The enthalpy of combustion of methane graphite and dihydrogen at $298 K$ are $- 890.3$ kJ $m o l^{- 1}$ and $- 285.8$ kJ $m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4} (g)$ will be:
$- 52.27$ $k J$ $m o l^{- 1}$
$- 78.8$ $k J$ $m o l^{- 1}$
$+ 74.8$ $k J$ $m o l^{- 1}$
$+ 52.26$ $k J$ $m o l^{- 1}$

Question

The enthalpy of combustion of methane graphite and dihydrogen at $298 K$ are $- 890.3$ kJ $m o l^{- 1}$ and $- 285.8$ kJ $m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4} (g)$ will be:
$- 52.27$ $k J$ $m o l^{- 1}$
$- 78.8$ $k J$ $m o l^{- 1}$
$+ 74.8$ $k J$ $m o l^{- 1}$
$+ 52.26$ $k J$ $m o l^{- 1}$

Toppr Admin · Accepted Answer

The correct option is  B -x2212-78-8 kJ -xA0-mol-x2212-1The combustion reactions are shown below-CH4-g-2O2-g-x2192-CO2-g-2H2O-l-x394-H-x2212-890-3kJ-mol-1-C-s-O2-g-x2192-CO2-g-x394-H-x2212-393-5kJ-mol-2-H2-g-0-5O2-g-x2192-H2-l-x394-H-x2212-285-8kJ-mol-3-The equation for the formation of methane is C-s-2H2-g-x2192-CH4-g-x394-H-4-Multiply equation 3 with 22H2-g-O2-g-x2192-2H2-l-x394-H-x2212-2-285-8-kJ-molSubtract equation 1 from equation -3-2H2-g-O2-g-x2192-2H2-l-x394-H-x2212-2-285-8-kJ-mol-CH4-g-2O2-g-x2192-CO2-g-2H2O-l-x394-H-x2212-890-3kJ-mol-2H2-g-CO2-g-2H2O-l-x2192-CH4-g-2H2-l-O2-g-x394-H-x2212-2-285-8-890-3kJ-mol-5- -xA0-add equation 2 to equation -5- to form equation 4-2H2-g-CO2-g-2H2O-l-x2192-CH4-g-2H2-l-O2-g-x394-H-x2212-2-285-8-890-3kJ-mol -C-s-O2-g-x2192-CO2-g-x394-H-x2212-393-5kJ-mol-C-s-2H2-g-x2192-CH4-g-x394-H-x2212-393-5-2-x2212-285-8-x2212-x2212-890-3-x2212-74-8kJ-mol

The enthalpy of combustion of methane graphite and dihydrogen at 298K are −890.3 kJ mol−1 and −285.8 kJ mol−1 respectively. Enthalpy of formation of CH4(g) will be:−52.27 kJ mol−1−78.8 kJ mol−1+74.8 kJ mol−1+52.26 kJ mol−1