Chemistry - Thermochemistry Question with Solution | TestHub
ChemistryThermochemistryEnthalpy of CombustionEasy2 minQB
ChemistryEasysingle choice
Passage / Comprehension
When 0.1 mole of C6H14(l) at 300 K is completely burned at constant pressure in some oxygen gas at 300 K, yielding products gaseous H2O, CO and CO2 at 700 K, the process yielding 75.16 kcal of heat to the surrounding. Given:
Δf H (kcal/mol): C6H14(l) = -40.0, CO2(g) = -94.0, CO(g) = -26.0, H2O(g) = -58.0
Molar heat capacity at constant pressure (cal/K-mol): CO(g) = 7.0, CO2(g) = 9.0, H2O(g) = 8.0
Assume that all Δr H are independent of temperature and R = 2 cal/K-mol.
The value of Δr H for the reaction:
C6H14(l) + 13/2O2(g) → 6CO(g) + 7H2O(g)
is (in kcal/mol)
Options:
Answer:
B
Solution:
The enthalpy change of the reaction, Δr H, is calculated as follows:
Δr H = 6(-26.0) + 7(-58.0) - (-40.0)
Δr H = -156 - 406 + 40
Δr H = -522 kcal/mol
This corresponds to option (b).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Combustion
⏱ 2mℹ️ Source: QB
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