Question

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

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

Step2: Identify the stoichiometric coefficients and $\Delta G_f^\circ$ values for each compound

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

• Reactants: $\ce{P4O10(s)}$ with stoichiometric coefficient 1, $\Delta G_f^\circ = - 2698\space kJ/mol$; $\ce{H2O(l)}$ with stoichiometric coefficient 6, $\Delta G_f^\circ=-237\space kJ/mol$.
• Products: $\ce{H3PO4(aq)}$ with stoichiometric coefficient 4, $\Delta G_f^\circ = - 1119\space kJ/mol$.

Step3: Calculate the sum of $\Delta G_f^\circ$ for products

For the product $\ce{H3PO4(aq)}$, the sum is $4\times(- 1119\space kJ/mol)=-4476\space kJ/mol$.

Step4: Calculate the sum of $\Delta G_f^\circ$ for reactants

For the reactants, the sum is $1\times(-2698\space kJ/mol)+6\times(-237\space kJ/mol)=-2698 - 1422=-4120\space kJ/mol$.

Step5: Calculate $\Delta G_{rxn}^\circ$

Using the formula $\Delta G_{rxn}^\circ=\sum n\Delta G_f^\circ(\text{products})-\sum m\Delta G_f^\circ(\text{reactants})$, we substitute the values:
$\Delta G_{rxn}^\circ=(-4476\space kJ/mol)-(-4120\space kJ/mol)=-4476 + 4120=-356\space kJ$

Answer:

-356