Energy Lecture 4

Hess’s Law & Review

Calculating Enthalpy Change

• A theoretical way to determine ∆H for a chemical reaction is provided by Hess’s law, which states that if two or more thermochemical equations can be added to produce a final equation for a reaction, then the enthalpy change for the final reaction equals the sum of the enthalpy changes for the individual reactions.

Hess’s Law

• There is an amount of heat associated with every chemical reaction.

Hess’s Law

• Often you know the heat for parts of the reaction, and you must add them together to find the heat for the total reaction.

Applying Hess’s Law

• Use 2 thermochemical reactions to determine ∆H for the oxidation of ethanol (C2H5OH) to form acetaldehyde (C2H4O) and water.

Here is the overall reaction:

Here are the two component reactions:

ethanol + oxygen gas acetaldehyde + water

Applying Hess’s Law

• For the overall reaction, Acetaldehyde should be on the right side of the equation, so reverse equation a.

• Note that you must change the sign of ∆H.

• The desired equation has two moles of ethanol, so double equation b and its ∆H.

Applying Hess’s Law

• Add these two equations, and cancel any terms common to both sides of the combined equation.

• Note that ∆H is negative, which means the reaction is exothermic. (releasing energy)

∆ H = -349kJ

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Basic Assessment QuestionsBasic Assessment Questions

Practice Hess’s Law!

Use reactions a and b to determine ∆H for this single-displacement reaction.

Cl2(g) + 2HBr(g) 2 HCl (g) + Br2(g)

a. H2(g) + Cl2(g) 2HCl (g) ∆H= -185

b. H2(g) + Br2(g) 2HBr (g) ∆H= -73

Practice Hess’s LawKeep equation “a” as written because HCl is on the right

in the total reaction:

a. H2(g) + Cl2(g) 2HCl (g) ∆H= -185kJ

Flip equation “b” because HBr needs to be on the left in the overall equation

b. H2(g) + Br2(g) 2HBr (g) ∆H= -73kJ

b. 2HBr (g) H2(g) + Br2(g) ∆H=73kJ

Now add the two equations together

a. H2(g) + Cl2(g) 2HCl (g) ∆H= -185 kJ

b. 2HBr (g) H2(g) + Br2(g) ∆H= 73kJ

Cl2(g) + 2HBr(g) 2 HCl (g) + Br2(g)

-185 + 73 = -112kJ

∆H = -112

Answer

∆Hrxn = –112 kJ

Practice using q=cm∆T

A 15.6-g sample of ethanol absorbs 868 J as it is heated. If the initial temperature of the ethanol was 21.5°C, what is the final temperature of the ethanol?

Hint:

solve for ∆T then add 21.5!

A 15.6-g sample of ethanol absorbs 868 J as it is heated. If the initial temperature of the ethanol was 21.5°C, what is the final temperature of the ethanol?

∆T = q . c m

= 868 J (2.44J/gºC ) (15.6g)

∆T= 22.8ºC Remember this is not your final answer. You are looking for the final temp… so add the initial temp to this number

22.8ºC + 21.5ºC = 44.3ºC = final temp

Practice q=mc∆T again!

If 335 g water at 65.5°C loses 9750 J of heat, what is the final temperature of the water?

Answer

58.5°C

Practice using q=mol x ∆H

How much heat is evolved when 24.9 g of propanol (C3H7OH) is burned? ∆Hcomb

= –2010 kJ/mol

Molar mass of C3H7OH is 60.1g/mol

Answer

833 kJ