enthalpy change for the first reaction? P4 + 6Cl2 → 4PCl3 ΔH = P4 + 10Cl2(g) → 4PCl5(s) ΔH = -1,777.5 PCl3(l) + Cl2 → PCl5(s) ΔH = -127.2?
回答 (1)
2017-10-20 9:43 pm
Rewrite the two given thermochemical equations as follows :
P₄ + 10Cl₂(g) → 4PCl₅(s) …… ΔH = -1,777.5 kJ
4PCl₅(s) → 4PCl₃(l) + 4Cl₂(g) …… ΔH = -4(-127.2) kJ = +408.8 kJ
Add the above two thermochemical equations, and cancel 4Cl₂(g) and 4PCl₅(s) on the both sides. Then,
P₄ + 6Cl₂(g) → 4PCl₃(l) …… ΔH = (-1777.5 kJ) + (+408 kJ) = -1368.7 kJ
P₄ + 10Cl₂(g) → 4PCl₅(s) …… ΔH = -1,777.5 kJ
4PCl₅(s) → 4PCl₃(l) + 4Cl₂(g) …… ΔH = -4(-127.2) kJ = +408.8 kJ
Add the above two thermochemical equations, and cancel 4Cl₂(g) and 4PCl₅(s) on the both sides. Then,
P₄ + 6Cl₂(g) → 4PCl₃(l) …… ΔH = (-1777.5 kJ) + (+408 kJ) = -1368.7 kJ
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