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Question
The heat of combustion of benzene in a bomb calorimeter (i.e., constant volume) was found to be 3263.9 kJmol−1 at 25oC. Calculate the heat of combustion of benzene at constant pressure.
The heat of combustion of benzene in a bomb calorimeter (i.e., constant volume) was found to be 3263.9 kJmol−1 at 25oC. Calculate the heat of combustion of benzene at constant pressure.
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Solution
The correct option is C −3267.6 kJmol−1
The reaction is:
C6H6(l)+152O2(g)⇌6CO2(g)+3H2O(l)
In this reaction, O2 is the only gaseous reactant and CO2 is the only gaseous product.
∴Δng=np−nr=6−152=−32
Also, given that ΔU=−3263.9 kJmol−1, T=25oC=25+273=298K,
R=8.314 JK−1mol−1
we know ΔH=ΔU+ΔngRT
ΔH=−3263.9−32×8.314×10−3×298
ΔH=−3263.9−3.7
ΔH=−3267.6 kJmol−1
The reaction is:
C6H6(l)+152O2(g)⇌6CO2(g)+3H2O(l)
In this reaction, O2 is the only gaseous reactant and CO2 is the only gaseous product.
∴Δng=np−nr=6−152=−32
Also, given that ΔU=−3263.9 kJmol−1, T=25oC=25+273=298K,
R=8.314 JK−1mol−1
we know ΔH=ΔU+ΔngRT
ΔH=−3263.9−32×8.314×10−3×298
ΔH=−3263.9−3.7
ΔH=−3267.6 kJmol−1
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