Question

Consider the following standard heats of formation:

P₄O₁₀(s) = -3110 kJ/mol
H₂O(l) = -286 kJ/mol
H₃PO₄(s) = -1279 kJ/mol
Calculate the change in enthalpy for the following process:

`P_4O_(10)(s) + 6H_2O(l) \rightarrow 4H_3PO_4(s)`

Answer 1

The Standard enthalpy of reaction:
`\Delta H_(r)^(\circ ) = -290\,kJ`

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

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

`\Rightarrow \Delta H_(r)^(\circ ) = [4 * (-1279\, kJ/mol)] - [1 * (-3110\, kJ/mol) + 6 * (-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`