A) 8.7 kcal
B) 13.7 kcal
C) 23.7 kcal
D) 30.7 kcal
Correct Answer: B
Solution :
The heat of neutralization for strong acid and strong base is constant and is equal to -13.7 kcal. It involves the combination of\[{{\text{H}}^{\text{+}}}\]and \[O{{H}^{-}}\]ions to form unionized water. \[H+C{{l}^{-}}+N{{a}^{+}}+O{{H}^{-}}\to N{{a}^{+}}+C{{l}^{-}}+{{H}_{2}}O\] or \[{{H}^{+}}+O{{H}^{-}}\to {{H}_{2}}O\] Hence it is also known as heat of formation of water. Hence, the correct option is (b).
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