objectives: today i will be able to: correctly manipulate thermochemical equations to predict the...
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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
€
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.
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