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ChemistryThermodynamics - IFirst law of ThermodynamicsMedium2 minPYQ_2022
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For combustion of one mole of magnesium in an open container at300 Kand1bar pressure,ΔCH=-601.70 kJ mol-1, the magnitude of change in internal energy for the reaction iskJ. (Nearest integer)
(Given :R=8.3 J K-1 mol-1)

Answer:
600
Solution:

Mgs+12O2GMgOSΔHc°=-601.70KJ/Mole

ΔH°=ΔU+ΔngRT

-601.70=ΔU+-12×8.3×300×10-3

-601.7=ΔU-1.245

ΔU=-599.455KJ

ΔU=599.455KJ600

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics
2mℹ️ Source: PYQ_2022

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