Question

Use the following data to calculate the standard enthalpy of formation of pentane, C5H12(l)?
C5H12(l) + 8 O2 (g) -> 5 CO2 (g) + 6 H2O (l) delta H=-3535.9 kJ/mol

delta Hf of CO2 (g) = -393.5 kJ/mol

delta Hf of H2O (l) = -285.8 kJ/mol



please show all work

Answer 1

Answer:

$\large \boxed{\text{-146.4 kJ/mol}}$

Explanation:

C₅H₁₂(l) + 8O₂(g) ⟶ 5CO₂(g) + 6H₂O(l); ΔᵣH = -3535.9 kJ·mol⁻¹

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

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

Let x = the heat of formation of pentane

                            C₅H₁₂(l) + 8O₂(g) ⟶ 5CO₂(g) + 6H₂O(l)

ΔH°f/kJ·mol⁻¹:            x                          -393.5      -285.8

$\begin{array}{rcl}-3535.9 & = & [5\times(-393.5) -6 \times(-285.8)] - x\\-3535.9 & = & -3682.3 - x\\x& = & \textbf{-146.4 kJ/mol}\\\end{array}\\\text{The enthalpy of formation of pentane is } \large \boxed{\textbf{-146.4 kJ/mol}}$