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# Calculating Enthalpy Change

Download Presentation ## Calculating Enthalpy Change

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1. Calculating Enthalpy Change 15.4: Pgs. 534 - 541

2. Main Idea… • The enthalpy change for a reaction is the enthalpy of the products minus the enthalpy of the reactants

3. Hess’s Law • Most reactions occur in steps • What if you need to know the heat change of a step, but you can’t measure it? • Hess’s law allows you to gather this information indirectly • Elemental carbon can be found as graphite and diamond at 25oC C(diamond)  C(graphite) • Takes millions of years to go from diamond to graphite • Reaction is too slow to measure the heat change

4. Hess’s Law • Hess’s Law of Heat Summation: • If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction • Example: Conversion of diamond to graphite C(diamond)  C(graphite) C(s, diamond) + O2 (g)  CO2 (g) ΔH = -395.4kJ C(s, graphite) + O2 (g)  CO2 (g) ΔH = -393.5kJ • REVERSE the above equation, so that we can show diamond being converted into graphite CO2 (g)  C(s, graphite) + O2 (g) ΔH = 393.5kJ

5. Add the equations together! C(s, diamond)  C(s, graphite) ΔH = -1.9 kJ • The overall ΔH is negative, so you can see this is an exothermic process!

6. Now you try some! • Try #32 and 33 on pg. 537 • 32 = -385.4 kJ • 33 = -521 kJ cuz direction of b is changed to get desired equation!

7. Standard Enthalpy (Heat) of Formation - standard heat of formation– the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 298 K (25oC) • Free elements at 298 K & 1 atm have an enthalpy of zero. • Examples, HONClBrIF • Other ΔHºf values have been calculated • ΔHºreaction can be calculated as follows:

8. Let’s do an example! • What is the standard heat of reaction (ΔHº) for the following equation: O2 (g) + 2CO (g)  2CO2 • From your text: • ΔHºf O2 (g) = 0 (free element) • ΔHºf CO (g) = -110.5 kJ/mol • ΔHºf CO2 (g) = -393.5 kJ/mol • Calculate the ΔHºf of the reactants: • 2 mol CO, 1 mol O • 2 mol of CO  2 mol x -110.5 kJ/mol = -221.0 kJ • 1 mol O2 = 0 KJ (free element) • TOTAL: 0 kJ + -221.0 kJ = -221 kJ (reactants) • Calculate the ΔHºf of the products: • 2 mol CO2 • 2 mol CO2 2 x -393.5 kJ/mol = -787.0 kJ • Use to find the total! ΔHº= -787.0 kJ – (-221.0) kJ ΔHº = -566.0 kJ

9. Try a Practice Problem 2NO (g) + O2 (g)  2NO2 (g) ΔHºf NO = 90.37 ΔHºf O2 = 0.0 ΔHºf NO2 = 33.85 2(33.85) – [2(90.37) + 0] = -113.04 kJ