Question

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Hf = –393.5 kJ/mol), and H2O(g) (Hf = –241.82) in the reaction:

2C4H10 (g) + 13O2 (g) = 8CO2 (g) + 10H2O (g)

What is the enthalpy of combustion, per mole, of butane?(A) -2,657.5 kJ/mol(B) -5315.0 kJ/mol(C) -509.7 kJ/mol(D) -254.8 kJ/mol

Answer 1

`\boxed{\text{(A) -2657.4 kJ/mol}}`

Step-by-step explanation:

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

`\Delta_{\text{r}}H^(\circ) = \sum \Delta_{\text{f}} H^(\circ) (\text{products}) - \sum\Delta_{\text{f}}H^(\circ) (\text{reactants})`

2C₄H₁₀ g) + 13O₂(g) ⇌ 8CO₂(g) + 10H₂O(g)

ΔH°f/kJ·mol⁻¹: -125.7 0 -393.5 -241.82

`\begin{array}{rcl}\Delta_{\text{c}}H^(\circ) & = & [8(-393.5) + 10(-241.82)] - 2(-125.7)\\& = & [-3148.0 - 2418.2] +251.4\\& = & -5566.2 + 251.4\\& = & -5314.8\\\end{array}\\\\\text{This is the value for 2 mol of butane.}\\\text{The enthalpy of combustion is } (-5314.8)/(2) = \boxed{\textbf{-2657.4 kJ/mol}}`