Question

what is $\delta g^\circ_{\text{rxn}}$ for the combustion of propane at 298 k?
$\ce{c3h8(g) + 5o2(g) -> 3co2(g) + 4h2o(g)}$
substance: $\delta g^\circ_\text{f}$(kj/mol):
$\ce{c3h8(g)}$: $-24.5$
$\ce{o2(g)}$: $0$
$\ce{co2(g)}$: $-394.4$
$\ce{h2o(g)}$: $-228.6$
options:

• $-2073.1$ kj/mol
• $+598.5$ kj/mol
• $-598.5$ kj/mol
• $+2073.1$ kj/mol
• $-1387.3$ kj/mol

Explanation:

Step1: Recall the formula for $\Delta G^\circ_{\text{rxn}}$

The formula for the standard Gibbs free energy change of a reaction ($\Delta G^\circ_{\text{rxn}}$) is the sum of the standard Gibbs free energies of formation ($\Delta G^\circ_f$) of the products minus the sum of the standard Gibbs free energies of formation of the reactants. Mathematically, it is:
$$\Delta G^\circ_{\text{rxn}} = \sum n \Delta G^\circ_f(\text{products}) - \sum m \Delta G^\circ_f(\text{reactants})$$
where $n$ and $m$ are the stoichiometric coefficients of the products and reactants, respectively.

Step2: Identify the reactants and products with their stoichiometric coefficients

The reaction is:

$$\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) ightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{g})$$

• Reactants: $\text{C}_3\text{H}_8$ (stoichiometric coefficient = 1), $\text{O}_2$ (stoichiometric coefficient = 5)
• Products: $\text{CO}_2$ (stoichiometric coefficient = 3), $\text{H}_2\text{O}$ (stoichiometric coefficient = 4)

Step3: List the $\Delta G^\circ_f$ values for each substance

From the given data:

• $\Delta G^\circ_f(\text{C}_3\text{H}_8(\text{g})) = -24.5\ \text{kJ/mol}$
• $\Delta G^\circ_f(\text{O}_2(\text{g})) = 0\ \text{kJ/mol}$ (standard state for elements)
• $\Delta G^\circ_f(\text{CO}_2(\text{g})) = -394.4\ \text{kJ/mol}$
• $\Delta G^\circ_f(\text{H}_2\text{O}(\text{g})) = -228.6\ \text{kJ/mol}$

Step4: Calculate the sum of $\Delta G^\circ_f$ for products

For the products ($3\text{CO}_2$ and $4\text{H}_2\text{O}$):
$$\sum n \Delta G^\circ_f(\text{products}) = 3 \times \Delta G^\circ_f(\text{CO}_2) + 4 \times \Delta G^\circ_f(\text{H}_2\text{O})$$
Substitute the values:
$$3 \times (-394.4) + 4 \times (-228.6)$$
First, calculate each term:

• $3 \times (-394.4) = -1183.2\ \text{kJ/mol}$
• $4 \times (-228.6) = -914.4\ \text{kJ/mol}$

Now, sum these two terms:
$$-1183.2 + (-914.4) = -1183.2 - 914.4 = -2097.6\ \text{kJ/mol}$$

Step5: Calculate the sum of $\Delta G^\circ_f$ for reactants

For the reactants ($1\text{C}_3\text{H}_8$ and $5\text{O}_2$):
$$\sum m \Delta G^\circ_f(\text{reactants}) = 1 \times \Delta G^\circ_f(\text{C}_3\text{H}_8) + 5 \times \Delta G^\circ_f(\text{O}_2)$$
Substitute the values:
$$1 \times (-24.5) + 5 \times 0$$
Simplify:
$$-24.5 + 0 = -24.5\ \text{kJ/mol}$$

Step6: Calculate $\Delta G^\circ_{\text{rxn}}$

Using the formula from Step1:
$$\Delta G^\circ_{\text{rxn}} = \sum n \Delta G^\circ_f(\text{products}) - \sum m \Delta G^\circ_f(\text{reactants})$$
Substitute the values from Step4 and Step5:
$$\Delta G^\circ_{\text{rxn}} = (-2097.6) - (-24.5)$$
Simplify the subtraction (subtracting a negative is adding the positive):
$$\Delta G^\circ_{\text{rxn}} = -2097.6 + 24.5 = -2073.1\ \text{kJ/mol}$$

Answer:

$\boldsymbol{-2073.1\ \text{kJ/mol}}$ (corresponding to the option: -2073.1 kJ/mol)