At `25^(@)C` and 1 atm pressure. the enthalpies of combustion are as given below:

The enthalpy of formation of ethane is

Options

A. `+54.0 " kJ mol"^(-1)`
B. `-68.0 " kJ mol"^(-1)`
C. `-86.0 " kJ mol"^(-1)`
D. `+97.0 " kJ mol"^(-1)`

Explanation

Given
(i) `C_2H_6(g)7/2O_2(g) to 2CO_2(g)+3H_2O(l)DeltaH_"comb"^@= -1560.0` kJ/mole
(ii) `C(s) +O_2(g) to CO_2(g) Delta_"comb"^(@)= -394.0` kJ/mole
(iii) `H_2(g) +1/2O_2(g) to H_2O(g) DeltaH_"comb"^@=-286.0` kJ/mole
Target `2C(s)+3H_2(g) to C_2H_6(g)DeltaH_(rx^n)^@ =DeltaH_f^@(C_2H_6,g)`
`Delta H_r^@ = Delta H_c^@ ("reactant") - Delta H_c^@ ("Product")`
`=2xx(-394) +3(-286) - (-1560)`
`= -788 -858 +1560`
`= -86.0` KJ/mole