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Lecture 4: Hess’s Law

Lecture 4: Hess’s Law. Reading: Zumdahl 9.5 Outline: Definition of Hess’ Law Using Hess’ Law (examples). Q: What is Hess’s Law ?. Recall (lecture 3) Enthalpy is a state function . As such, D H for going from some initial state to some final state is pathway independent.

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Lecture 4: Hess’s Law

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  1. Lecture 4: Hess’s Law Reading: Zumdahl 9.5 Outline: Definition of Hess’ Law Using Hess’ Law (examples)

  2. Q: What is Hess’s Law ? • Recall (lecture 3) Enthalpy is a state function. As such, DH for going from some initial state to some final state is pathway independent. • Hess’s Law: DH for a process involving the transformation of reactants into products is not dependent on pathway. (This means we can calculate DH for a reaction by a single step, or by multiple steps)

  3. Using Hess’s Law When calculating DH for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine DH for our “single step” reaction.

  4. The reaction of interest is: N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ This reaction can also be carried out in two steps: N2 (g) + O2 (g) 2NO(g) DH = +180 kJ 2NO (g) + O2(g) 2NO2(g) DH = -112 kJ

  5. If we take the previous two reactions and add them, we get the original reaction of interest: N2 (g) + O2 (g) 2NO(g) DH = +180 kJ 2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ N2 (g) + 2O2 (g) 2NO2(g) DH = + 68 kJ

  6. Note: the important things about this example is that the sum of DH for the two reaction steps is equal to the DH for the reaction of interest. Big idea: We can combine reactions of known DH to determine the DH for the overall “combined” reaction.

  7. Hess’s Law: An Important Detail One can always reverse the direction of a reaction when making a combined reaction. When you do this, the sign of DH changes. N2(g) + 2O2(g) 2NO2(g) DH = +68 kJ 2NO2(g) N2(g) + 2O2(g) DH = - 68 kJ

  8. One more detail: • The magnitude of DH is directly proportional to the quantities involved. (This means DH is an “extensive” quantity). • So, if the coefficients of a reaction are multiplied by a number, the value of DH is also multiplied by the same number. N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ 2N2(g) + 4O2(g) 4NO2(g) DH = 136 kJ

  9. Using Hess’s Law: tips • When trying to combine reactions to form a reaction of interest, it is usually best to work backwards from the reaction of interest. • Example: What is DH for the following reaction? 3C (gr) + 4H2 (g) C3H8 (g)

  10. 3C (gr) + 4H2 (g) C3H8 (g) DH = ? You’re given the following reactions: C (gr) + O2 (g) CO2 (g) DH = -394 kJ C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ

  11. Step 1. Only reaction 1 has C (gr). Therefore, we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation. (x3) C (gr) + O2 (g) CO2 (g) DH = -394 kJ 3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ

  12. Step 2: To get C3H8 on the product side of the reaction, we need to reverse reaction 2, and change the sign of DH. C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ 3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ

  13. Step 3: Add two “new” reactions together to see what remains: 3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ 3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ 2 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

  14. Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction: 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ 3C (gr) + 4H2 (g) C3H8 (g) Need to multiply second reaction by 4

  15. Example (cont.) • Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction: 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ 4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ 3C (gr) + 4H2 (g) C3H8 (g)

  16. 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ 4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ 3C (gr) + 4H2 (g) C3H8 (g) DH = -106 kJ Which is the one step reaction of interest

  17. Another Example: • Calculate DH for the following reaction: H2(g) + Cl2(g) 2HCl(g) Given the following: NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ N2 (g) + 3H2 (g) 2NH3 (g) DH = - 92 kJ N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = - 629 kJ

  18. Step 1: Only the first reaction contains the product of interest (HCl), but as a reactant. Therefore, reverse this reaction and multiply by 2 to get stoichiometry correct. NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ 2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = +352 kJ

  19. Step 2: Need Cl2 as a reactant, therefore, add reaction 3 to result from step 1 and see what is left. 2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ N2(g) + 4H2(g) + Cl2(g) 2NH4Cl(s) DH = -629 kJ N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ

  20. Step 3: Use remaining known reaction in combination with the result from Step 2 to get final reaction. N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ ? ( N2 (g) + 3H2(g) 2NH3(g) DH = -92 kJ) H2(g) + Cl2(g) 2HCl(g) DH = ? Key: need to reverse the middle reaction

  21. Step 3. Use remaining known reaction in combination with the result from Step 2 to get final reaction. N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ 2NH3(g) 3H2 (g) + N2 (g) DH = +92 kJ 1 H2(g) + Cl2(g) 2HCl(g) DH = -185 kJ This is the desired reaction and resultant DH!

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