A) \[-571.6\text{ }kJ\]
B) \[+571.6\text{ }kJ\]
C) \[-1143.2\text{ }kJ\]
D) \[+285.8kJ\]
Correct Answer: D
Solution :
:\[{{H}_{2}}{{O}_{(l)}}\to {{H}_{2(g)}}+\frac{1}{2}{{O}_{2(g)}};\,\,\Delta H=\frac{571.6}{2}=285.8kJ\] Since, the forward reaction is exothermic, backward has to be endothermic.
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