Given the standard enthalpy changes for the reactions:
P₄(s) + 3 O₂(g) -> P₄O₆(s) ΔHº = -1640 kJ mol-1
P₄(s) + 5 O₂(g) -> P₄O₁₀(s) ΔHº = -2940 kJ mol-1
Calculate the standard enthalpy change ∆Hº for the
following reaction:
P₄O₆(s) + 2 O₂(g) -> P₄O₁₀(s)
(A) -4.58 x 10 kJ
(B) -1.30 x 10'KJ
(C) 1.79 kJ
(D) 4.82 x 10 kJ