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 $kJ mol^{-1}$

M manish

$CH_4+2O_{2}\rightarrow CO_{2}+2H_{2}O\ \Delta H$................. -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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