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

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

Step-by-step 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 \\u .\Delta H_(f)^(\circ )(products)-\sum \\u .\Delta H_(f)^(\circ )(reactants)`

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

`\Rightarrow \Delta H_(r)^(\circ ) = [1 * (+66.4\,kJ/mol) + 4 * (-285.8\,kJ/mol) ] - [1 * (-904.6\,kJ/mol) + 4 * (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`