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Q.

ΔfHH2O=−68K cal/moland ∆H of neutralization is −13.7 k.cal/mol then heat of OH-formation of is

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a

−68.K.cal.mol−1

b

−54.3K.cal.mol−1

c

54.3K.cal.mol−1

d

−71.7K.cal.mol−1

answer is B.

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Detailed Solution

H(eq)++OHeq-⇌H2O(l);ΔH=-13.7K.calΔNHH+/OH-°=ΔfHH2O°-ΔfHH+°+ΔfHOH-°ΔfHOH-°=-68-(0-13.7)=-54.3k.cal/molΔfHH+0=0( convension )

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