10-3 changes in enthalpy during chemical...

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10-3 Changes in Enthalpy During Chemical Reactions 1) Explain the principles of calorimetry 2) Define and use Hess’s law and standard enthalpies of formation to calculate ΔH

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Page 1: 10-3 Changes in Enthalpy During Chemical Reactionsvanguardchemistry.weebly.com/.../4/7/92475196/10-3_changes_in_e… · 10-3 Changes in Enthalpy During Chemical Reactions •1) Explain

10-3 Changes in Enthalpy During Chemical Reactions

• 1) Explain the principles of calorimetry

• 2) Define and use Hess’s law and standard enthalpies of formation to calculate ΔH

Page 2: 10-3 Changes in Enthalpy During Chemical Reactionsvanguardchemistry.weebly.com/.../4/7/92475196/10-3_changes_in_e… · 10-3 Changes in Enthalpy During Chemical Reactions •1) Explain

Application of Enthalpy

• ΔH used to determine calories in food

• Food burned in calorimeter

• Heat captured (enthalpy change) calories in food

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Standard Enthalpy of Formation, ΔH0

f

= ΔH to synthesize 1 mol of a compound from elements in standard states

Example:

• C + 2H2 → CH4 ΔH0f = -74.9 kJ/mol

Table 2 (pg. 355) and Appendix A

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Hess’s Law

Total energy change ΔH = sum of each of the steps in the overall reaction

Example:

You run the 5K in cross country. Your 1K splits add up to your final time for the race.

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Hess’s Law and Formation of NaCl

Sum of all steps = total enthalpy change (ΔH)

Na(s) + ½Cl2(g) → NaCl(s) ΔH = -411.3 kJ/mol

Vaporize Na: Na(s) -> Na(g) 107.3 kJ/mol

Ionize Na: Na(g) -> Na+(g) + e- 495.8 kJ/mol

Dissociate C2 bonds: ½ Cl2(g) -> Cl(g) 121.7 kJ/mol

Electron Affinity Cl: Cl(g) + e- -> Cl-(g) -348.8 kJ/mol

Lattice Energy NaCl: Na+(g) + Cl-

(g) -787.3 kJ/mol

-411.3 kJ/mol

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Why Use Hess’s Law?

• Some energy changes are difficult to measure directly

• 2 C + O2 2 CO ΔH = ? Hard to do in lab!

• Can measure ΔH for: • C + CO CO2

• CO + O2 CO2

• Use Hess’s Law to solve ΔH for formation of CO

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Hess’s Law and Algebra

Chemistry and algebra follow many of same rules!

Reverse direction of reaction:

2C + O2 → 2CO ΔH = -221 kJ

2CO → 2C + O2 ΔH = 221 kJ (note sign change!)

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Hess’s Law and Algebra

Addition

2C + O2 → 2CO ΔH = -221 kJ

CO2 → C + O2 ΔH = 393 kJ

2C + O2 + CO2 → 2CO + C + O2 ΔH = 172 kJ

Cancellation

2C + O2 + CO2 → 2CO + C + O2 ΔH = 172 kJ

C + CO2 → 2CO ΔH = 172 kJ

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Sample Problem – Using Hess’s Law to Calculate ΔH

The enthalpy of reaction for the combustion of C to CO2 is -393.5 kJ, and the enthalpy for the combustion of CO to CO2 is -566.0 kJ:

(1) C(s) + O2(g) CO2(g) ΔH1 = -393.5 kJ

(2) 2 CO(g) + O2(g) 2 CO2(g) ΔH2 = -566.0 kJ

Using these date, calculate the enthalpy for the combustion of C to CO:

(3) 2 C(s) + O2(g) 2 CO(g) ΔH3 = ?

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Example

Given:

SO2(g) + NO2(g) → SO3(g) + NO(g) ΔH = ?

How do we solve this?

• Find standard enthalpy of formation for reactants and products

ΔHtotal = ΔHproducts - ΔHreactants

-41.8 kJ

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Example

Given:

C + CO2 → 2CO ΔH = ?

ΔHtotal = ΔHproducts - ΔHreactants

(Elements have ΔH0f = 0)

ΔH = 172.5 kJ

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Practice

Calculate enthalpy change for

NO(g) + ½O2(g) → NO2(g)

ΔH = -57.2 kJ

Calculate ΔH for

C2H6(g) + 3.5 O2(g) → 2CO2(g) + 3H2O(g)

Is it exothermic or endothermic?

-1428.6 kJ; exothermic

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Guided Practice

Calculate the change of enthalpy, ΔH, for the formation of pentane, C5H12, using the info below:

C(s) + O2(g) CO2(g) ΔH = -393.5 kJ

H2(g) + 1/2O2(g) H2O(l) ΔH =-285.8 kJ

C5H12(g) + 8O2(g) 5CO2(g) +6H2O(l) ΔH = -3535.6 kJ

Ans:

5C(s) + 6H2(g) C5H12(g) ΔH = -145.7 kJ