communicating enthalpy change. method 1: molar enthalpies of reaction, Δ r h m to communicate a...
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Communicating Enthalpy Change
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Method 1: Molar Enthalpies of Reaction, ΔrHm
To communicate a molar enthalpy, both the substance and the reaction must be specified.
When reactants and products are in their standard state, they are at a pressure of 100 kPa, an aqueous concentration of 1.0 mol/L. and liquids and solids are in their pure state.
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12 2 32C(s) + 2 H (g) + O (g) CH OH(l)
Δf Hm° = –239.2 kJ/mol
When 1 mol of methanol is formed from its elements when they are in their standard states at SATP, 239.2 kJ of energy is released.
33 2 2 22CH OH(l) + O (g) CO (g) + 2 H O(l)
Formation Reaction
Combustion Reaction Δc Hm° = –725.9 kJ/mol
CH3OH
CH3OH
The complete combustion of 1 mol of methanol releases 725.9 kJ of energy.
Note that the above reactions are balanced for one mole of the compound.
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Method 2: Enthalpy Changes, ΔrH
Write an enthalpy change (Δr H) beside the chemical equation.
CO(g) + 2 H2(g) → CH3OH(l) Δr H = –725.9 kJ
The enthalpy change is not a molar value, so does not require the “m” subscript and is not in kJ/mol.
12 2 32SO (g) + O (g) SO (g) Δc H° = –98.9 kJ
2 2 32 SO (g) + O (g) 2 SO (g) Δc H° = –197.8 kJ
When 2 moles of sulfur dioxide are burned, twice as much heat energy is released as when 1 mole of sulfur dioxide is burned.
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2 SO2(g) + O2 (g) 2 SO3(g)
Then get the chemical amount of sulfur dioxide from its coefficients in the balanced equation and use ΔcH° = n ΔcH°m
ΔcH° = n ΔcH°m
= 2 mol x (-98.9 kJ)/1 mol
= -197.8 kJ
Finish it off by communicating the enthalpy next to a balanced equation
2 SO2(g) + O2 (g) 2 SO3(g) ΔcH° = -197.8 kJ
Sulfur dioxide and oxygen react to form sulfur trioxide. The standard molar enthalpy of combustion of sulfur dioxide, in this reaction, is -98.9 kJ/mol. What is the enthalpy change for this reaction?
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Another example...Wild natural gas wells are sometimes lit on fire to eliminate the very toxic hydrogen sulfide gas. The standard molar enthalpy of combustion of hydrogen sulfide is -518.0 kJ/mol. Express this value as a standard enthalpy change for the following:
2 H2S (g) + 3 O2(g) 2 H2O (g) + 2 SO2(g) ΔcH°= ?
ΔcH= n ΔcH°
= 2 mol x -518.0 kJ/mol
= -1 036.0 kJ
2 H2S (g) + 3 O2(g) 2 H2O (g) + 2 SO2(g) ΔcH°= -1 036.0 kJ
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Method 3: Energy Terms in Balanced Equations
reactants → products + energy
reactants + energy → products
For endothermic reactions, the energy is listed along with the reactants.
For exothermic reactions, the energy is listed along with the products.
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Method 4: Chemical Potential Energy Diagrams
During an exothermic reaction, the enthalpy of the system decreases. Heat flows out of the system and into the surroundings and we observe a temperature increase.
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Method 4: Chemical Potential Energy Diagrams
During an endothermic reaction, the enthalpy of the system increases. Heat flows into the system from the surroundings and we observe a temperature decrease.
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Read pgs. 495 – 500
Read over the Communication Example Problem 4 on page 500
pg. 501 Section 11.3 Questions #’s 1 – 7
Handout