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"?
1 Answer
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 andR 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
H = U + PV
For
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),
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)))