Question

formation of butane from carbon dioxide and water gibbs free energy the formation of butane and oxygen from carbon dioxide and water occurs according to the equation below. 8co₂(g) + 10h₂o(g) → 2c₄h₁₀(g) + 13o₂(g) δgᵣₓₙ =? using the equation and data table, calculate the gibbs free energy for the butane reaction. -606 kj -5408 kj 5408 kj 606 kj compound | δgᵒ_f (kj/mol) c₄h₁₀ (g) | -17 co₂(g) | -394 h₂o(g) | -229 o₂(g) | 0

Explanation:

Step1: Recall the formula for Gibbs Free Energy change

The formula for the Gibbs Free Energy change of a reaction ($\Delta G_{rxn}$) is given by the sum of the Gibbs Free Energy of formation ($\Delta G_f^\circ$) of the products minus the sum of the Gibbs Free Energy of formation of the reactants, multiplied by their respective stoichiometric coefficients. Mathematically, it is:
$\Delta G_{rxn} = \sum (n \times \Delta G_f^\circ)_{\text{products}} - \sum (n \times \Delta G_f^\circ)_{\text{reactants}}$

Step2: Identify the products and their stoichiometric coefficients

The products are $\ce{C4H10(g)}$ and $\ce{O2(g)}$.

• For $\ce{C4H10(g)}$: Stoichiometric coefficient $n = 2$, $\Delta G_f^\circ = -17\ \text{kJ/mol}$
• For $\ce{O2(g)}$: Stoichiometric coefficient $n = 13$, $\Delta G_f^\circ = 0\ \text{kJ/mol}$ (since $\ce{O2}$ is an element in its standard state)

Calculate the sum of $(n \times \Delta G_f^\circ)$ for products:
$\sum (n \times \Delta G_f^\circ)_{\text{products}} = (2 \times -17) + (13 \times 0) = -34 + 0 = -34\ \text{kJ}$

Step3: Identify the reactants and their stoichiometric coefficients

The reactants are $\ce{CO2(g)}$ and $\ce{H2O(g)}$.

• For $\ce{CO2(g)}$: Stoichiometric coefficient $n = 8$, $\Delta G_f^\circ = -394\ \text{kJ/mol}$
• For $\ce{H2O(g)}$: Stoichiometric coefficient $n = 10$, $\Delta G_f^\circ = -229\ \text{kJ/mol}$

Calculate the sum of $(n \times \Delta G_f^\circ)$ for reactants:
$\sum (n \times \Delta G_f^\circ)_{\text{reactants}} = (8 \times -394) + (10 \times -229)$
First, calculate $8 \times -394 = -3152\ \text{kJ}$
Then, calculate $10 \times -229 = -2290\ \text{kJ}$
Sum: $-3152 + (-2290) = -5442\ \text{kJ}$

Step4: Calculate $\Delta G_{rxn}$

Using the formula from Step1:
$\Delta G_{rxn} = (-34) - (-5442) = -34 + 5442 = 5408\ \text{kJ}$

Answer:

5408 kJ