Question 6.PS.14: Standard Molar Enthalpy of Formation from Enthalpy of Combus......

Strategy and Explanation Write a balanced equation for the target reaction whose ΔH° you want to calculate. Also write a balanced equation for combustion of octane, for which you know the standard enthalpy change. By studying these two equations, decide what additional information is needed to set up a Hess’s law calculation that will yield the standard molar enthalpy of formation of octane.

( target reaction)   8 C( s ) + 9 H_2(g) → C_8H_{18}(\ell)          ΔH_f^\circ = ??  kJ/mol

(a) C_8H_{18}(\ell) + \frac{25}{2} O_2(g) → 8 CO_2(g)  +  9 H_2O(g)     ΔH_{combustion}^\circ = – 5116.0  kJ

Notice that the combustion equation involves 1 mol C_8H_{18}(\ell) as a reactant and the target equation (for enthalpy of formation) involves 1 mol C_8H_{18}(\ell) as a product. Therefore, it seems reasonable to reverse the combustion equation and see where that leads.

(a’) 8 CO_2(g) + 9 H_2O(g) → C_8H_{18}(\ell)  +  \frac{25}{2} O_2(g)    ΔH° = + 5116.0  kJ

On the reactant side of the target equation we have 8 C(s) and 9 H_2(g). These elements, combined with O_2(g), are on the left side of equation (a’), so perhaps it would be reasonable to use the equations corresponding to standard molar enthalpies of formation of carbon dioxide and water. From Table 6.2, we have

(b)  C( s ) + O_2(g) → CO_2(g)            ΔH_f^\circ = – 393.509  kJ/mol

(c)  H_2(g) + \frac{1}{2} O_2(g) → H_2O(g)            ΔH_f^\circ = + 241.818 kJ/mol

Multiplying equation (b) by 8 and equation (c) by 9 gives the correct number of moles of C(s) and of H_2(g) on the reactant side of the target equation. This gives

(a’) 8 CO_2(g) + 9 H_2O(g) → C_8H_{18}(\ell)  +  \frac{25}{2} O_2(g)       ΔH° = + 5116.0  kJ

(b’) 8 C( s ) + 8  O_2(g) → 8  CO_2(g)                         ΔH° = – 3148.072  kJ

(c’) \underline{9 H_2(g) + \frac{9}{2}  O_2(g) → 9 H_2O(g)                                        ΔH° =- 2176.362  kJ/mol}

 8 C( s ) + 9 H_2(g) → C_8H_{18}(\ell)        ΔH_f^\circ = ΔH_{a'}^\circ + ΔH_{b'}^\circ + ΔH_{c'}^\circ = – 208.4  kJ/mol