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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

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Best answer

According to the question, 

Thus, the desired equation is the one that represents the formation of CH4 (g) i.e., 

Enthalpy of formation of CH4(g) = –74.8 kJ mol–1 Hence, 

alternative (i) is correct. 

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