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Chapter 16 Hess’s Law

Chapter 16 Hess’s Law. HESS’S LAW. If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps. Hess’s law can be used to determine the enthalpy change for a reaction that cannot be

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Chapter 16 Hess’s Law

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  1. Chapter 16 Hess’s Law

  2. HESS’S LAW • If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps.

  3. Hess’s law can be used to determine the enthalpy change for a reaction that cannot be measured directly!

  4. HNET =  H1 + H2 ADD THEM UP ALEGBRAICALLY N2(g) + O2(g)  2NO(g) ΔH1= +181kJ 2NO(g) + O2(g)  2NO2(g) ΔH2= -113kJ

  5. First, add up the chemical equations. N2(g) + O2(g)  2NO(g) ΔH1= +181kJ 2NO(g) + O2(g)  2NO2(g) ΔH2= -113kJ N2(g) + 2O2(g) 2NO(g) + 2NO(g)  + 2NO2(g)

  6. Notice that 2NO(g) is on both the reactants and products side and can be cancelled out. N2(g) + O2(g)  2NO(g) ΔH1= +181kJ 2NO(g) + O2(g)  2NO2(g) ΔH2= -113kJ N2(g) + 2O2(g) + 2NO(g)  2NO(g) + 2NO2(g)

  7. Write the net equation: N2(g) + 2O2(g) + 2NO(g)  2NO(g) + 2NO2(g) N2(g) + 2O2(g)  2NO2(g)

  8. HNET =  H1 + H2 Apply Hess’s Law to calculate the enthalpy for the reaction. ΔH1= +181kJ ΔH2= -113kJ

  9. HNET =  H1 + H2 Overall, the formation of NO2 from N2 and O2 is an endothermic process, although one of the steps is exothermic. ΔHNET= (+181kJ) + (-113kJ) ΔHNET= +68kJ

  10. 2NO(g) + O2(g) ΔH1 = +181kJ ΔH2 = -113kJ ΔH 2NO2(g) N2(g) + 2O2(g) ΔHNET = +68kJ Reaction Progress

  11. RULES for Hess’s Law Problems • If the coefficients are multiplied by a factor, then the enthalpy value MUST also be multiplied by the same factor. • 2.If an equation is reversed, the sign of ΔH MUST also be reversed.

  12. Practice Problem: #1 C(s) + ½O2(g)  CO(g) ΔH1= -110.5kJ CO(g) + ½O2(g)  CO2(g) ΔH2= -283.0kJ C(s) + O2(g) + CO(g)  CO(g) + CO2(g) Net Equation C(s) + O2(g)  CO2(g) HNET =  H1 + H2 HNET = (-110.5kJ) + (-283.0kJ) HNET = -393.5kJ

  13. Practice Problem: #3 C2H5OH(l)+ 3O2(g)  2CO2(g) + 3H2O(g) ΔH1= -1234.7kJ CH3OCH3(l)+ 3O2(g)  2CO2(g) + 3H2O(g) ΔH2= -1328.3kJ You have to REVERSE equation 2 to get the NET equation. DON’T forget to change the sign Of ΔH2

  14. Practice Problem: #3 C2H5OH(l)+ 3O2(g)  2CO2(g) + 3H2O(g) ΔH1= -1234.7kJ 2CO2(g) + 3H2O(g)  CH3OCH3(l)+ 3O2(g) ΔH2= +1328.3kJ C2H5OH(l)+ 3O2(g) + 2CO2(g) + 3H2O(g)  2CO2(g) + 3H2O(g) + CH3OCH3(l)+ 3O2(g) Net Equation C2H5OH(l) CH3OCH3(l)

  15. Net Equation C2H5OH(l) CH3OCH3(l) HNET =  H1 + H2 HNET = (-1234.7kJ) + (+1328.3kJ) HNET = +93.6kJ

  16. Practice Problem: #5 H2(g) + F2(g)  2HF(g) ΔH1= -542.2kJ 2H2(g) + O2(g)  2H2O(g) ΔH2= -571.6kJ You have to REVERSE equation 2 to get the NET equation. DON’T forget to change the sign Of ΔH2

  17. Practice Problem: #5 H2(g) + F2(g)  2HF(g) ΔH1= -542.2kJ 2H2O(g)  2H2(g) + O2(g) ΔH2= +571.6kJ You will need to multiply the first equation by 2. DON’T forget to multiply the ΔHby 2 also.

  18. Practice Problem: #5 2H2(g) + 2F2(g)  4HF(g) ΔH1= -1084.4kJ 2H2O(g)  2H2(g) + O2(g) ΔH2= +571.6kJ 2H2(g) + 2F2(g) + 2H2O(g)  4HF(g) + 2H2(g) + O2(g) 2F2(g) + 2H2O(g)  4HF(g) + O2(g) Net Equation HNET =  H1 + H2 HNET = (-1084.4kJ) + (+571.6kJ) HNET = -512.8kJ

  19. Hess’s Law Start Finish Enthalpy is Path independent. Both lines accomplished the same result, they went from start to finish. Net result = same.

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