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Calculating Enthalpies of Reaction

Calculating Enthalpies of Reaction. PbO (s) + CO (g)  Pb (s) + CO 2 (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.

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Calculating Enthalpies of Reaction

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  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

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