1.Calculate the change in enthalpy, in kJ, for the combustion of 6.75 moles of C
ID: 829276 • Letter: 1
Question
1.Calculate the change in enthalpy, in kJ, for the combustion of 6.75 moles of C6H14 in oxygen. Report your answer in scientific notation (X.XXEX) with three significant figures. Do not include units in your answer. The enthalpy of formation of C6H14 is -198.7 kJ mol-1. 2.Calculate the change in enthalpy, in kJ, for the combustion of 4.25 moles of C2H6 in oxygen. Report your answer in scientific notation (X.XXEX) with three significant figures. Do not include units in your answer. The enthalpy of formation of C2H6 is -84.7 kJ mol-1. * It would be great if you guys showed the work clearly so I know how to do it later on for reference. :)
Explanation / Answer
a)C6H14(l) + (19/2)O2(g) ---> 6CO2(g) + 7H2O(l)
O2 (g)
zero KJ/mole
CO2 (g)
-393.5 KJ/mole
H2O (l)
-285.8 KJ/mole
Hrxn=6Hf(CO2)+7Hf(H2O)-[Hf(hexane)+9.5Hf(O2)]
Hrxn=6(-393.5)+7(-285.8)-[-198.7 +9.5(0)]
Hrxn=-2361-2000.6+198.7=-4162.9=-4.162E3KJ/mole
combustion of 6.75moles of C6H14=2.498E+4KJ/mole
b) C2H6(l) + (7/2)O2(g) ---> 2CO2(g) + 3H2O(l)
Hrxn=2Hf(CO2)+3Hf(H2O)-[Hf(ethane)+7/2Hf(O2)]
Hrxn=2(-393.5)+3(-285.8)-[Hf(ethane)+7/2Hf(O2)]
Hrxn=-787-857.4+84.7=-1559.7=1.559E+3KJ/mole
Combustion 4.25moles of C2H6=-6.628E+3KJ/mole
O2 (g)
zero KJ/mole
CO2 (g)
-393.5 KJ/mole
H2O (l)
-285.8 KJ/mole
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