slide 1 of 57 7-7 indirect determination of h: hess’s law h is an extensive property. –...
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Slide 1 of 57
7-7 Indirect Determination of H:Hess’s Law
H is an extensive property.–Enthalpy change is directly proportional to the amount of substance in a system.
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7
N2(g) + O2(g) → 2 NO(g) H° = +180.50 kJ
½N2(g) + ½O2(g) → NO(g) H° = +90.25 kJ
• H changes sign when a process is reversed
NO(g) → ½N2(g) + ½O2(g) H° = -90.25 kJ
Hess’s Law and Enthalpies• The ideas on the previous slide are familiar.
∆H is an intensive quantity. → The bigger the steak we want to barbecue the more moles (or grams) of propane we’ll have to burn.
• H changes sign when a process is reversed. An analogy here would be climbing up or down a ladder. Climbing up our gravitational potential energy increases – climbing down our potential energy decreases (mgh rule!).
All ∆H’s Need Not Be Measured• Elemental nitrogen and oxygen react to form
a variety of oxides. We need not measure every ∆H experimentally – some can be calculated using data for a subset of all possible reactions. Reactions forming NXOY molecules can be exothermic or endothermic.
• N2(g) + O2(g) → 2 NO(g) Endothermic
• In the atmosphere lightning supplies the energy needed to form NO(g).
Nature’s Synthesis of NO(g)!
• The reaction of N2(g) and O2(g) to give NO(g) is an endothermic process. In nature, lightning storms supply the needed energy!
All ∆H’s Need Not Be Measured
• The reaction of NO(g) with additional oxygen is exothermic.
• NO(g) + ½ O2(g) → NO2(g) Exothermic
• Knowing the ∆H for this rxn and the one on the previous slide (by experiment) we can calculate ∆H for the rxn
• ½ N2(g) + O2(g) → NO2(g) (red brown, LAX)
Smog in Southern California
• The formation of NO2(g) is one factor in smog formation.
Slide 7 of 57
Hess’s Law Schematically
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Slide 8 of 57
• Hess’s Law of Constant Heat Summation
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7
If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.
½N2(g) + O2(g) → NO2(g) H° = +33.18 kJ
½N2(g) + O2(g) → NO(g) + ½ O2 (g) H° = +90.25 kJ
NO(g) + ½O2(g) → NO2(g) H°= -57.07 kJ
Slide 9 of 57
7-8 Standard Enthalpies of Formation,
•The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states.
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7
Hof
The standard enthalpy of formation of a pure element in its reference state is 0.
Slide 10 of 57
Liquid bromine vaporizing
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7
Br2(l) Br2(g) Hf° = 30.91 kJ
Slide 11 of 57
Diamond and graphite
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7
Element 79!
• Gold (Au), like most metals, is a shiny metal at room temperature (standard state!).
You can Mix a Metal and a Nonmetal
• If the “chemistry” is right! (One standard state?)
Enthalpies of Formation• Heats of combustion of hydrocarbons are
always –ve (exothermic reactions). Heats of formation of compounds can be +ve or –ve (endothermic and exothermic reactions).
• ½ H2(g) + ½ Cl2(g) → HCl(g) ∆Hof = -92.3 kJ
• ½ H2(g) + ½ I2(g) → HI(g) ∆Hof = +26.5 kJ
• If we knew the ∆Hof for every substance we
could calculate the ∆Ho value for every reaction.
Slide 15 of 57Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7
Slide 16 of 57
Some standard enthalpies of formation at 298.15 KFIGURE 7-18
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7
number of carbons
Combining Heats of Formation
• The process for calculating a general enthalpy change from heat of formation relies on the fact that enthalpy, H, is a state function. Thus, we can imagine proceeding from the reactants to the products by a one step (direct process) or by a two step process:
• Step 1: Reactants → Constituent Elements• Step 2: Constituent Elements → Products
REACTANTS PRODUCTS
CONSTITUENTELEMENTS
Heats of Formation
• The previous slide outlines a simple process for calculating Heats of Reactions (ΔHo’s) if the relevant Heats of Formation for all reactants and all products are known.
• ΔHoRxn = ΣΔHo
f(Products) – ΣΔHof(Reactants)
• In practice a particular temperature is specified for the Heats of Formation (usually 298K). Care must be taken to ensure that the correct phase for reactants and products has been specified.
Slide 20 of 57
Standard Enthalpies of Reaction
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7
FIGURE 7-20
•Computing heats of reaction from standard enthalpies of formation
Hoverall = -2Hf°NaHCO3
+ Hf°Na2CO3
+ Hf
°CO2
+ Hf°H2O
Slide 21 of 57
Diagramatic representation of equation (7.21)FIGURE 7-21
Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7
∆H° = ∑p∆Hf°(products) - ∑r∆Hf°(reactants) (7.21)
Examples: Use of Heats of Formation
• Class examples: Show how to calculate the standard enthalpy change for each of the following reactions using Heats of Formation data.
• (a) CaCO3(s) → CaO(s) + CO2(g)
• (b) 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)
• (c) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
Combustion Reactions – History• Combustion reactions for hydrocarbons are
important sources of energy for many processes. One could easily put together a spreadsheet to calculate the enthalpies of combustion for all hydrocarbons – if the heats of formation of the hydrocarbons were known. Historically this is backwards as enthalpies of combustion are used in most cases to determine heats of formation for hydrocarbons. n-octane will be used as an example.
8CO2(g) + 9H2O(l)
H(Enthalpy) (kJ mol∙ -1)
“Elements”
Products
Combustion of Liquid n-Octane – C8H18(l)
Reactants
Combustion of n-octane
• Class example: The standard enthalpy of combustion of n-octane was determined by experiment to be -5470.3 kJ mol∙ -1 at 298K. Use heat of formation data for the products of combustion to determine the heat of formation of n-octane.
Combining Thermochemical Equations• Unknown ΔH (and ΔU) values can be
calculated by combining known thermochemical equations which are often more complex than those written for ΔHo
f’s.
• It is often necessary to combine several thermochemical equations to obtain a ΔHo value not known from experiments. This requires that we compare the coefficients of substances appearing only once in the given and desired thermochemical equations.
Combining Themochemical Eqtns - Hydrazine
• Example: Determine ∆Ho for the reaction• N2H4(l) + 2 H2O2(l) → N2(g) + 4 H2O(l)
• Given:• N2H4(l) + O2(g) → N2(g) + 2 H2O(l) ∆Ho = - 622.2 kJ
• H2(g) + ½ O2(g) → H2O(l) ∆Ho = - 285.8 kJ
• H2(g) + ½ O2(g) → H2O2(l) ∆Ho = - 187.8 kJ
Combining Thermochemical Equations
Results of Combining EquationsEquation New Balanced Thermochemical Equations
-1 x Eq (1) N2H4(l) + O2(g) → N2(g) +2 H2O(l) ΔHo = -622.2 kJ
-2 x Eq (3) 2 H2O2(l) → 2 H2(g) + 2 O2(g) ) ΔHo = +375.6 kJ
+2 x Eq (2) 2 H2(g) + O2(g) ) → 2 H2O(l) ΔHo = - 571.6 kJ
SUM N2H4(l) + 2 H2O2(l) → N2(g) +4 H2O(l) ΔHo = - 818.2 kJ