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AlyssaYeh_1B wrote:From the data the question gives, I did the calculation: -1500 - (-1300 + 2(-286)) = 312kJ/mol. However, the answer key shows that it's -312kJ/mol. What did I do wrong?
If you add the standard enthalpies up, it should be -1300 +1560 +2(-286) which comes out to -312
I think that the enthalpy given in the problem is enthalpy of combustion. Meaning that the enthalpy of formation of each respective molecule would be -enthalpy of combustion. Thus, when you do your calculations, it would be dHrxn = 1560 kJ/mol - 1300 kJ/mol - 2(286 kJ/mol). Which equals -312 kJ/mol.
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