Question

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

Answer 1

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!)

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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.

That indicates the water formed should be a gas, as the container is not rigid and closed, and condensation is not induced.

(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)`

And you were given

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

which is the same as the enthalpy of reaction for a combustion reaction, forming ONE mol of nitromethane. This is given in general 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 realize that the reaction utilizes TWO mols of nitromethane, that's going to actually give you, for this 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.)