Question

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 ______.

विकल्प

• - 74.8 kJ mol^-1

• - 52.27 kJ mol^-1

• + 74.8 kJ mol^–1

• + 52.26 kJ mol^-1

Answer 1

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 - 74.8 kJ mol^-1.

Explanation:

According to the question,

1) \[\ce{CH_{4(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H_2O_{(g)}}\]

`triangle" H" = - 890.3 " kJ" " mol"^(-1)`

2) \[\ce{C_{(s)} + O_{2(g)} -> CO_{2(g)}}\]

`triangle" H" = - 393.5 " kJ mol"^(-1)`

3) \[\ce{2H_{2(g)} + O_{2(g)} -> 2H_2O_{(g)}}\]

`triangle" H" = - 285.8 " kJ mol"^(-1)`

Thus, the desired equation is the one that represents the formation of CH_4(g) i.e.,

\[\ce{C_{(s)} + O_{2(g)} -> CO_{2(g)}}\]

`triangle_"f""H"_("CH"_4) = triangle_"c""H"_"c" + 2 triangle_"c" "H"_("H"_2) - triangle_"c""H"_("CO"_2)`

= [- 393.5 + 2(- 285.8) - (- 890.3)] kJ mol^-1

= - 74.8 kJ mol^-1

∴ Enthalpy of formation of CH_4(g) = - 74.8 kJ mol^-1