERSE the above equation, so that we can show diamond being converted into graphite CO2 (g)  C(s, graphite) + O2 (g) ΔH = 393.5kJ

  5. Add the equations together! C(s, diamond)  C(s, graphite) ΔH = -1.9 kJ • The overall ΔH is negative, so you can see this is an exothermic process!

  6. Now you try some! • Try #32 and 33 on pg. 537 • 32 = -385.4 kJ • 33 = -521 kJ cuz direction of b is changed to get desired equation!

  7. Standard Enthalpy (Heat) of Formation - standard heat of formation– the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 298 K (25oC) • Free elements at 298 K & 1 atm have an enthalpy of zero. • Examples, HONClBrIF • Other ΔHºf values have been calculated • ΔHºreaction can be calculated as follows:

  8. Let’s do an example! • What is the standard heat of reaction (ΔHº) for the following equation: O2 (g) + 2CO (g)  2CO2 • From your text: • ΔHºf O2 (g) = 0 (free element) • ΔHºf CO (g) = -110.5 kJ/mol • ΔHºf CO2 (g) = -393.5 kJ/mol • Calculate the ΔHºf of the reactants: • 2 mol CO, 1 mol O • 2 mol of CO  2 mol x -110.5 kJ/mol = -221.0 kJ • 1 mol O2 = 0 KJ (free element) • TOTAL: 0 kJ + -221.0 kJ = -221 kJ (reactants) • Calculate the ΔHºf of the products: • 2 mol CO2 • 2 mol CO2 2 x -393.5 kJ/mol = -787.0 kJ • Use to find the total! ΔHº= -787.0 kJ – (-221.0) kJ ΔHº = -566.0 kJ

  9. Try a Practice Problem 2NO (g) + O2 (g)  2NO2 (g) ΔHºf NO = 90.37 ΔHºf O2 = 0.0 ΔHºf NO2 = 33.85 2(33.85) – [2(90.37) + 0] = -113.04 kJ

More Related