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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-2220.1 kJ

-2044.1 kJ

-2396.1 kJ

-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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