Question

or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, under standard conditions. C 8 H 18 ( g ) + 25 2 O 2 ( g ) ⟶ 8 CO 2 ( g ) + 9 H 2 O ( g ) Δ H ∘ rxn = − 5104.1 kJ / mol What is the standard enthalpy of formation of this isomer of C 8 H 18 ( g ) ?

Answer 1

The standard enthalpy of formation of this isomer of
`C_(8)H_(18)` is -220.1 kJ/mol.

Step-by-step explanation:

The given chemical reaction is as follows.

`C_(8)H_(18)(g)+ (25)/(2)O_(2)(g) \rightarrow 8CO_(2)(g)+9H_(2)O(g)`

`\Delta H^(o)_(rxn)= -5104.1kJ/mol`

The expression for the entropy change for the reaction is as follows.

`\Delta H^(o)_(rxn)=[8\Delta H^(o)_(f)(CO_(2)) +9\Delta H^(o)_(f)(H_(2)O)]-[\Delta H^(o)_(f)(C_(8)H_(18))+ (25)/(2)\Delta H^(o)_(f)(O_(2))]`

`\Delta H^(o)_(f)(H_(2)O)= -241.8kJ/mol`

`\Delta H^(o)_(f)(CO_(2))= -393.5kJ/mol`

`\Delta H^(o)_(f)(O_(2))= 0kJ/mol`

Substitute the all values in the entropy change expression.

`-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^(o)_(f)(C_(8)H_(18))+ (25)/(2)(0)kJ/mol]`

`-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^(o)_(f)(C_(8)H_(18))`

`\Delta H^(o)_(f)(C_(8)H_(18)) =-5324.2kJ/mol +5104.1kJ/mol`

`=-220.1kJ/mol`

Therefore, The standard enthalpy of formation of this isomer of
`C_(8)H_(18)` is -220.1 kJ/mol.