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Class 11
Chemistry
Thermodynamics
the enthalpy of combustion of methane graphite an...

Question:
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

Answer:

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 CH_4(g) is -74.8 kJ Mol^-1

That means answer is (i).

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