# 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"?

Jan 2, 2018

I got:

$\Delta {H}_{r x n}^{\circ} = - {\text{318.4 kJ/mol OF}}_{2} \left(g\right)$
$\Delta {U}_{r x n}^{\circ} = - {\text{320.9 kJ/mol OF}}_{2} \left(g\right)$

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

$\text{OF"_2(g) + "H"_2"O"(g) -> "O"_2(g) + 2"HF} \left(g\right)$,

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:

$\overline{\underline{|}} \stackrel{\text{ ")(" "DeltaH_(rxn)^@ = sum_"Products" n_P DeltaH_(f,P)^@ - sum_"Reactants" n_R DeltaH_(f,R)^@" }}{|}$

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

So,

$\Delta {H}_{r x n}^{\circ} = \left[\text{1 mol" cdot "0 kJ/mol" + "2 mol" cdot -"268.6 kJ/mol"] - ["1 mol" cdot "23.0 kJ/mol" + "1 mol" cdot -"241.8 kJ/mol}\right]$

$= - \text{318.4 kJ}$,

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

INTERNAL ENERGY

For the standard internal energy change of reaction $\Delta {U}_{r x n}^{\circ}$, we begin from the definition of:

$H = U + P V$

For $\Delta$ changes, which are not small,

$\Delta H = \Delta U + \Delta \left(P V\right)$

$= \Delta U + {P}_{2} {V}_{2} - {P}_{1} {V}_{1}$

$= \Delta U + \left({P}_{2} + {P}_{1} - {P}_{1}\right) \left({V}_{2} + {V}_{1} - {V}_{1}\right) - {P}_{1} {V}_{1}$

$= \Delta U + \left({P}_{1} + \Delta P\right) \left({V}_{1} + \Delta V\right) - {P}_{1} {V}_{1}$

$= \Delta U + {P}_{1} \Delta V + {V}_{1} \Delta P + \Delta P \Delta V$

Since the atmospheric pressure is constant in this reaction,

$\overline{\underline{|}} \stackrel{\text{ ")(" "DeltaH_(rxn)^@ = DeltaU_(rxn)^@ + PDeltaV_(rxn)" }}{|}$

or

$\Delta {H}_{r x n}^{\circ} = {q}_{r x n}$, the heat flow,
,

with ${P}_{1} \equiv P$.

Assuming only ideal gases are involved in the reaction (which is open to the air), $P \Delta {V}_{r x n} \approx \Delta {n}_{\text{gas" cdot RT_"room}}$.

Therefore,

$\Delta {U}_{r x n}^{\circ} \approx \Delta {H}_{r x n}^{\circ} - \Delta {n}_{\text{gas" cdot RT_"room}}$

$= - \text{318.4 kJ" - [n_("HF"(g)) + n_("O"_2(g)) - (n_("OF"_2(g)) + n_("H"_2"O"(g)))] RT_"room}$

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

$= - \text{320.9 kJ}$,

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