Homework Answers
sol: The balanced combustion reaction of 1 mol of ethylene is shown below.
C2H2(g) + 5/2 O2 (g) -> 2CO2(g) + H2O (l)
delta Hcombustion = summation of heat of formation of product - summation of heat of formation of reactant
= (2XHf(CO2)+Hf(H2O))-(Hf(C2H2)+5/2Hf(O2))
= (2X(-393.5) + (-295.8))-(226.73 + 5/2X0)mol x KJ/mol
= -1309.5 KJ
Hence for one mol of gas, the heat of combustion is -1309.5 KJ/molgas
When the gas in balloon, the cumbustion reaction is :
2C2H2(g) + 5 O2 (g) -> 4CO2(g) + 2H2O (l)
delta Hrxn = (4x(-393.5)+2x(-295.8))-(2x(226.73) + 5x0)molxKJ/mol
= - 2619 KJ
hence heat of reaction for the balloon is - 2619 KJ.
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