How do I calculate the standard enthalpy of formation (ΔH∘f) for nitromethane?

Question

Ellie Andrews · Accepted Answer

I got #color(blue)(-"47 kJ/mol")#, assuming a constant-pressure calorimeter. So, I answer the question as-written...

The "correct" answer marked by MasteringChemistry was #-"401.6 kJ/mol"#, but that is based on a #DeltaH_C^@# that doesn't account for how the reaction given here uses #2# mols of nitromethane, rather than #1#.

(Note: The NIST reference I cited below uses a constant-volume calorimeter with a small difference between #DeltaE_C^@# and #DeltaH_C^@#, so it is a different situation!)

I think your question isn't totally clear. Some other websites claim the water formed is a liquid, but in order to get #DeltaH_C^@#, you need a constant-pressure calorimeter, which is open to the atmosphere.

Given that the container is not closed and rigid and that condensation is not induced, it follows that the water that forms should be a gas.

(If it were, then the question should have asked for the standard internal energy of combustion, #DeltaE_C^@#, to be more accurate.)

#2CH_3NO_2(l)+3/2O_2(g)→2CO_2(g)+3H_2Ocolor(red)((g))+N_2(g)#

Additionally, you were given

#DeltaH_C^@ = -"709.2 kJ/mol"#,

which is typically provided by:

#CH_3NO_2(l)+3/4O_2(g)→CO_2(g)+3/2H_2O(g)+1/2N_2(g)#

#DeltaH_(rxn)^@ = sum_(P) nu_P DeltaH_(f,P)^@ - sum_(R) nu_R DeltaH_(f,R)^@#

#= -"709.2 kJ/mol"#,

where #P# and #R# stand for products and reactants, respectively, and #nu# is the stoichiometric coefficient.

And here, you are asked to find #DeltaH_(f,CH_3NO_2(l))^@#. From your appendix, you can look these up:

#DeltaH_(f,O_2(g))^@ = "0 kJ/mol"#

#DeltaH_(f,CO_2(g))^@ = -"393.5 kJ/mol"#

#DeltaH_(f,H_2O(g))^@ = -"241.8 kJ/mol"#

#DeltaH_(f,N_2(g))^@ = "0 kJ/mol"#

When you consider that TWO mols of nitromethane are used in this reaction, you will actually obtain the following for the reaction as written:

#2(-"709.2 kJ/mol") = overbrace([2DeltaH_(f,CO_2(g))^@ + 3DeltaH_(f,H_2O(g))^@ + 1(0)])^"Products" - overbrace([2DeltaH_(f,CH_3NO_2(l))^@ + 3/2 (0)])^"Reactants"#

Now just solve for #DeltaH_(f,CH_3NO_2(l))^@# as the only unknown.

#color(blue)(DeltaH_(f,CH_3NO_2(l))^@) = {2xx(-"709.2 kJ/mol") - [2DeltaH_(f,CO_2(g))^@ + 3DeltaH_(f,H_2O(g))^@]}/(-2)#

#= color(blue)(???)#

I get #color(blue)(-"47 kJ/mol")#. The factor of #2# in front of the #-709.2# makes the difference between #-47# and #-401.6# #"kJ/mol"#.

(If you had used the #DeltaH_f^@# for liquid water, you would get #-"113 kJ/mol"#, which is in the NIST reference, but the reference uses a bomb calorimeter, which is a constant-volume system that enforces formation of liquid water, not water vapor.)

Serenity Allen · Answer

undefined To calculate the standard enthalpy of formation (ΔH∘f) for nitromethane (CH3NO2), you need to subtract the standard enthalpies of formation of the reactants from the standard enthalpies of formation of the products, using Hess's law or tabulated values. The standard enthalpy of formation of nitromethane is -125.5 kJ/mol.