Home kim 11 2.4 2P(s) + O2(g) + 3Cl2(g) → 2POCl3(g) ∆H = -1.150 kJ H2(g) + Cl2(g) → 2HCl(g) ∆H = -184 kJ 2P(s) + 5Cl2(g) → 2PCl5(g) ∆H = -640 kJ 2H2(g) + O2(g) → 2H2O(g) ∆H = -482 kJ Hitunglah ∆H untuk reaksi: PCl5(g) + H2O(g) → POCl3(g) + 2HCl(g)

Tuesday, 17 March 2020

2P(s) + O2(g) + 3Cl2(g) → 2POCl3(g) ∆H = -1.150 kJ H2(g) + Cl2(g) → 2HCl(g) ∆H = -184 kJ 2P(s) + 5Cl2(g) → 2PCl5(g) ∆H = -640 kJ 2H2(g) + O2(g) → 2H2O(g) ∆H = -482 kJ Hitunglah ∆H untuk reaksi: PCl5(g) + H2O(g) → POCl3(g) + 2HCl(g)

2P(s) + O2(g) + 3Cl2(g) → 2POCl3(g) ∆H = -1.150 kJ H2(g) + Cl2(g) → 2HCl(g) ∆H = -184 kJ 2P(s) + 5Cl2(g) → 2PCl5(g) ∆H = -640 kJ 2H2(g) + O2(g) → 2H2O(g) ∆H = -482 kJ Hitunglah ∆H untuk reaksi: PCl5(g) + H2O(g) → POCl3(g) + 2HCl(g)
2P(s) + O2(g) + 3Cl2(g) → 2POCl3(g)          ∆H = -1.150 kJ
H2(g) + Cl2(g) → 2HCl(g)                  ∆H = -184 kJ
2P(s) + 5Cl2(g) → 2PCl5(g)               ∆H = -640 kJ
2H2(g) + O2(g) → 2H2O(g)               ∆H = -482 kJ
Hitunglah ∆H untuk reaksi:
PCl5(g) + H2O(g) → POCl3(g) + 2HCl(g)

PEMBAHASAN


BACA SELENGKAPNYA : Pembahasan Kimia Erlangga Kelas 11 Termokimia


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