Calculating Enthalpies of Reaction

1. Calculating Enthalpies of Reaction PbO (s) + CO (g) Pb (s) + CO2(g) • What is the enthalpy (or change in energy) of this reaction? • To find the answer, figure out… • 1) How much energy is needed the break the reactants apart • 2) How much energy is released when the products form

2. What is the change in enthalpy of this reaction? Figure out… 1) How much energy is needed & 2) How much energy is released PbO (s) + CO (g) Pb (s) + CO2(g) • Step 1: Look at the reactants (they break apart) This is #1 (How much energy is needed) PbO (s) Pb (s) + O2(g) Requires 217.3 kJ *How do I know it is 217.3 kJ? I look up ΔH of PbO on p. 345. This tells me how much energy is required to break down PbO. + CO (g) C + O2(g) Requires 110.5 kJ ________ Total e! needed: 327.8 kJ *How do I know it is 110.5 kJ? I look up ΔH of CO on p. 344-345.

3. What is the change in enthalpy of this reaction? Figure out… 1) How much energy is needed & 2) How much energy is released PbO (s) + CO (g) Pb (s) + CO2(g) • Step 2: Look at the products (they are formed) This is #2 (How much energy is released) Pb (s) Requires 0 kJ *How do I know it is 0 kJ? The enthalpy of formation for any element is zero. + CO2(g) Releases 393.5kJ -393.5 kJ ________ Total: -393.5 kJ *How do I know 393.5kJ are released? Look up ΔH of CO2 on p. 344-345. I write -393.5 kJ because heat is released (not required).

4. What is the change in enthalpy of this reaction? Figure out… 1) How much energy is needed & 2) How much energy is released PbO (s) + CO (g) Pb (s) + CO2(g) Now find the total… 1) Requires: 327.8 kJ (enthalpy for breakdown of reactants) 2) Releases: 393.5 kJ (enthalpy for formation of products) Enthalpy of reaction = 327.8 kJ + -393.5 kJ = -65.7 kJ *Why did we use a negative sign and write -393.5 kJ? It is because energy released is the opposite of energy required. • A negative Enthalpy of reaction (like -65.7) means the reaction is exothermic

5. Homework: CRQ 22abc & Prelab