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# Calculating Δ H using molar heats of formation

Calculating Δ H using molar heats of formation. Chem 12. If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation,  H o f . Standard conditions (standard state): 1 atm and 25 o C (298 K).

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## Calculating Δ H using molar heats of formation

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1. If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Hof . • Standard conditions (standard state): 1 atm and 25 oC (298 K). • Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.

2. Examples – write formation reactions for each: • *Remember the compounds are formed directly from their elements. • H2SO4 • NH4Cl

3. You want only one mole of the product being formed. • Look up the ΔHf on the table • H2 + S + 2O2 H2SO4 ΔHf = -814 kJ/ mol 2. ½ N2 + 2H2 + ½ Cl2 NH4Cl ΔHf = -314.4 kJ/ mol

4. Enthalpies of Formation

5. Using Enthalpies of Formation for Calculating Enthalpies of Reaction • For a reaction

6. By definition, the enthalpy of formation of an element in its standard state is zero. Example, oxygen (O2) and chlorine (Cl2) both have Hof of zero.

7. Sample Problem 1 • Calculate ΔH for the following reaction using standard molar heats of formation, ΔH°f. • 2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g) ΔH = ?

8. ΔH°ffor NH3(g) = -45.9 kJ/mol • ΔH°ffor HCl(g) = -92.3 kJ/mol • ΔH°ffor Cl2(g) and N2(g) is 0 • ΔHrxn= Σ nΔH°f(product) - Σ nΔH°f(reactant) • ΔHrxn= (0 + 6(-92.3 kJ)) - (2(-45.9 kJ) + 0) = (-553 kJ) - (-91.8 kJ) = -461 kJ

9. Examples • Use the tables in the back of your book to calculate ΔH for the following reactions: • 4 CuO (s)  2 Cu2O (s) + O2 (g) • C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) • NH3 (g) + HCl (g)  NH4Cl (s) • 1. +292 kJ 2. -2220. kJ 3. -176.2 kJ

10. Practice: • Page 685 #15 • Page 687 #19-21 • Page 691 # 5

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