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Chemistry Question on Thermodynamics terms

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

A

–74.8 kJmol–1–74.8\ kJ mol^{–1}

B

–52.27 kJmol–1–52.27 \ kJ mol^{–1}

C

+74.8 kJmol–1+74.8 \ kJ mol^{–1}

D

+52.26 kJmol–1+52.26\ kJ mol^{–1}

Answer

–74.8 kJmol–1–74.8\ kJ mol^{–1}

Explanation

Solution

According to the question,
(i) CH4(g)+2O2(g)→CO2(g)+H2O(g);CH_4(g) + 2O_2(g) → CO_2(g) + H_2O(g); △H=−890.3 kJmol−1△H = -890.3\ kJmol^{-1}
(ii) C(s)+O2(g)→CO2(g);C(s)+O_2(g) → CO_2(g); △H=−393.5 kJmol−1△H = -393.5\ kJmol^{-1}
(iii) 2H2(g)+O2(g)→H2O(g);2H_2(g) + O_2(g) → H_2O(g); △H=−285.8 kJmol−1△H = -285.8\ kJmol^{-1}
Thus, the desired equation is the one that represents the formation of CH4(g)CH_4 (g) i.e.,
C(s)+2H2(g)→CH4(g)C(s)+2H_2(g) → CH_4(g)
△fHCH4=△CHC+2△CHH2−△CHCO2△_fH_{CH4} = △_CH_C+ 2△_CH_{H_2} - △_CH_{CO_2}
= [−393.5+2(−285.8)−(−890.3)] kJmol−1[-393.5 + 2(-285.8)-(-890.3)]\ kJmol^{-1}
= −74.8 kJmol−1-74.8\ kJ mol^{-1}
Enthalpy of formation of CH4(g)=–74.8 kJmol–1CH_4(g) = –74.8\ kJ mol^{–1}.

Hence, the correct option is (A): –74.8 kJmol–1–74.8\ kJ mol^{–1}.