The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 , –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be
(i) –74.8 kJ mol–1
(ii) –52.27 kJ mol–1
(iii) +74.8 kJ mol–1
(iv) +52.26 kJ mol–1
According to the question,
Thus, the desired equation is the one that represents the formation of CH4(g) i.e.,
[-393.5 + 2(-285.8) - (-890.3)] kJ Mol-1
= -74.8 kJ Mol-1
So,Enthalpy of formation of CH4(g) is -74.8 kJ Mol-1
That means answer is (i).
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 5: The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol&nda....
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