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Question

If heat of dissociation of CHCl2COOH is 0.7 kcal/mole then ΔH for the reaction:
CHCl2COOH+KOHCHCl2COOK+H2O

A
-13 kcal/mol
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B
+ 13 kcal/mol
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C
-14.4 kcal/mol
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D
-13.7 kcal/mol
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Solution

The correct option is A -13 kcal/mol
Heat of neutralization of strong acid by a strong base,
H++OHH2O(l)
ΔH=13.7kcal.......(i)
For neutralization of weak acid by a strong base some of the heat is consumed in dissociation of the acid.
CHCl2COOHCHCl2COO+H+ΔH
= 0.7 kcal ..........(ii)
Hence, heat of neutralization of CHCl2COOH by a strong base is obtained by adding (i) and (iii)
CHCl2COOH+OH
CHCl2COO+H2O(l)ΔH=?
Adding equation (i) and (ii)ΔH = -13 kcal.

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