Nitromethane burns as fuel according to the following balanced combustion equation: 2 CH3NO2(l) + 3/2 O2(g) --> 2 CO2(g) + 3 H2O(g) + N2(g)

asked by @catherinea12 • 11 months ago • Chemistry • 5 pts

DHrxn = -1418 kJ. The standard enthalpy of combustion for nitromethane is -709.2 kJ/mol. Calculate the standard enthalpy of formation for nitromethane. Answer in kJ/mol.

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Hey there! To calculate the enthalpy of formation of nitromethane, we need to know the enthalpy of formations of other compounds in this reaction.

We know that enthalpy of the reaction equals to : deltaHrx = H formation of products - H formation of reactants.

Because compounds in their natural state have an H of formation of zero, we do not need to include them in the equation. O2 and N2 occur naturally.

You are given the deltaH of rxn, -1418kJ. Notice that the standard enthalpy of combustion of nitromethane is just half of the deltaH of rxn, that's because we are combusting 2 moles of nitromethane, and enthalpy for one mole is -709.2kJ.

After plugging in all the values, you get the standard enthalpy of formation for nitromethane.

Let us know if you have any questions about the solution. enter image description here

answered by @dashab2 • 11 months ago