Question

p₄o₁₀(s) + 6h₂o(l) → 4h₃po₄(aq)
calculate the gibbs free energy change for
the reaction from the thermodynamic data.

compound	δg°ᵥ(kj/mol)
p₄o₁₀(s)	-2698
h₂o(l)	-237
h₃po₄(aq)	-1119

δg°ᵣₓₙ = ? kj
enter either a + or - sign.

Explanation:

Step1: Recall the formula for Gibbs Free Energy change

The formula for the standard Gibbs Free Energy change of a reaction ($\Delta G^{\circ}_{rxn}$) is the sum of the Gibbs Free Energy of formation ($\Delta G^{\circ}_{f}$) of the products minus the sum of the Gibbs Free Energy of formation of the reactants, considering their stoichiometric coefficients. Mathematically, it is:
$$\Delta G^{\circ}_{rxn} = \sum n\Delta G^{\circ}_{f}(\text{products}) - \sum m\Delta G^{\circ}_{f}(\text{reactants})$$
where $n$ and $m$ are the stoichiometric coefficients of the products and reactants respectively.

Step2: Identify the stoichiometric coefficients and $\Delta G^{\circ}_{f}$ values

For the given reaction: $\ce{P4O10(s) + 6H2O(l) -> 4H3PO4(aq)}$

• Reactants: $\ce{P4O10(s)}$ with $m = 1$ and $\Delta G^{\circ}_{f} = -2698\ \text{kJ/mol}$; $\ce{H2O(l)}$ with $m = 6$ and $\Delta G^{\circ}_{f} = -237\ \text{kJ/mol}$.
• Products: $\ce{H3PO4(aq)}$ with $n = 4$ and $\Delta G^{\circ}_{f} = -1119\ \text{kJ/mol}$.

Step3: Calculate the sum of $\Delta G^{\circ}_{f}$ for products

$$\sum n\Delta G^{\circ}_{f}(\text{products}) = 4\times(-1119\ \text{kJ/mol}) = -4476\ \text{kJ/mol}$$

Step4: Calculate the sum of $\Delta G^{\circ}_{f}$ for reactants

$$\sum m\Delta G^{\circ}_{f}(\text{reactants}) = 1\times(-2698\ \text{kJ/mol}) + 6\times(-237\ \text{kJ/mol})$$
First, calculate $6\times(-237)$: $6\times(-237) = -1422$
Then, add to $-2698$: $-2698 + (-1422) = -4120\ \text{kJ/mol}$

Step5: Calculate $\Delta G^{\circ}_{rxn}$

$$\Delta G^{\circ}_{rxn} = \sum n\Delta G^{\circ}_{f}(\text{products}) - \sum m\Delta G^{\circ}_{f}(\text{reactants})$$
Substitute the values:
$$\Delta G^{\circ}_{rxn} = (-4476\ \text{kJ/mol}) - (-4120\ \text{kJ/mol})$$
$$\Delta G^{\circ}_{rxn} = -4476 + 4120 = -356\ \text{kJ}$$

Answer:

-356