YR12 IB CHEM Lesson 3 Hess's Law.pptx

Hess's Law

1. Hess’s Law / Cycles How would you find out the enthalpy change of a reaction that you cannot measure directly?

2. Hess’s Law “the enthalpy change for a chemical reaction is the same, no matter what route is taken from the reactants to products.”

3. Hess’s Law ΔH f θ , ΔH c θ and bond enthalpy values are commonly used in thermodynamic cycles. These are in your data books!

4. Put a box around the part of the triangle where all arrows are going away from (this is the start). C (graph) + 2H 2(g) + 2O 2(g) START CO 2(g) + 2H 2 O CH 4(g) + 2O 2(g) ΔH f CH 4(g) -394 2 x -286 = -572 -890.4 Hess’s Cycle

5. Put a box around where all the arrows are going to (this is the finish) C (graph) + 2H 2(g) + 2O 2(g) START CO 2(g) + 2H 2 O FINISH CH 4(g) + 2O 2(g) ΔH f CH 4(g) -394 2 x -286 = -572 -890.4

6. There are two ways of going from the start to the finish (routes 1 and 2) C (graph) + 2H 2(g) + 2O 2(g) START CO 2(g) + 2H 2 O FINISH CH 4(g) + 2O 2(g) ΔH f CH 4(g) -394 2 x -286 = -572 -890.4 ROUTE 1 ROUTE 2

7. The enthalpy changes for routes 1 and 2 are the same C (graph) + 2H 2(g) + 2O 2(g) START CO 2(g) + 2H 2 O FINISH CH 4(g) + 2O 2(g) ΔH f CH 4(g) -394 2 x -286 = -572 -890.4 ROUTE 1 ROUTE 2

8. Hess’s Cycle Reactants Products Intermediate A B C ΔH (Route A) = ΔH (Route B) – ΔH (Route C)

9. ΔH r = Σ ΔH c(reactants) – Σ ΔH c(products) Reactants Products Combustion products ΔH r Σ ΔH c(reactants) Σ ΔH c(products) Using combustion data

10. You are provided with the following enthalpy changes of combustion Determine the enthalpy change for the following reactions: 4C(s) + 5H 2 (g) → C 4 H 10 (g) Substance C(s) H 2 (g) C 4 H 10 (g) ΔH c / kJ mol -1 -394 -286 -2877 Example

11. 4C(s) + 5H 2 (g) → C 4 H 10 (g) Example Combustion products Substance C(s) H 2 (g) C 4 H 10 (g) ΔH c / kJ mol -1 -394 -286 -2877

12. 4C(s) + 5H 2 (g) C 4 H 10 (g) Example ΔH 1 ΔH 2 ΔH r ΔH r = ΔH 1 – ΔH 2 Combustion products Substance C(s) H 2 (g) C 4 H 10 (g) ΔH c / kJ mol -1 -394 -286 -2877

13. 4C(s) + 5H 2 (g) C 4 H 10 (g) Example ΔH 1 = 4 x (-394) + 5 x (-286) ΔH r ΔH r = ΔH 1 – ΔH 2 ΔH 2 = -2877 Combustion products Substance C(s) H 2 (g) C 4 H 10 (g) ΔH c / kJ mol -1 -394 -286 -2877

14. 4C(s) + 5H 2 (g) C 4 H 10 (g) Example ΔH 1 = 4 x (-394) + 5 x (-286) Δ r H Δ r H = [4(-394) + 5(-286) ] – (-2877) = -129 kJ mol -1 ΔH 2 = -2877 Pay attention to: Units! Directions of arrows! Coefficients! Combustion products

15. You are provided with the following enthalpy changes of combustion Determine the enthalpy change for the following reactions: 2C(s) + 3H 2 (g) + ½ O 2 (g) → C 2 H 5 OH(l) Substance C(s) H 2 (g) C 2 H 5 OH(l) ΔH c / kJ mol -1 -394 -286 -1367 Try it yourself

16. Answer 2C(s) + 3H 2 (g) + ½ O 2 (g) C 2 H 5 OH(l) Combustion products ΔH r 2ΔH c (C) + 3ΔH c (H 2 ) ΔH c (C 2 H 5 OH) ΔHr = [(2 x -394) + (3 x -286)] – (-1367) = -279 kJ mol -1 Substance C(s) H 2 (g) C 2 H 5 OH(l) ΔH c / kJ mol -1 -394 -286 -1367

17. Using formation data ΔH r = - Σ ΔH f(reactants) + Σ ΔH f(products) Reactants Products Elements ΔH r Σ ΔH f(reactants) Σ ΔH f(products)

18. Example You are provided with the following enthalpy changes of formation Determine the enthalpy change for the following reaction: 2NO (g) + O 2 (g) → 2NO 2 (g) Substance NO 2 (g) NH 3 (g) NO(g) H 2 O(l) HNO 3 (l) ΔH f / kJ mol -1 33 -46 90 -286 -174

19. Example 2NO(g) + O 2 (g) 2NO 2 (g) Elements ΔH r 2ΔH f (NO) 2ΔH f (NO 2 ) ΔH r = -(2 x 90) + (2 x 33) = -114 kJ mol -1 Substance NO 2 (g) NH 3 (g) NO(g) H 2 O(l) HNO 3 (l) ΔH f / kJ mol -1 33 -46 90 -286 -174

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