Question

For the gas-phase reaction `"OF"_2(g) + "H"_2"O"(g) -> "O"_2(g) + 2"HF"(g)`, the nonzero enthalpies of formation are `23.0`, `-241.8`, and `-"268.6 kJ/mol"`, respectively. What are the `DeltaH_(rxn)^@` and `DeltaU_(rxn)^@` in `"kJ/mol"`?

Answer 1

I got:

`DeltaH_(rxn)^@ = -"318.4 kJ/mol OF"_2(g)`
`DeltaU_(rxn)^@ = -"320.9 kJ/mol OF"_2(g)`

In this case we find that the change in internal energy is larger (more negative) than the change in enthalpy, since the mols of gas increased in this exothermic reaction.

---

For

`"OF"_2(g) + "H"_2"O"(g) -> "O"_2(g) + 2"HF"(g)`,

with

`DeltaH_f("OF"_2(g))^@ = "23.0 kJ/mol"`,
`DeltaH_f("H"_2"O"(g))^@ = -"241.8 kJ/mol"`,
`DeltaH_f("HF"(g))^@ = -"268.6 kJ/mol"`,

we assume the reaction occurs in a coffee-cup calorimeter , so that every gas here stays uncondensed at constant pressure (rather than constant volume in a bomb calorimeter).

ENTHALPY

The standard enthalpy change of reaction is then:

`barul|stackrel(" ")(" "DeltaH_(rxn)^@ = sum_"Products" n_P DeltaH_(f,P)^@ - sum_"Reactants" n_R DeltaH_(f,R)^@" ")|`

where `n` is the mols of a substance. `P` and `R` stand for product or reactant.

So,

`DeltaH_(rxn)^@ = ["1 mol" cdot "0 kJ/mol" + "2 mol" cdot -"268.6 kJ/mol"] - ["1 mol" cdot "23.0 kJ/mol" + "1 mol" cdot -"241.8 kJ/mol"]`

`= -"318.4 kJ"`,

or `color(blue)(DeltaH_(rxn)^@ = -"318.4 kJ"/("mol OF"_2(g)))`.

INTERNAL ENERGY

For the standard internal energy change of reaction `DeltaU_(rxn)^@`, we begin from the definition of:

`H = U + PV`

For `Delta` changes, which are not small,

`DeltaH = DeltaU + Delta(PV)`

`= DeltaU + P_2V_2 - P_1V_1`

`= DeltaU + (P_2 + P_1 - P_1)(V_2 + V_1 - V_1) - P_1V_1`

`= DeltaU + (P_1 + DeltaP)(V_1 + DeltaV) - P_1V_1`

`= DeltaU + P_1DeltaV + V_1DeltaP + DeltaPDeltaV`

Since the atmospheric pressure is constant in this reaction,

`barul|stackrel(" ")(" "DeltaH_(rxn)^@ = DeltaU_(rxn)^@ + PDeltaV_(rxn)" ")|`

or

`DeltaH_(rxn)^@ = q_(rxn)`, the heat flow,
,

with `P_1 -= P`.

Assuming only ideal gases are involved in the reaction (which is open to the air), `PDeltaV_(rxn) ~~ Deltan_"gas" cdot RT_"room"`.

Therefore,

`DeltaU_(rxn)^@ ~~ DeltaH_(rxn)^@ - Deltan_"gas" cdot RT_"room"`

`= -"318.4 kJ" - [n_("HF"(g)) + n_("O"_2(g)) - (n_("OF"_2(g)) + n_("H"_2"O"(g)))] RT_"room"`

`= -"318.4 kJ" - [2xx"HF" + 1xx"O"_2 - (1xx"OF"_2 + 1xx"H"_2"O") cancel"mols ideal gas"] cdot "0.008314472 kJ/"cancel"mol gases"cdotcancel"K" cdot 298.15 cancel"K"`

`= -"320.9 kJ"`,

or `color(blue)(DeltaU_(rxn)^@ = -"320.9 kJ"/("mol OF"_2(g)))`