Chemistry - Thermochemistry Question with Solution | TestHub

ChemistryThermochemistryEnthalpy of ReactionEasy2 minPYQ_2017
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Given:

Cgraphite+O2g→CO2g ;ΔrHo=−393.5 kJ mol−1

H2g+12O2g→H2Ol;ΔrHo=-285.8 kJ mol-1

CO2g+2H2Ol→CH4g+2O2g;ΔrHo=+890.3 kJ mol-1

Based on the above thermochemical equations, the value ofΔrHoat298 Kfor the reaction

Cgraphite+2H2g→CH4gwill be:

Options:

Answer:
B
Solution:

The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.

The enthalpy of combustion of a substance is defined as the heat energy given out when one mole of a substance burns completely in oxygen.

The formation reaction of methane is 

Cgraphite+2H2g→CH4g

The heat of combustion data for C(graphite), H2g and CH4g

∆Hfo=∑∆HCombustionreactants-∑∆HCombustionproducts∆HfoCH4=∆HCoCgraphite+2×∆HCoH2g-∆HCoCH4g

∆HfoCH4=-393.5-2×285.8+890.3
ΔfHo of CH4 (g)=−74.8 kJ/mol

Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction
⏱ 2mℹ️ Source: PYQ_2017

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