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Q.

The standard heat of combustion of propane is -2220.1 kJ mol-1. The standard heat of vaporisation of liquid water is 44.0 kJ mol-1. What is ∆H0  ofC3H8g+5O2g → 3CO2g+4H2Og

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a

-2220.1  kJ

b

-2044.1  kJ

c

-2396.1  kJ

d

-2176.1  kJ

answer is B.

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Detailed Solution

∆Hvap0  of H2O  = 44.0 kJ mol-1 ∴ ∆Hvap0  of 4 mol H2O  =176.0 kJ ∴ ∆HRxn0 = ∆Hcomb0 C3H8 + ∆Hvap0  4H2O =-2220.1 + 176.0 =-2044.1 kJ
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