Chemistry - Thermochemistry Question with Solution | TestHub
ChemistryThermochemistryHess's Law of Constant Heat SummationMedium2 minQB
ChemistryMediumsingle choice
Consider the equations:
HOCH2CHBrCOOH + 2KOH → P + KBr + 2H2O; ΔH = -16.9 kcal
HOCH2CHBrCOOK + KOH → P + KBr + H2O; ΔH = -2.35 kcal
where P is the epoxide CH2(O)CHCOOK shown below.
ΔH for the neutralization of HCl and KOH is -13.7 kcal/eq. The enthalpy of ionization of HOCH2CHBrCOOH is
Options:
Answer:
C
Solution:
Subtract the second equation from the first: HOCH2CHBrCOOH + KOH → HOCH2CHBrCOOK + H2O, ΔH = -16.9 - (-2.35) = -14.55 kcal. For a weak acid, ΔHneut = ΔHion + (-13.7). So ΔHion = -14.55 + 13.7 = -0.85 kcal/mol. This is option (c).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Hess's Law of Constant Heat Summation
⏱ 2mℹ️ Source: QB
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