Question

Calculate Horxn for the following reaction:H3AsO4(aq) + 4 H2(g) --> AsH3(g) + 4 H2O(l)(Hof [AsH3(g)] = 66.4 kJ/mol; Hof [H3AsO4(aq)] = -904.6 kJ/mol; Hof [H2O(l)] = -285.8 kJ/mol)

Answer 1

Answer:

The Standard enthalpy of reaction: $\Delta H_{r}^{\circ } = (-172.2) \, kJ$

Explanation:

Given- Standard Heat of Formation:

$\Delta H_{f}^{\circ } [H_{3}AsO_{4}(aq)]$ = -904.6 kJ/mol

$\Delta H_{f}^{\circ } [H_{2}(g)]$ = 0 kJ/mol,

$\Delta H_{f}^{\circ } [AsH_{3}(g)]$ = +66.4 kJ/mol

$\Delta H_{f}^{\circ } [H_{2}O(l)]$ = -285.8 kJ/mol

   

Given chemical reaction: H₃AsO₄(aq) + 4H₂(g) → AsH₃(g) + 4H₂O(l)

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

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

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

$\Delta H_{r}^{\circ } = [1 \times \Delta H_{f}^{\circ } [AsH_{3} (g)] + 4 \times \Delta H_{f}^{\circ } [H_{2}O(l)]] - [1 \times \Delta H_{f}^{\circ } [H_{3}AsO_{4}(aq)] + 4 \times \Delta H_{f}^{\circ } [H_{2}(g)]$

$\Rightarrow \Delta H_{r}^{\circ } = [1 \times (+66.4\,kJ/mol) + 4 \times (-285.8\,kJ/mol) ] - [1 \times (-904.6\,kJ/mol) + 4 \times (0\,kJ/mol)]$

$\Rightarrow \Delta H_{r}^{\circ } = [-1076.8\, kJ] - [-904.6\,kJ]$

$\Rightarrow \Delta H_{r}^{\circ } = (-172.2 \, kJ)$

Therefore, the Standard enthalpy of reaction: $\Delta H_{r}^{\circ } = (-172.2) \, kJ$