1 / 8

Hess's Law

Hess's Law. Hess’s law. Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps. For example: C + O 2  CO 2 The book tells us that this can occur as 2 steps C + ½O 2  CO H = – 110.5 kJ CO + ½O 2  CO 2 H = – 283.0 kJ.

jabari
Download Presentation

Hess's Law

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. Hess's Law

  2. Hess’s law • Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps. • For example: C + O2 CO2 The book tells us that this can occur as 2 steps C + ½O2  CO H = – 110.5 kJ CO + ½O2  CO2 H = – 283.0 kJ C + CO + O2  CO + CO2 H = – 393.5 kJ • I.e. C + O2  CO2 H = – 393.5 kJ • Hess’s law allows us to add equations. • We add all reactants, products, & H values. • We can also show how these steps add together via an “enthalpy diagram” …

  3. Steps in drawing enthalpy diagrams • Balance the equation(s). • Sketch a rough draft based on H values. • Draw the overall chemical reaction as an enthalpy diagram (with the reactants on one line, and the products on the other line). • Draw a reaction representing the intermediate step by placing the relevant reactants on a line. • Check arrows: Start: two leading away Finish: two pointing to finish Intermediate: one to, one away • Look at equations to help complete balancing (all levels must have the same # of all atoms). • Add axes and H values.

  4. C + O2  CO2 H = – 393.5 kJ C + O2 C + ½O2  CO H = – 110.5 kJ CO + ½O2  CO2 H = – 283.0 kJ Reactants H = – 110.5 kJ CO + ½O2 Intermediate Enthalpy H = – 393.5 kJ H = – 283.0 kJ CO2 Products Note: states such as (s) and (g) have been ignored to reduce clutter on these slides. You should include these in your work.

  5. Practice Exercise with Diagram C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H= –1411.1 kJ 2CO2(g) + 3H2O(l)  C2H5OH(l) + 3O2(g) H= +1367.1 kJ Draw the related enthalpy diagram. C2H4(g) + H2O(l)  C2H5OH(l) H= – 44.0 kJ C2H4(g) + H2O(l) + 3O2(g) Reactants Products C2H5OH(l) + 3O2(g) H= –44.0 kJ H = –1411.1 kJ H = +1367.1 kJ Enthalpy 2CO2(g) + 3H2O(l) Intermediate

  6. GeO(s)  Ge(s) + ½ O2(g) H= + 255 kJ Ge(s) + O2(g)  GeO2(s) H= – 534.7 kJ GeO(s) + ½ O2(g)  GeO2(s) H= – 279.7 kJ Ge(s) + O2(g) Intermediate H = +255 kJ GeO(s) + ½ O2(g) H = – 534.7 kJ Reactants Enthalpy H= –280 kJ GeO2(s) Products

  7. NO(g)  ½ N2(g) + ½ O2(g) H= – 90.37 kJ ½ N2(g) + O2(g)  NO2(g) H= + 33.8 kJ H= – 56.57 kJ NO(g) + ½ O2(g)  NO2(g) NO + ½ O2(g) Reactants H= –56.6 kJ NO2(g) H = –90.37 kJ Enthalpy Products H = +33.8 kJ ½ N2(g) + O2(g) Intermediate

  8. Hess’s law We may need to manipulate equations further: 2Fe + 1.5O2  Fe2O3 H=?, given Fe2O3 + 3CO  2Fe + 3CO2 H= – 26.74 kJ CO + ½O2  CO2 H= –282.96 kJ 1: Align equations based on reactants/products. 2: Multiply based on final reaction. 3: Add equations. 2Fe + 3CO2  Fe2O3 + 3CO H= + 26.74 kJ CO + ½O2  CO2 H= –282.96 kJ 3CO + 1.5O2  3CO2 H= –848.88 kJ 2Fe + 1.5O2  Fe2O3 H= –822.14 kJ Don’t forget to add states For more lessons, visit www.chalkbored.com

More Related