Question #fb528

1 Answer
Truong-Son N.
Jan 7, 2017

#DeltaH > DeltaE# for #color(red)((2))#, and #DeltaH ~~ DeltaE# for #color(red)((1))#.

For ideal gases, #DeltaH = DeltaE# for #color(red)((1))#.

Recall that #DeltaH# is the change in enthalpy and #DeltaE# is the change in internal energy.

In an open container, we should recognize the condition of constant pressure (the phrase "open to the air", or "coffee-cup calorimeter", imply constant pressure as well). As a result, we can use this equation:

#DeltaH = DeltaE + Delta(PV)#

#= DeltaE + PDeltaV + cancel(VDeltaP + DeltaPDeltaV)#

#= DeltaE + PDeltaV#

where the result is something you should recognize (it was given in your textbook). Note that #E# should not be confused with the total energy.

Your two reactions were:

#"H"_2(g) + "Br"_2(g) -> 2"HBr"(g)# #" "" "" "" "" "" "color(red)((1))#

#"C"(s) + 2"H"_2"O"(g) -> 2"H"_2(g) + "CO"_2(g)# #" "" "color(red)((2))#

If we assume ideal gases, then in #color(red)((1))# we have two #"mol"#s of gases going to two #"mol"#s of gases, and in #color(red)((2))#, we have two #"mol"#s of gases going to three #"mol"#s of gases.

Note that the change in volume due to gas formation is significantly more than due to liquid formation, for instance (gases take up way more space, since they generally have a density over 1000 times as small as liquids or solids).

In this approximation, #color(red)((2))# has a nonnegative in volume, since #Deltan_(gas) > 0#, and #PDeltaV = Deltan_(gas)RT# for a process at constant pressure containing ideal gases.

Therefore, #DeltaH > DeltaE# for #color(red)((2))#, and #DeltaH ~~ DeltaE# for #color(red)((1))#.

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