Question
Standard enthalpy of combustion of $$C{H_4}$$ is $$ - 890\,kJ\,mo{l^{ - 1}}$$ and standard enthalpy of vaporisation of water is $$40.5\,kJ\,mo{l^{ - 1}}.$$ The enthalpy change of the reaction $$C{H_4}\left( g \right) + 2{O_2}\left( g \right) \to C{O_2}\left( g \right) + {H_2}O\left( g \right)$$
A.
$$ - 809.5\,kJ\,mo{l^{ - 1}}$$
B.
$$ - 890\,kJ\,mo{l^{ - 1}}$$
C.
$$809\,kJ\,mo{l^{ - 1}}$$
D.
$$ - 971\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 809.5\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& C{H_4}\left( g \right) + 2\,{O_2}\left( g \right) \to C{O_2}\left( g \right) + 2{H_2}O\left( l \right) \cr
& \Delta H = - 890\,kJ\,\,\,....\left( {\text{i}} \right) \cr
& 2\,{H_2}O\left( l \right) \to 2\,{H_2}O\left( g \right); \cr
& \Delta H = 2 \times 40.5 = 81\,kJ\,\,\,...\left( {{\text{ii}}} \right) \cr
& {\text{From}}\,\,\left( {\text{i}} \right) + \left( {{\text{ii}}} \right),{\text{we get}} \cr
& C{H_4}\left( g \right) + 2\,{O_2}\left( g \right) \to C{O_2}\left( g \right) + 2{H_2}O\left( g \right) \cr
& \Delta H = - 890 + 81 = - 809\,kJ \cr} $$