In: Chemistry
For which of the following reactions is ?H?rxn equal to ?H?f of the product(s)?
You do not need to look up any values to answer this question. Check all that apply.
SO(g)+12O2(g)?SO2(g)
2Na(s)+F2(g)?2NaF(s)
Na(s)+12F2(g)?NaF(s)
S(s)+O2(g)?SO2(g)
Na(s)+12F2(l)?NaF(s)
SO3(g)?12O2(g)+SO2(g)
1) SO(g)+1/2O2(g)→SO2(g)
ΔH∘rxn = ΔH∘f(SO2) - (ΔH∘fSO + 0.5ΔH∘fO2)
ΔH∘rxn = ΔH∘f(SO2) - (ΔH∘fSO)
so here ΔH∘rxn is not equal to ΔH∘f of the product(s)
2) 2Na(s)+F2(g)→2NaF(s)
ΔH∘rxn = 2 X ΔH∘f (NaF) - (2 X ΔH∘f (Na) + ΔH∘f F2)
ΔH∘f Na = ΔH∘f F2 = 0
Therefore
ΔH∘rxn = 2 X ΔH∘f (NaF)
Hence ΔH∘rxn is not equal to ΔH∘f of the product(s)
3) Na(s)+1/2F2(g)→NaF(s)
ΔH∘rxn = ΔH∘f (NaF) - ( ΔH∘f (Na) + 0.5ΔH∘f F2)
ΔH∘f (Na) = ΔH∘f F2 = 0
ΔH∘rxn = ΔH∘f (NaF)
Hence ΔH∘rxn equal to ΔH∘f of the product(s)
4) S(s)+O2(g)→SO2(g)
ΔH∘rxn = ΔH∘f (SO2) - ( ΔH∘f (S) + ΔH∘f O2)
ΔH∘f (S) = ΔH∘f O2 = 0
ΔH∘rxn = ΔH∘f (SO2)
Hence ΔH∘rxn equal to ΔH∘f of the product(s)
5) Na(s)+1/2F2(l)→NaF(s)
ΔH∘rxn = ΔH∘f (NaF) - ( ΔH∘f (Na) + 0.5ΔH∘f F2(l))
ΔH∘rxn = ΔH∘f (NaF) - ( 0.5ΔH∘f F2(l))
hence ΔH∘rxn is not equal to ΔH∘f of the product(s)
6) SO3(g)→1/2O2(g)+SO2(g)
ΔH∘rxn = (ΔH∘f SO2 + 0.5ΔH∘f O2) -( ΔH∘f (SO3)
ΔH∘rxn = (ΔH∘f SO2 ) -( ΔH∘f (SO3)
Hence ΔH∘rxn is not equal to ΔH∘f of the product(s)