

4.2 a/b/c: Enthalpy and Hess' Law
Presentation
•
Chemistry, Science
•
10th - 12th Grade
•
Easy
Erin Hannan
Used 28+ times
FREE Resource
14 Slides • 4 Questions
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Syllabus dot points
a | explain the enthalpy changes in a reaction in terms of breaking and reforming bonds, and relate this to: – the law of conservation of energy |
b |
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c |
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3
Enthalpy = measure of total energy (potential) that is in the bonds of a system
We calculate in enthalpy changes or ΔH
endothermic rxn = positive ΔH
exothermic rxn = negative ΔH
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law of conservation of energy - can't be created or destroyed, only changed in form
requires energy to break bonds - endothermic
releases energy when forming bonds - exothermic
we determine net release or absorption of energy
Hess' Law brings all of this + 4.1 together
5
in a reaction, the enthalpy changes is always constant regardless of the steps within the reaction
directly relates to law of conservation of energy
reactions may have several pathways, but the overall enthalpy change will always be the same - because the amount of energy to break and form bonds will always be the same
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these are two pathways (though incomplete and complete combustion)
Reaction 1 = complete combustion, and Reaction 1 = Reaction 2 + Reaction 3
The enthalpy change data also supports this
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Multiple Choice
Hess' Law makes use of which principle to calculate the enthalpy change of a reaction?
The law of conservation of energy
The law of conservation of matter
The law that you will always find a lost item in the last place you look for it
Murphy's law
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aka heat of formation/reaction
ΔH°f
the increase in enthalpy when 1 mole of compound in its standard state is formed from elements in their standard state
standard state = 100kPa, 1mol/L, temp depends on state
forming an element means that ΔH°f = 0
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can either be done with just the data/equations, or with an energy cycle diagram
both can display the same information, and often it depends on the question - or in industry it's personal preference
both are just displaying ΔH° = ΣΔH°f (products) - ΣΔH°f (reactants)
must have 1 mole of product! Means you can have fractions on the left hand side, but means that you can use the data.
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Multiple Choice
Using the equations below:
C(s) + O2(g) → CO2(g) ∆H = –390 kJ
Mn(s) + O2(g) → MnO2(s) ∆H = –520 kJ
what is ∆H (in kJ) for the following reaction?
MnO2(s) + C(s) → Mn(s) + CO2(g)
910
130
-130
-910
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negative of the enthalpy change for the combustion process
because enthalpy change is heat gained, and heat of combustion is heat released when a substance undergoes combustion
calculating is simple; use ΔH°C = ΣΔH°f (products) - ΣΔH°f (reactants)
Either asked to calculate it out right, or calculate the heat of formation of a reac/prod
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both are multistep reactions, but are simplified into one step
they are also the reverse of one another
except photosynthesis = endothermic, and respiration = exothermic
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Multiple Choice
How do you calculate the Enthalpy of Reaction?
ΔH = ΔHproducts - ΔHreactants
ΔG = ΔH -TΔS
ΔT = q / mC
E = mc2
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Open Ended
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