Question

When `"32.4 mL"` of liquid benzene (`C_6H_6`, `rho = "0.879 g/mL"`) reacts with `"81.6 L"` of oxygen gas, measured at `"1.00 atm"` and `"25°C"`, `1.19xx10^3 "kJ"` of heat is released at const. pressure. What is `DeltaH^@` for the following reaction?

(`R = "0.0821 L"cdot"atm/K"cdot"mol"`)
`2"C"_6"H"_6(l) + 15"O"_2(g) -> 12"CO"_2(g) + 6"H"_2"O"(l)`

Answer 1

`DeltaH_(rxn)^@ = -3.264 xx 10^3 "kJ/mol"`

Why is it important (i.e. useful!) to use `"kJ/mol"` instead of `"kJ"`?

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At constant pressure, the heat released IS the `DeltaH` of the reaction. The combusted quantity is

`32.4 cancel"mL" xx "0.879 g"/cancel"mL" = "28.48 g benzene"`

`28.48 cancel"g benzene" xx ("1 mol benzene")/(78.1134 cancel"g benzene") = "0.3646 mols benzene"`
vs. `"2.735 mols O"_2` needed

and we can probably tell that the `"O"_2` is in excess... assuming ideal gases,

`n = (PV)/(RT) = ("1.00 atm" cdot "81.6 L")/("0.08206 L"cdot"atm/mol"cdot"K" cdot "298.15 K")`

`= "3.335 mols O"_2`

Indeed, it is in excess by `"0.6006 mols"`. We base the mols off of the limiting reactant benzene.

The reaction apparently releases `"1190 kJ"` of heat at constant pressure, so

`color(blue)(DeltaH_(rxn)^@) = -"1190 kJ"/"0.3646 mols benzene"`

`= color(blue)(-3.264 xx 10^3 "kJ/mol")`