For which of the following is deltaH equal to deltaE(internal energy)?
2"CO"(g)+"O"_2(g) -> 2"CO"_2(g)
"H"_2(g)+"Br"_2(g) ->2"HBr"(g)
"C"(s)+2"H"_2"O"(g) ->2"H"_2+"CO"_2
"PCl"_5(g) ->"PCl"_3(g)+"Cl"_2(g)
2"CO"(g)+"O"_2(g) -> 2"CO"_2(g) "H"_2(g)+"Br"_2(g) ->2"HBr"(g) "C"(s)+2"H"_2"O"(g) ->2"H"_2+"CO"_2 "PCl"_5(g) ->"PCl"_3(g)+"Cl"_2(g)
1 Answer
Well, only
2"CO"(g)+"O"_2(g) -> 2"CO"_2(g) "H"_2(g)+"Br"_2(g) ->2"HBr"(g) "C"(s)+2"H"_2"O"(g) ->2"H"_2+"CO"_2 "PCl"_5(g) ->"PCl"_3(g)+"Cl"_2(g)
By definition,
H = E + PV where
H is enthalpy,E is internal energy, andP andV 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
Hence, we can say