题目内容
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已知 :O2(g)=O2+(g)+e- ΔH1=+1175.7 kJ·mol-1PtF6-(g)=PtF6(g)+e- ΔH2=+771.1 kJ·mol-1 O2+(g)+PtF6-(g)=O2PtF6(S) ΔH3=-482.2 kJ·mol-1 则反应 O2(g)+PtF6(g)=O2PtF6(s)的ΔH是 | |
| [ ] | |
A. |
77 .6 kJ |
B. |
-77 .6 kJ·mol-1 |
C. |
+77 .6 kJ·mol-1 |
D. |
-886 .8 kJ·mol-1 |
答案:B
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