Question

Question #fb528

Answer 1

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

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

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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))`.