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