A) \[Xe(g)+2{{F}_{2}}(g)\xrightarrow[{}]{{}}Xe{{F}_{4}}(g)\]
B) \[2CO(g)+{{O}_{2}}(g)\xrightarrow[{}]{{}}2C{{O}_{2}}(g)\]
C) \[{{N}_{2}}(g)+{{O}_{3}}(g)\xrightarrow[{}]{{}}2C{{O}_{2}}(g)\]
D) \[C{{H}_{4}}(g)+2C{{l}_{2}}(g)\xrightarrow[{}]{{}}C{{H}_{2}}C{{l}_{2}}(\ell )+2HCl\,(g)\]
Correct Answer: A
Solution :
\[\Delta H_{react}^{o}=\Delta H_{f}^{o}\] for the reaction \[Xe(g)+2{{F}_{2}}(g)=Xe{{F}_{4}}(g)\] Because in this reaction one mole of \[Xe{{F}_{4}}\]is formed from its constituent elements.
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