Calculate the standard enthalpy of formation of n- butane. Given : standard enthalpy of combustion of n- butane, C (graphite) and `H_(2)(g) " are " -2878.5 " kJ mol"^(-1), -393.5 " kJ mol"^(-1) and -285.8 " kJ mol"^(-1)` respectively.

Explanation

`C_(4)H_(10)(g)+(13)/(2)O_(2)(g) to 4 CO_(2)(g)+5H_(2)O (l) ,`
`Delta H^(0)= - 2878.5 kJ. "mol"^(-1)`
`C("graphite,s")+O_(2)(g)to CO_(2)(g)`,
`Delta H^(0)= - 393.5 " kJ mol"^(-1)`
`H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l)`,
`Delta H^(0)+ - 285.8 " kJ mol"^(-1)`
`4xx " equation"[2] +5xx "equation" [3]- "equation"[1]`, we have 4 C(graphite,s)`+5 H_(2)(g) to C_(4)H_(10)(g)`
`Delta H^(0)= [4(-393.5)+5(-285.8)-(-2878.5)]kJ`
= `-124.5 kJ`