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

2 Answers
Mar 21, 2018

#"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)#

Mar 21, 2018

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")#.