For which of the following is deltaH equal to deltaE(internal energy)?

  1. 2"CO"(g)+"O"_2(g) -> 2"CO"_2(g)
  2. "H"_2(g)+"Br"_2(g) ->2"HBr"(g)
  3. "C"(s)+2"H"_2"O"(g) ->2"H"_2+"CO"_2
  4. "PCl"_5(g) ->"PCl"_3(g)+"Cl"_2(g)

1 Answer
Mar 3, 2018

Well, only (2), but only if we assume constant atmospheric pressure...


  1. 2"CO"(g)+"O"_2(g) -> 2"CO"_2(g)
  2. "H"_2(g)+"Br"_2(g) ->2"HBr"(g)
  3. "C"(s)+2"H"_2"O"(g) ->2"H"_2+"CO"_2
  4. "PCl"_5(g) ->"PCl"_3(g)+"Cl"_2(g)

By definition,

H = E + PV

where H is enthalpy, E is internal energy, and P and V are known from the ideal gas law.

Taking the change, we get:

color(green)(DeltaH) = DeltaE + Delta(PV)

= DeltaE + P_2V_2 - P_1V_1

= DeltaE + (P_2 - P_1 + P_1)(V_2 - V_1 + V_1) - P_1V_1

= DeltaE + (DeltaP + P_1)(DeltaV + V_1) - P_1V_1

= DeltaE + P_1DeltaV + V_1DeltaP + DeltaPDeltaV + cancel(P_1V_1 - P_1V_1)

= color(green)(DeltaE + P_1DeltaV + V_1DeltaP + DeltaPDeltaV)

Now, we have assumed that pressure is constant, so:

DeltaH = DeltaE + P_1DeltaV

Under this assumption, if DeltaV = 0 too, then DeltaH = DeltaE. Liquids are generally incompressible, and so are solids. Thus, if the mols of gas don't change, the volume of the system pretty much doesn't change.

Hence, we can say color(blue)(DeltaH ~~ DeltaE) color(blue)" if " color(blue)(Deltan_"gas" = 0).