Use the standard reaction enthalpies given below to determine
H° for the following reaction: P4(g) + 10Cl2(g) → 4PCl5(s)
H° = ?
Given:
PCl5(s) → PCl3(g) + Cl2(g)
H°= +157 kJ
P4(g) + 6Cl2(g) → 4PCl3(g)
H° = -1207 kJ
A) -1835 kJ mol-1
B) -1364 kJ mol-1
C) -1050 kJ mol-1
D) -1786 kJ mol-1
E) -2100 kJ mol-1
Correct Answer:
Verified
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