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Thermodynamics

Question
CBSEENCH11006073

The heat of combustion of glucose (C6H12,O6) is 2840 kJ mol–1. What is its calorific value?

Solution

Molar mass of glucose (C6H12,O6) = 180
therefore space space Calorific space value space equals space 2840 over 180 kJg to the power of negative 1 end exponent space equals space 15.78 space kJg to the power of negative 1 end exponent space space space space