Thermochemisty (Enthalpy) and Hess’s Law

Chapter 10Sections 10.6-10.7

Enthalpy• Change in enthalpy (ΔHp = qp): the

amount of heat exchanged when heat exchange occurs under conditions of constant pressure

• Enthalpy is a state function – State function is defined as a property that changes

independent of pathway

• ΔH is independent of the path taken

At constant pressure, 890kJ of energyper mole of CH4 is produced as heat

Qp = ΔHp = -890 kJ exothermic

If 5.8 g of methane is burned, how much heat will be released?

5.8g CH4 x 1mol CH4 = 0.36 mol CH4 16.0 g CH4

0.36 mol CH4 x -890 kJ = -320 kJ mol CH4

Calorimeter• Calorimetry the

process of measuring heat flow in the form of temperature changes in a system

• Calorimeter an insulated device used to measure temperature changes

Hess’s Law• Hess’s Law

– ΔH for a multi-step process is the sum of ΔH for the individual steps

– Hess’s law allows us to calculate ΔH for new reactions from previously determined values

Germain Henri Hess

Altitude

San Francisco, CA0 ft

Salt Lake City, UT4,266 ft

Denver, CO5,280 ft

Flight 1

Flight 2

Flight 1: Δ Alt = -4266Flight 2: Δ Alt = +5280SL City-Denver Δ Alt = +1014

What is the change in altitude (Δ Alt) between Salt Lake City and Denver?

?

4266 ft

1014 ft

5280 ft

Hess’s Law

• Applying Hess’s Law– When reversing a reaction, change the

sign of ΔH. – When multiplying a reaction by a new

coefficient, multiply ΔH

Hess’s LawHess’s Law“In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

(products) (reactants)o oreaction p f r fH n H n H

Calculate the enthalpy change for a partial combustion of 1 mol

of graphite to diamond C(s, graphite) C(s, diamond) ΔH = ?

Given:C(s, graphite) + O2(g) CO2(g) ΔH = -393.5 KJC(s, diamond) + O2(g) CO2(g) ΔH = -395.4 KJ

Answer:C(s, graphite) + O2(g) CO2(g) ΔH = -393.5 kJCO2(g) C(s, diamond) + O2(g) ΔH = +395.4 kJC(s, graphite) C(s, diamond) ΔH = +1.9 kJ

Hess’s Law

Hess’s Law Example Hess’s Law Example ProblemProblemCalculate H for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O    Reaction Ho  

C + 2H2 CH4 -74.80 kJ C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

CH4 C + 2H2 +74.80 kJ

Calculate H for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O    Reaction Ho  

C + 2H2 CH4 -74.80 kJ C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ

Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

C + O2 CO2 -393.50 kJ

Calculate H for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O    Reaction Ho  

C + 2H2 CH4 -74.80 kJ C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ CH4 C + 2H2 +74.80 kJ

C + O2 CO2 -393.50 kJ

Step #3: Multiply reaction #2 by 2

2H2 + O2 2 H2O -571.66 kJ

Calculate H for the combustion of methane, CH4: CH4 + 2O2 CO2 + 2H2O

    Reaction Ho   C + 2H2 CH4 -74.80 kJ C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ

2H2 + O2 2 H2O -571.66 kJ

Step #4: Sum up reaction and HCH4 + 2O2 CO2 + 2H2O

-890.36 kJ

Calculation of Heat of Reaction Calculation of Heat of Reaction Calculate H for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2OHrxn = Hf(products) - Hf(reactants)

    Substance Hf  

CH4 -74.80 kJO2 0 kJ

CO2 -393.50 kJ

H2O -285.83 kJ

Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]Hrxn = -890.36 kJ