Question

Calculate the standard enthalpy change and standard internal energy change for the following reaction at $300{{K}}$:
${{O}}{{{F}}_{2\left( {{g}} \right)}} + {{{H}}_2}{{{O}}_{\left( {{g}} \right)}} \to {{{O}}_2}_{\left( {{g}} \right)} + 2{{H}}{{{F}}_{\left( {{g}} \right)}}$
Given the standard enthalpy of formation of ${{O}}{{{F}}_2},{{{H}}_2}{{O}}$ and ${{HF}}$ are $23{{kJ}}.{{mo}}{{{l}}^{ - 1}}, - 241{{kJ}}.{{mo}}{{{l}}^{ - 1}}$ and $- 268.6{{kJ}}.{{mo}}{{{l}}^{ - 1}}$ respectively.

Answer 1

Enthalpy is a thermodynamic quantity equivalent to the total heat content of a system. It is denoted by ${{H}}$. Standard enthalpy change of reaction is the enthalpy change when the amount of reactants react to give products under standard condition. Standard enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its elements under standard conditions.

Complete step by step answer:
Standard enthalpy changes can be represented by the symbol $\Delta {{{H}}^ \circ }$. This refers that the reaction is carried out under standard conditions.
Consider the chemical reaction given below:
${{O}}{{{F}}_{2\left( {{g}} \right)}} + {{{H}}_2}{{{O}}_{\left( {{g}} \right)}} \to {{{O}}_2}_{\left( {{g}} \right)} + 2{{H}}{{{F}}_{\left( {{g}} \right)}}$
It is given that $\Delta {{{H}}_{{{f}}\left( {{{O}}{{{F}}_2}} \right)}} = 23{{kJ}}.{{mo}}{{{l}}^{ - 1}}$
$\Delta {{{H}}_{{{f}}\left( {{{{H}}_2}{{O}}} \right)}} = - 241.8{{kJ}}.{{mo}}{{{l}}^{ - 1}}$
$\Delta {{{H}}_{{{f}}\left( {{{HF}}} \right)}} = - 268.6{{kJ}}.{{mo}}{{{l}}^{ - 1}}$
Temperature, ${{T}} = 300{{K}}$
Therefore, $\Delta {{{H}}_{{R}}} = \Delta {{{H}}_{{{f}}\left( {{{products}}} \right)}} - \Delta {{{H}}_{{{f}}\left( {{{products}}} \right)}}$, where $\Delta {{{H}}_{{{f}}\left( {{{products}}} \right)}}$ is the standard enthalpy change of formation of products and $\Delta {{{H}}_{{{f}}\left( {{{reactants}}} \right)}}$ is the standard enthalpy change of formation of reactants.
Substituting the values, we get
$\Delta {{{H}}_{{R}}} = 2 \times \left( { - 268.6{{kJ}}.{{mo}}{{{l}}^{ - 1}}} \right) - \left( {23{{kJ}}.{{mo}}{{{l}}^{ - 1}} + - 241.8{{kJ}}.{{mo}}{{{l}}^{ - 1}}} \right)$
$\Delta {{{H}}_{{R}}} = - 537.2{{kJ}}.{{mo}}{{{l}}^{ - 1}} - 218.8{{kJ}}.{{mo}}{{{l}}^{ - 1}} = - 756{{kJ}}.{{mo}}{{{l}}^{ - 1}}$
Thus we can say that the standard enthalpy change is $- 756{{kJ}}.{{mo}}{{{l}}^{ - 1}}$
Enthalpy change, $\Delta {{H}} = \Delta {{U}} + {{P}}\Delta {{V}}$, where $\Delta {{U}}$ is the change in internal energy, ${{P}}$ is the pressure and $\Delta {{V}}$ is the change in volume.
From ideal gas equation, ${{P}}\Delta {{V}} = \Delta {{nRT}}$, where $\Delta {{n}}$ is the change in number of moles and ${{R}}$ is the gas constant $\left( {8.314 \times {{10}^{ - 3}}{{kJ}}.{{{K}}^{ - 1}}.{{mo}}{{{l}}^{ - 1}}} \right)$
Thus substituting this equation in the above equation, we get
$\Delta {{H}} = \Delta {{U}} + \Delta {{nRT}}$
i.e. $\Delta {{U}} = \Delta {{H - }}\Delta {{nRT}}$
Change in number of moles, $\Delta {{n = }}\Delta {{{n}}_{{p}}} - \Delta {{{n}}_{{r}}}$, where $\Delta {{{n}}_{{p}}},\Delta {{{n}}_{{r}}}$ are the change in the number of moles of products and reactants respectively.
i.e. $\Delta {{n = }}\left( {2 + 1} \right) - \left( {1 + 1} \right) = 3 - 2 = 1$
Substituting all the values in the equation of internal energy gives,
$\Delta {{U}} = \left( { - 756{{kJ}}.{{mo}}{{{l}}^{ - 1}}} \right) - 1 \times 8.314 \times {10^{ - 3}}{{kJ}}.{{{K}}^{ - 1}}.{{mo}}{{{l}}^{ - 1}} \times 300{{K}}$
On simplification, we get
$\Delta {{U}} = - 756 - 2.494 = - 753.506{{kJ}}.{{mo}}{{{l}}^{ - 1}}$
Hence the enthalpy change and the change in internal energy are $- 756{{kJ}}.{{mo}}{{{l}}^{ - 1}}$ and $- 753.506{{kJ}}.{{mo}}{{{l}}^{ - 1}}$.

Note:
In case of solids and liquids reactants, there is no appreciable change in volume. Thus the enthalpy change will be equal to the change in internal energy. Enthalpy is used to quantify the heat flow into or out of the system in a process that occurs at constant pressure. Since it varies with conditions, standard values are used.