A) \[\Delta {{H}_{1}}<\Delta {{H}_{2}}\]
B) \[\Delta {{H}_{1}}+\Delta {{H}_{2}}=0\]
C) \[\Delta {{H}_{1}}>\Delta {{H}_{2}}\]
D) \[\Delta {{H}_{1}}=\Delta {{H}_{2}}\]
Correct Answer: A
Solution :
The enthalpy of \[{{H}_{2}}O(l)\] is lees than that of \[{{H}_{2}}O(g)\], hence more energy will be released when \[{{H}_{2}}O(l)\] is formed, therefore \[\Delta {{H}_{1}}<\Delta {{H}_{2}}\]You need to login to perform this action.
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