Question

If `DeltaH_c^@` of solid benzoic acid at 27°C is -x kcal/mol, then `DeltaE_c^@` (in kcal/mol) is?

1. -x + 0.9
2. -x + 0.3
3. -x - 0.9
4. -x - 0.3

EDIT: In a bomb calorimeter.
- Truong-Son

Answer 1

`"2. -x + 0.3"`

Explanation:

Combustion of 1 mol of benzoic acid

`"C"_6"H"_5"COOH"_"(s)" + 15/2"O"_"2(g)" -> "7CO"_("2(g)") + "3H"_2"O"_"(l)"`

`Δ"n"_"g" = 7 - 15/2 = -0.5`

`Δ"H"_"c"^0 = Δ"E"^0 + Δ"n"_"g" "RT"`

`Δ"E"^0 = Δ"H"_c^0 - Δ"n"_"g""RT"`

`Δ"E"^0 = -"x kcal/mol" - [-0.5 × 1.987 × 10^-3 "kcal"//"(mol K)" × "300 K"]`

`Δ"E"^0 = -"x" + 0.298 ≈ color(blue)(-"x" + 0.3)`

Answer 2

By definition, at constant temperature and pressure, the change in enthalpy is given by:

`DeltaH = DeltaE + Delta(PV) = DeltaE + PDeltaV`

`= DeltaE + Deltan cdot RT`

where `DeltaE` is the change in internal energy, `Deltan` is the change in mols of ideal gas, and `R` and `T` are known from the ideal gas law.

So then, for a combustion, the change in internal energy is given by:

`DeltaE_c^@ = DeltaH_c^@ - Deltan_"gas" cdot RT`

For the combustion of benzoic acid in a bomb calorimeter CLOSED to the atmosphere, we have a constant-volume vessel that condenses the water vapor back to a liquid from the initial non-equilibrium gaseous state:

`"C"_6"H"_5"COOH"(s) + 15/2"O"_2(g) -> 7"CO"_2(g) + 3"H"_2"O"(l)`

Of course, if you did this open to the atmosphere, you would get `"H"_2"O"(g)` instead, which would change the whole problem... For `"1 mol"` of benzoic acid then, you get

`Deltan_"gas" = 7 - 15/2 = -"0.5 mols ideal gas"`

If this was done open to the atmosphere (which is a condition that was aptly unstated in the problem!), then `Deltan_"gas" = color(red)(+"2.5 mols")`...

Hence, in a bomb calorimeter,

`DeltaE_c^@ = overbrace(-x " kcal/mol")^(DeltaH_c^@)`

`- [overbrace(-0.5 cancel"mols ideal gas")^(Deltan_"gas") xx overbrace(8.314472 cancel"J""/mol"cdotcancel"K" xx cancel("1 cal")/(4.184 cancel"J") xx "1 kcal"/(1000 cancel"cal"))^(R) xx overbrace((27 + 273.15 cancel"K"))^(T)]`

`= -x + "0.298 kcal/mol"`

And from the given answers, we are closest to

`color(blue)(DeltaE_c^@ ~~ -x + "0.3 kcal/mol")`.