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)))#
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Answer 2

DeltaH_(rxn)^@ = 92.4 kJ/mol, DeltaU_(rxn)^@ = 95.6 kJ/mol.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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