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ChemistryThermochemistryEnthalpy of ReactionMedium2 minPYQ_2022
ChemistryMediummatching list

For the reaction H2F2gH2g+F2g

ΔU=-59.6 kJ mol-1 at 27 °C

The enthalpy change for the above reaction is -____ kJmol-1 (nearest integer)

(Given : R=8.314JK-1 mol-1) .

Answer:
57
Solution:

For the reaction H2F2gH2g+F2g

ΔH=ΔU+ΔngRT

Given ΔU=-59.6 kJ mol-1

Δng=2-1

=-59.6+1×8.3141000×300

=-57.11  kJ mol-1

Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction
2mℹ️ Source: PYQ_2022

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