Objectives: Today I will be able to:

Correctly manipulate thermochemical equations to predict the enthalpy of reaction (Hess’s Law)

Informal assessment – monitoring student interactions and questions as they complete the practice problems

Formal assessment – analyzing student responses to the practice and exit ticket

Common Core Connection Make sense of problems and persevere in solving

them Build strong content knowledge Reason abstractly and quantitatively

Lesson SequenceEvaluate: Warm Up

Explain: Thermochemical Equations

Elaborate: Thermochemical Equations Practice

Explain: Hess’s Law

Elaborate: Hess’s Law Practice

Evaluate: Exit Ticket

Warm Up350 J are released as ice ( Specific Heat = 2.1

J / (g oC) ) cools from - 5.0 oC to -32 oC. What is the mass of ice?

ObjectivesToday I will be able to:

Correctly manipulate thermochemical equations to predict the enthalpy of reaction (Hess’s Law)

HomeworkFinish practice problems

AgendaWarm Up

Thermochemical Equations Notes

Thermochemical Equations Practice

Hess’s Law Notes

Hess’s Law Practice

Exit Ticket

Thermochemical Equations

Balanced Equation with two additions:Enthalpy (heat) accompanying

the reactionCoefficients represent moles – it

is possible to have fractions – Example: ½ means half a mole of the substance

State of matter is specified

Laws of Thermochemistry

Δ H is directly proportional to the amount of substance produced or reacting in a reaction

H2(g) + ½ O2(g) H2O(l)

Δ H = -285.8 kJ

2 H2(g) + O2(g) 2 H2O(l)

Δ H = -571.6 kJ

Laws of Thermochemistry

Δ H for a reaction is equal in magnitude but opposite in sign from the reverse reaction

HgO(s) Hg(l) + ½ O2(g)

Δ H = 90.7 kJ

Hg(l) + ½ O2(g) HgO(s)

Δ H = -90.7 kJ

Hess’s LawΔ H is independent of the

number of steps involved

If a reaction occurs in several steps, the sum of the enthalpy changes must equal Δ H for the overall reaction

Hess’s Law ExampleDetermine ∆H of the reaction

Sn(s) + 2 Cl2(g) SnCl4(l) using the information provided below.

Sn(s) + Cl2(g) SnCl2(s) ∆H= -83.6 kJ

SnCl2(s) + Cl2(g) SnCl4(l) ∆H = -46.7 kJ

Answer∆H = -130.3 kJ

Hess’s Law Example 2

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2 C (s) + 2 H2O (g) → CH4 (g) + CO2(g)

Determine the standard enthalpy change for this reaction from the following

standard enthalpies of reaction :

(1) C(s) + H2O (g) → CO (g) + H2 (g) ΔH° = 131.3 kJ

(2) CO (g) + H2O (g) → CO2 (g) + H2 (g) ΔH° = - 41.2 kJ

(3) CH4 (g) + H2O (g) → 3 H2 (g) + CO (g) ΔH° = 206.1 kJ

Answer+ 15.3 kJ

Exit TicketWhich question on the Hess’s Law practice did

you find the most challenging? We will start by reviewing this question tomorrow.