Question

Consider the following standard heats of formation:

Answer 1

Answer:

The Standard enthalpy of reaction: $\Delta H_{r}^{\circ } = -290\,kJ$            

Explanation:

Given- Standard Heat of Formation:

$\Delta H_{f}^{\circ } [P_{4}O_{10}(s)]$ = -3110 kJ/mol,

$\Delta H_{f}^{\circ } [H_{2}O(l)]$ = -286 kJ/mol,

$\Delta H_{f}^{\circ } [H_{3}PO_{4}(s)]$ = -1279 kJ/mol

Given chemical reaction: P₄O₁₀(s) + 6H₂O → 4H₃PO₄

The standard enthalpy of reaction: $\Delta H_{r}^{\circ }$ = ?

To calculate the Standard enthalpy of reaction ($\Delta H_{r}^{\circ }$), we use the equation:

$\Delta H_{r}^{\circ } = \sum \nu .\Delta H_{f}^{\circ }(products)-\sum \nu .\Delta H_{f}^{\circ }(reactants)$

$\Delta H_{r}^{\circ } = [4 \times \Delta H_{f}^{\circ } [H_{3}PO_{4}(s)]] - [1 \times \Delta H_{f}^{\circ } [P_{4}O_{10}(s)] + 6 \times \Delta H_{f}^{\circ } [H_{2}O(l)]]$

$\Rightarrow \Delta H_{r}^{\circ } = [4 \times (-1279\, kJ/mol)] - [1 \times (-3110\, kJ/mol) + 6 \times (-286\, kJ/mol)]$

$\Rightarrow \Delta H_{r}^{\circ } = [-5116\, kJ] - [-3110\, kJ -1716\, kJ]$

$\Rightarrow \Delta H_{r}^{\circ } = [-5116\, kJ] - [-4826\, kJ] = -290\,kJ$

Therefore, the Standard enthalpy of reaction: $\Delta H_{r}^{\circ } = -290\,kJ$