Answer:
the overall reaction is
P4(s) + 10Cl2(g) => 4PCl5(g)
and its ΔH is
ΔH overall = (-1608 KJ)
Explanation:
given that
P4(s) + 6Cl2(g) => 4PCl3(g) , ΔH1 = –1148 kJ
4PCl3(g) + 4Cl2(g) => 4PCl5(g) , ΔH2 =–460 kJ
summing both equation gives
P4(s) + 6Cl2(g) +4PCl3(g) + 4Cl2(g) => 4PCl3(g) + 4PCl5(g)
but 4PCl3(g) is in both sides , so is cancelled out
P4(s) + 6Cl2(g)+ 4Cl2(g) => 4PCl5(g)
summing the Cl2(g)
P4(s) + 10Cl2(g) => 4PCl5(g) → overall reaction
(can be verified that the reaction is balanced)
since we sum both equations the ΔH of the overall reaction is also the sum of the ΔH of the individual reactions.
ΔH overall = ΔH1 + ΔH2 = (–1148 kJ) + (–460 kJ) = (-1608 KJ)
