Calculating Heats of Reaction
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Transcript of 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
• C(diamond) → C(graphite)
• 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
• C(diamond) → C(graphite)
• If we rearrange the first equation ...
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
Hess’s Law
• C(diamond) → C(graphite)
• If we rearrange the first equation ...
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
Hess’s Law
• C(diamond) → C(graphite)
• ... add them together ...
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
Hess’s Law
• C(diamond) → C(graphite)
• ... add them together ...
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
• CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)
Hess’s Law
• C(diamond) → C(graphite)
• ... and simplify ...
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
• CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)
Hess’s Law
• C(diamond) → C(graphite)
• ... and simplify ...
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
• CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)
Hess’s Law
• C(diamond) → C(graphite)
• ... and simplify ...
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
• C(diamond) → C(graphite)
Hess’s Law
• C(diamond) → C(graphite)
• ... we get the desired equation.
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
• C(diamond) → C(graphite)
Hess’s Law
• C(diamond) → C(graphite)
• Now, we add the heats of reaction.
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
• C(diamond) → C(graphite)
Hess’s Law
• C(diamond) → C(graphite)
• Now, we add the heats of reaction.
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
• C(diamond) → C(graphite) ∆H = +393.5 kJ − 395.4 kJ
Hess’s Law
• C(diamond) → C(graphite)
• Now, we add the heats of reaction.
• CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ
• C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
• 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:
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• Notice that C(graphite) is on the left of both equations.
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• This means that we will write the equation with C(graphite) the same way in our final equations.
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• Notice that CO(g) is on the right above and on the left of below.
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• This means that we will write the equation with CO(g) the opposite way in our final equations.
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• This means that we will write the equation with CO(g) the opposite way in our final equations.
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• Next, we add everything together ... spacer spacer spacer
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• Next, we add everything together ... spacer spacer spacer
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• 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
• C(graphite) + ½ O2(g) → CO(g)
• ... and simplify ... spacer spacer spacer spacer spacer
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• 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
• C(graphite) + ½ O2(g) → CO(g)
• ... and simplify ... spacer spacer spacer spacer spacer
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ
• C(graphite) + ½ O2(g) → CO(g)
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• ... to get the equation that we want. spacer spacer spacer
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ
• C(graphite) + ½ O2(g) → CO(g)
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• Now, we add the heats of reaction together. spacer spacer spacer
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ
• C(graphite) + ½ O2(g) → CO(g)
Hess’s Law
• C(graphite) + ½ O2(g) → CO(g)
• Now, we add the heats of reaction together. spacer spacer spacer
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• 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
• C(graphite) + ½ O2(g) → CO(g)
• Now, we add the heats of reaction together. spacer spacer spacer
• C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
• 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)
• Next, we add everything together ... 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
Hess’s Law
• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
• Next, we add everything together ... 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 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)
• ... and simplify ... spacer spacer 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 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)
• ... and simplify ... spacer spacer 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)
• ... 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)
• Now, we add the heats of reaction together. 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)
• Now, we add the heats of reaction together. 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) ∆H = −175.8 kJ − 574 kJ
Hess’s Law
• 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
• Now, we add the heats of reaction together. 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) ∆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.
Standard Heats of Reaction
• 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
Standard Heats of Reaction
• 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)
Standard Heats of Reaction
• 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
Standard Heats of Reaction
• For example, to calculate the standard heat of reaction of methane burned in air.
• CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
• ∆Ho = ∆Hfo(products) − ∆Hf
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
Standard Heats of Reaction
• For example, to calculate the standard heat of reaction of methane burned in air.
• CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
• ∆Ho = −965.1 kJ − (−74.86 kJ) = −890.2 kJ
• This is a highly exothermic reaction.
Standard Heats of Reaction
• 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
Standard Heats of Reaction
• Calculate the standard heat of reaction of :
• SO2(g) + ½ O2(g) → SO3(g)
• ∆Ho = ∆Hfo(products) − ∆Hf
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
Standard Heats of Reaction
• Calculate the standard heat of reaction of :
• SO2(g) + ½ O2(g) → SO3(g)
• ∆Ho = −395.7 kJ − (−296.8 kJ) = −98.9 kJ
• This is a highly exothermic reaction.