1. Hess’s Law

2. Several reactions in chemistry occur in a series of steps, rather than just one step. For example, the following reaction explains the combustion (burning) of carbon: C (s) + O 2 (g) → CO 2(g) ΔH = -393.5 kJ However, sometimes the series below actually takes place in order to burn the carbon: Step 1: C (s) + ½ O 2(g) → CO (g) ΔH = -110.5 kJ Step 2: CO (g) + ½ O 2(g) → CO 2 (g) ΔH = -283.0 kJ

3. Hess’s Law If we add these two reactions together, namely recognize that CO in step 2 can be replaced by the reaction that first produced it, we get: Step 2: CO (g) + ½ O 2(g) → CO 2 (g) Then: C (s) + ½ O 2(g) + ½ O 2(g) → CO 2(g) Which we can simplify to: C (s) + O 2(g) → CO 2(g)

4. Hess’s Law An easier way to show this addition is by putting all the steps in a row and doing the following: C (s) + ½ O2(g) → CO (g) CO (g) + ½ O2(g) → CO2 (g) We can add the ½ O2 molecules together to make 1 mole of O2. In summary: 1. Cancel off like terms on opposite sides of the equations. 2. Add like terms on same side of equations.

5. Hess’s Law C (s) + ½ O 2(g) → CO (g) ΔH = -110.5 kJ CO (g) + ½ O 2(g) → CO 2 (g) ΔH = -283.0 kJ C (s) + O 2(g) → CO 2 (g)

6. Hess’s Law Now, if we add the heat of the first reaction to the heat of the second reaction, we should get the heat of reaction for the overall reaction. ΔH 1 + ΔH 2 = ΔHtotal -110.5 kJ + (-283.0 kJ) = -393.5 kJ Therefore, the end result does not matter which path it takes. The carbon could combust right away, or it could take the path we just investigated. Either way, the enthalpy change in the end was the same.

7. Hess’s Law C (s) + ½ O 2(g) → CO (g) ΔH = -110.5 kJ CO (g) + ½ O 2(g) → CO 2 (g) ΔH = -283.0 kJ C (s) + O 2(g) → CO 2 (g) ΔH = -393.5 kJ

8. Hess’s Law Hess’s Law states that the enthalpy change for any reaction only depends on the energy states of the final products and initial reactants and is independent on the number of steps (or the pathway) in between.

9. Hess’s Law Ex) Given the intermediate steps in the production of tetraphosphorus decaoxide, P 4 O 10, calculate ΔH°f for P 4 O 10 given the following reactions: 1. 4 P + 3 O 2 → P 4 O 6 ΔH = -1640 kJ 2. P 4 O 6 + 2 O 2 → P 4 O 10 ΔH = -1344 kJ

10. Tips Tips for using Hess’s Law: 1. If you need to flip an equation, the energy term must change sign, or it must be on the opposite side of the equation. i.e. 4 P + 3 O 2 → P 4 O 6 ΔH = -1640 kJ P 4 O 6 → 4 P + 3 O 2 ΔH = + 1640 kJ

11. Tips 2. If you change the balancing coefficients by multiplying by a number, multiply the energy value by the same number. i.e 4 P + 3 O 2 → P 4 O 6 ΔH = -1640 kJ if you wanted to produce 2 moles of P4O6 then: 8 P + 6 O2 → 2P4O6 ΔH = -3280 kJ

12. Example The combination of coke and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. If we assume coke can be represented by graphite, the equation for the production of coal gas is: 2C (s) + 2H 2 O (g)  CH 4 (g) + CO 2 (g) Determine the overall enthalpy change for the above reaction based on the following intermediate steps: C (s) + H 2 O (g)  CO (g) + H 2 (g)∆H = 131.3kJ CO (g) + H 2 O (g)  CO 2 (g) + H 2 (g) ∆H = -41.2 kJ CH 4 (g) + H 2 O (g)  3H 2 (g) + CO (g) ∆H = 206.1 kJ