Asked by adrian
Given these two reactions, how do I calculate
ΔH for the third one?
P4(s) + 6 Cl2(g) -> 4 PCl3(g) ΔH= -1148 kJ
P4(s) + 10 Cl2(g) -> 4 PCl5(g) ΔH = -1500 kJ
PCl3(g) + Cl2(g) -> PCl5(g) ΔH = ?
ΔH for the third one?
P4(s) + 6 Cl2(g) -> 4 PCl3(g) ΔH= -1148 kJ
P4(s) + 10 Cl2(g) -> 4 PCl5(g) ΔH = -1500 kJ
PCl3(g) + Cl2(g) -> PCl5(g) ΔH = ?
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