Calculating Heats of Reaction

Calculating Heats of Reaction

Hess’s Law and Heats of Formation

Introduction

• If we want to measure the heat of reaction of methane, we just burn a known amount of methane gas in oxygen and measure the heat evolved by using a calorimeter.

• But, sometimes we don’t want consume the material when we are interested in the heat of reaction.

• What is the heat of formation of a diamond?

Hess’s Law

• There is a way to measure the heat of reaction of a compound in a way that does not require destroying the compound.

• We use Hess’s Law:

• If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

Hess’s Law

• The law is easier to use than it is to say.

• For example:

• Find the heat of reaction to go from diamond to graphite.

• C(diamond) → C(graphite)

Hess’s Law


• We can’t measure this directly, because the process is too slow.

• But we do know the following information:

• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ

• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

Hess’s Law


• If we rearrange the first equation ...



Hess’s Law


• If we rearrange the first equation ...

• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ


Hess’s Law


• ... add them together ...



Hess’s Law


• ... add them together ...



• CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)

Hess’s Law


• ... and simplify ...




Hess’s Law






Hess’s Law






Hess’s Law


• ... we get the desired equation.




Hess’s Law


• Now, we add the heats of reaction.




Hess’s Law





• C(diamond) → C(graphite) ∆H = +393.5 kJ − 395.4 kJ

Hess’s Law





• C(diamond) → C(graphite) ∆H = − 1.9 kJ

• This is an exothermic process.

Hess’s Law

• Let’s try another one.

• Find ∆H for the reaction:

• C(graphite) + ½ O2(g) → CO(g)

• given:


• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

Hess’s Law


• Notice that C(graphite) is on the left of both equations.


• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

Hess’s Law


• This means that we will write the equation with C(graphite) the same way in our final equations.


• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

Hess’s Law


• Notice that CO(g) is on the right above and on the left of below.


• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

Hess’s Law


• This means that we will write the equation with CO(g) the opposite way in our final equations.


• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

Hess’s Law


• This means that we will write the equation with CO(g) the opposite way in our final equations.


• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ

Hess’s Law


• Next, we add everything together ... spacer spacer spacer


• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ

Hess’s Law




• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ

• C(graphite) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½ O2(g)

Hess’s Law


• ... and simplify ... spacer spacer spacer spacer spacer


• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ

• C(graphite) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½ O2(g) ½ O2(g)

Hess’s Law




• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ


Hess’s Law


• ... to get the equation that we want. spacer spacer spacer


• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ


Hess’s Law


• Now, we add the heats of reaction together. spacer spacer spacer


• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ


Hess’s Law




• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ

• C(graphite) + ½ O2(g) → CO(g) ∆H = −393.5 kJ + 283.0 kJ

Hess’s Law




• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ

• C(graphite) + ½ O2(g) → CO(g) ∆H = −110.5 kJ

•This is an exothermic process.

Hess’s Law

• Let’s try another one.

• Find ∆H for the reaction:

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• given:

• PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• Notice that P(s) is on the left of both equations. spacer spacer

• PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• This means that we will write the equation with P(s) the same way in our final equations.

• PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• Notice that PCl5(s) is on the right above and on the left of below.

• PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• This means that we will write the equation with PCl5(s) the opposite way in our final equations.

• PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• This means that we will write the equation with PCl5(s) the opposite way in our final equations.

• PCl3(g) + Cl2(g) → PCl5(s) ∆H = −87.9 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• Also notice that PCl5(g) is has a coefficient of “2” above and “1” of below.

• PCl3(g) + Cl2(g) → PCl5(s) ∆H = −87.9 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• This means that we need to multiply the equation below by 2.

• PCl3(g) + Cl2(g) → PCl5(s) ∆H = −87.9 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• This means that we need to multiply the equation below by 2.

• 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)


• 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)


• 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

• 2 PCl3(s) + 2 Cl2(g) + 2 P(s) + 3 Cl2(g) → 2 PCl5(s) + 2 PCl3(g)

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)


• 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

• 2 PCl3(s) + 2 Cl2(g) + 2 P(s) + 3 Cl2(g) → 2 PCl5(s) + 2 PCl3(g)

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)


• 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

• ... to get the equation that we want. spacer spacer spacer

• 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)


• 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)


• 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ − 574 kJ

Hess’s Law

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)


• 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ

• 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

• 2 P(s) + 5 Cl2(g) → 2 PCl5(s) ∆H = −749.8 kJ = −750. kJ

• This is a highly exothermic reaction.

Standard Heats of Reaction

• Changes in enthalpy (∆H) are usually specified at a set of standard conditions.

• Standard temperature = 25°C = 298 K

• Standard pressure = 1 atm = 101.3 kPa

• The change in enthalpy that accompanies the formation of one mole of a compound from its elements under standard conditions is called the standard heat of formation, ∆Hf

o.


• For example:

• H2(g) + ½ O2(g) → H2O(l) ∆Hfo = −285.8 kJ

• Ca(s) + ½ O2(g) → CaO(s) ∆Hfo = −635 kJ

• C(s) + 2 H2(g) → CH4(g) ∆Hfo = −74.86 kJ

• ½ N2(g) + ½ O2(g) → NO(g) ∆Hfo = 90.37

kJ


• For a reaction occurring at standard conditions, you can calculate the heat of reaction by using the standard heats of formation of the reactants and the products.

• ∆Ho = ∆Hfo(products) − ∆Hf

o(reactants)


• For example, to calculate the standard heat of reaction of methane burned in air.

• CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

• ∆Hfo(CH4, g) = −74.86 kJ

• ∆Hfo(O2, g) = 0.0 kJ

• ∆Hfo(CO2, g) = −393.5 kJ

• ∆Hfo(H2O, l) = −282.8 kJ



• CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)


o(reactants)

• ∆Hfo(products) = ∆Hf

o(CO2) + 2 ∆Hfo(H2O)

• = −393.5 kJ + 2(−285.8 kJ) = −965.1 kJ

• ∆Hfo(reactants) = ∆Hf

o(CH4) + 2 ∆Hfo(O2)

• = −74.86 kJ + 2(0.0 kJ) = −74.86 kJ



• CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

• ∆Ho = −965.1 kJ − (−74.86 kJ) = −890.2 kJ



• Let’s do another one.

• Calculate the standard heat of reaction of :

• SO2(g) + ½ O2(g) → SO3(g)

• ∆Hfo(SO2, g) = −296.8 kJ

• ∆Hfo(O2, g) = 0.0 kJ

• ∆Hfo(SO3, g) = −395.7 kJ



• SO2(g) + ½ O2(g) → SO3(g)


o(reactants)

• ∆Hfo(products) = ∆Hf

o(SO3)

• = −395.7 kJ

• ∆Hfo(reactants) = ∆Hf

o(SO2) + ½ ∆Hfo(O2)

• = −296.8 kJ + ½(0.0 kJ) = −296.8 kJ



• SO2(g) + ½ O2(g) → SO3(g)

• ∆Ho = −395.7 kJ − (−296.8 kJ) = −98.9 kJ


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Calculating Heats of Reaction