Answer:
The Standard enthalpy of reaction: [tex]\Delta H_{r}^{\circ } = -290\,kJ[/tex]
Explanation:
Given- Standard Heat of Formation:
[tex]\Delta H_{f}^{\circ } [P_{4}O_{10}(s)][/tex] = -3110 kJ/mol,
[tex]\Delta H_{f}^{\circ } [H_{2}O(l)][/tex] = -286 kJ/mol,
[tex]\Delta H_{f}^{\circ } [H_{3}PO_{4}(s)][/tex] = -1279 kJ/mol
Given chemical reaction: P₄O₁₀(s) + 6H₂O → 4H₃PO₄
The standard enthalpy of reaction: [tex]\Delta H_{r}^{\circ }[/tex] = ?
To calculate the Standard enthalpy of reaction ([tex]\Delta H_{r}^{\circ }[/tex]), we use the equation:
[tex]\Delta H_{r}^{\circ } = \sum \nu .\Delta H_{f}^{\circ }(products)-\sum \nu .\Delta H_{f}^{\circ }(reactants)[/tex]
[tex]\Delta H_{r}^{\circ } = [4 \times \Delta H_{f}^{\circ } [H_{3}PO_{4}(s)]] - [1 \times \Delta H_{f}^{\circ } [P_{4}O_{10}(s)] + 6 \times \Delta H_{f}^{\circ } [H_{2}O(l)]][/tex]
[tex]\Rightarrow \Delta H_{r}^{\circ } = [4 \times (-1279\, kJ/mol)] - [1 \times (-3110\, kJ/mol) + 6 \times (-286\, kJ/mol)][/tex]
[tex]\Rightarrow \Delta H_{r}^{\circ } = [-5116\, kJ] - [-3110\, kJ -1716\, kJ][/tex]
[tex]\Rightarrow \Delta H_{r}^{\circ } = [-5116\, kJ] - [-4826\, kJ] = -290\,kJ[/tex]
Therefore, the Standard enthalpy of reaction: [tex]\Delta H_{r}^{\circ } = -290\,kJ[/tex]
