Hess’s Law SECTION 5.3. Hess’s Law The enthalpy change of a physical or chemical process...
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Transcript of Hess’s Law SECTION 5.3. Hess’s Law The enthalpy change of a physical or chemical process...
Hess’s LawSECTION 5.3
Hess’s Law
The enthalpy change of a physical or chemical process depends only on the initial and final conditions of the process.
The enthalpy change of a multistep process is the sum of the enthalpy changes of its individual steps.
Problem
Iron can be obtained from the following reaction:
Fe2O3(s) + 3CO(g) → 3CO2 + 2Fe(s)
Determine the enthalpy change of reaction, given the following:
1. CO(g) + ½ O2 → CO2 ∆H°=-283.0kJ
2. 2Fe(s) + 3/2 O2 → Fe2O3(s) ∆H°=- 824.3kJ
Solution
Manipulate equations to make them match the overall equation:
#1 equation proceeds in the correct direction but must be multiplied by 3.
So:
CO(g) + ½ O2 → CO2 ∆H°=-283.0kJ X 3
3CO(g) + 3/2 O2 → 3CO2 ∆H°=-849.0 kJ
#2 equation needs to be reversed but the coefficients are ok.
So:
2Fe(s) + 3/2 O2 → Fe2O3(s) ∆H°= - 824.3kJ becomes:
Fe2O3(s) → 2Fe(s) + 3/2 O2 ∆H°= 824.3kJ (note the sign is reversed)
To get the overall equation: Since oxygen is the same on both sides of the
equation, cross it out and add the individual enthalpy changes:
Fe2O3(s) + 3CO(g) → 3CO2 + 2Fe(s) ∆H°= -24.7 kJ
Practice Problems – p. 316 – all questions
Standard Molar Enthalpies of Formation
Change in enthalpy when 1 mole of a compound is formed directly from its elements in their most stable state at SATP (25°C & 100kPa).
It’s a synthesis reaction when the compound is formed directly from its elements and not from any other compound.
Eg. C(s) + O2(g) → CO2 (g)
Coefficients are often fractions because there must be only 1 mole of product formed.
See Table 5.5 and Appendix B
Thermal Stability
The ability of a substance to resist decomposition when heated.
The greater the enthalpy change of a decomposition reaction, the greater its thermal stability.
Enthalpies of Formation and Hess’s Law
∆H°r =∑(n∆H°f products) - ∑(n ∆H°f reactants)
n= coefficient in equation ∑ = “sum of”
∆H°r = enthalpy change of reaction
Use standard molar enthalpies of formation on Appendix B
Determine the enthalpy of formation for the
following reaction:
CH4(g) + 2O2 → CO2 + 2H2O
∆H°r =∑(n∆H°f products) - ∑(n ∆H°f reactants)
**Use Appendix B
∆H°r =((1)(-393.2) + (2)(-241.9)) - ((1)(-74.6)(2)(0)) ** since oxygen is an element in its most stable state, standard molar enthalpy is 0.
= (-877.1kJ)-(-74.6kJ) = -802.5 kJ
Videos/Resources to study at home
Hess’s Law:
http://www.ausetute.com.au/hesslaw.html
https://www.youtube.com/watch?v=u7aTBxA7sL8
Hess’s Law and Enthalpies of Formation
http://www.chemguide.co.uk/physical/energetics/sums.html
https://www.youtube.com/watch?v=iQuy2mgbV9o
Practice Problems – p. 323 – #51, 52, 54, 55, 56, 58.
Research Application Questions
Choose from:
P. 291: #4, 10, 12
P. 311: #8, 12
P. 335: #2, 5, 6, 7, 8
P. 349: #62, 63, 65, 66, 67, 69, 70
P. 351: #23
P. 408-409: #68, 69