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Question

The reaction of 1 mol of benzene take place at 298 K and 1 atm as-
C6H6(l)+152O2(g)6CO2(g)+3H2O(l)
In this reaction CO2(g) and H2O(l) are produced and 3267.0 kJ heat is liberated. calculate the ΔfHo for benzene.
ΔfHo for CO2(g) and H2O(l) are 393.5 kJmol1 and 285.83 kJmol1 respectively.

A
68.51 Jmol1
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B
70.51 Jmol1
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C
55.51 kJmol1
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D
48.51 kJmol1
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Solution

The correct option is D 48.51 kJmol1
Given, ΔfHoCO2(g)=393.5 kJmol1,ΔfHoH2O(l)=285.83 kJmol1

Reaction is:
C6H6(l)+152O2(g)6CO2(g)+3H2O(l)
ΔrH=3267.0 kJ

so,
ΔfHo for benzene =6ΔfHo(CO2(g))+3ΔfHo(H2O(l)ΔrH

putting all the values,
=3267.0+[6×(393.5)3×(285.83)]
=3267.0+[2361857.49]
=48.51 kJmol1
Hence, enthalpy of formation of benzene is 48.51 kJmol1

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