Add the two equations to obtain the following:
P4 (s) + 6 Cl2 (g) → 4 PCl3 (g)
4 PCl5 (g) → P4 (s) + 10 Cl2 (g).
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P4 (s) + 6 Cl2 (g) + 4 PCl5 (g) ==> 4 PCl3 (g) + P4 (s) + 10 Cl2 (g)
Now cancel P4 on left and right. 6 Cl2 on left and 6 Cl2 on right leaving 4 Cl2 on right. This leaves:
4PCl5 ==> 4PCl3 + 4 Cl2 with dH being the sum of the two (-2439 + 3438 = ?)
That reaction is just four times the target reaction so divide the final reaction by 4 to obtain PCl5 ==> PCl3 + Cl2 and of course divide the sum above by 4 to get delta H for the target reaction.
For each problem, find the ΔH for the target reaction below, given the following step reactions and subsequent ΔH values:
Target Reaction: PCl5(g) → PCl3(g) + Cl2(g)
Step Reactions: P4 (s) + 6 Cl2 (g) → 4 PCl3 (g) ΔH = -2439 kJ
4 PCl5 (g) → P4 (s) + 10 Cl2 (g). ΔH = 3438 kJ
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