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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 CH 4(g) will be

6.5  The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.KJ mol ^{-1} , -393.5 KJ mol ^{-1} \: \: and -285 .8 KJ mol ^{-1}respectively. Enthalpy  of formation of  CH_4 (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}

Answers (1)

best_answer

CH_4+2O_{2}\rightarrow CO_{2}+2H_{2}O\ \Delta H................. -890.3 kJ/mol
C+2O_{2}\rightarrow CO_{2}                                 \Delta H..................-393.5 kJ/mol
2H_{2}+O_{2}\rightarrow 2H_{2}O                             \Delta H..................-258.8 kJ/mol

So, the required equation is to get the formation of CH_4 (g) by combining these three equations-

Thus, C+2H_{2}\rightarrow CH_{4}

\Delta _fH_{CH_{4}} = \Delta _cH_c+2\Delta _cH_{H_2}
                    = [-393.5 + 2(-285.8) + 890.3]
                    = -74.8 kJ/mol

therefore, the enthalpy of formation of methane is -74.8 kJ/mol. So, the correct option is (i)

 

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