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ChemistryThermochemistryEnthalpy of ReactionMedium2 minPYQ_2022
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At 25 °C and 1 atm pressure, the enthalpies of combustion are as given below:

SubstanceH2C (graphite)C2H6g
ΔcHΘkJmol-1-286.0-394.0-1560.0

The enthalpy of formation of ethane is

Options:

Answer:
C
Solution:

Given

(i) C2H6g+72O2g2CO2g+3H2OlΔHcomb°=-1560.0 kJ/mole

(ii) Cs+O2gCO2gΔHcomb°=-394.0 kJ/mole

(ii) H2g+12O2gH2OgΔHcomb°=-286.0 kJ/mole

Target 2Cs+3H2gC2H6gΔHrx°=ΔHf°C2H6, g

ΔHr°=ΔHc°reactant-ΔHc°Product

=2×-394+3-286--1560

=-788-858+1560

=-86.0KJ/mole

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

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