Question

If $\Delta {{\text{H}}^{\text{0}}}_{\text{f}}$ for ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}{\text{O}}$ are $- 188{\text{kJ/mole}}$ and $- 286{\text{kJ/mole}}$ . What will be the enthalpy change of the reaction ${\text{2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{l}} \right) \to {\text{2}}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right) + {{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$ ?

Answer 1

The standard enthalpy or heat of formation is defined as the change in enthalpy or heat when one mole of a substrate is formed from its elements in their standard state.
The standard enthalpy of any reaction is generally considered to be equal to the difference between the standard enthalpy of formation values of all the products and the standard enthalpy of formation values of all the reactants.

Complete step by step answer:
The standard enthalpy or heat of formation is usually represented by the symbol $\Delta {{\text{H}}^{\text{0}}}_{\text{f}}$ .
The standard enthalpy of formation of elements is considered to be equal to 0.
The enthalpy of reactions can be determined by using the enthalpy of formation values. The mathematical expression which relates the standard enthalpy of a reaction to the enthalpy of formation values is:
$\Delta {{\text{H}}^{\text{0}}} = \Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{products}}} \right) - \Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{reactants}}} \right)$
Here, $\Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{products}}} \right)$ represents the enthalpy of formation of all the products and $\Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{reactants}}} \right)$ represents the enthalpy of formation of all the reactants.
Now, according to the given question, the enthalpy of formation of hydrogen peroxide is $- 188{\text{kJ/mole}}$ and the enthalpy of formation of water is $- 286{\text{kJ/mole}}$ . We need to find out the enthalpy change of the given reaction:
${\text{2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{l}} \right) \to {\text{2}}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right) + {{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
In this reaction, the products are 2 moles of water and 1 mole of oxygen and the reactant is 2 moles of hydrogen peroxide. So,
$\Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{products}}} \right) = 2 \times \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)} \right) + \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{{\text{O}}_2}\left( {\text{g}} \right)} \right)$
But, $\Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)} \right)$ is $- 286{\text{kJ/mole}}$ and $\Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{{\text{O}}_2}\left( {\text{g}} \right)} \right)$ is 0 as oxygen gas is an element in standard state. So, we have:
$\Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{products}}} \right) = 2 \times \left( { - 286} \right) + 0 \\\ \Rightarrow \Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{products}}} \right) = - 572{\text{kJ/mole}} \\\$
Again, $\Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{reactants}}} \right) = 2 \times \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{{\text{H}}_{\text{2}}}{{\text{O}}_2}\left( {\text{l}} \right)} \right)$ .
But, $\Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{{\text{H}}_{\text{2}}}{{\text{O}}_2}\left( {\text{l}} \right)} \right)$ is $- 188{\text{kJ/mole}}$ .
So,
$\Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{reactants}}} \right) = 2 \times \left( { - 188} \right) \\\ \Rightarrow \Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{reactants}}} \right) = - 376{\text{kJ/mole}} \\\$
Thus, the enthalpy change of the given reaction is:
$\Delta {{\text{H}}^{\text{0}}} = \Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{products}}} \right) - \Sigma \Delta {{\text{H}}^{\text{0}}}_{\text{f}}\left( {{\text{reactants}}} \right) \\\ \Rightarrow \Delta {{\text{H}}^{\text{0}}} = \left( { - 572} \right) - \left( { - 376} \right){\text{kJ/mole}} \\\ \Rightarrow \Delta {{\text{H}}^{\text{0}}} = - 196{\text{kJ/mole}} \\\$

Note:
Some other types of enthalpy of reactions which are common are enthalpy of combustion and enthalpy of hydration.
The enthalpy of combustion is defined as the enthalpy change when one mole of a substance undergoes complete combustion. For example, the heat of combustion of carbon is $- 393.5{\text{kJ/mole}}$ .
${\text{C}}\left( {\text{s}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right);\Delta {\text{H}} = - 393.5{\text{kJ/mole}}$
The enthalpy of hydration is defined as the enthalpy change when one mole of an anhydrous or partially hydrated salt combines with water to form its hydrate. For example:
${\text{CuS}}{{\text{O}}_4}\left( {\text{s}} \right){\text{ + 5}}{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right) \to {\text{CuS}}{{\text{O}}_4}{\text{.5}}{{\text{H}}_2}{\text{O}}\left( {\text{s}} \right);\Delta {\text{H}} = - 78.2{\text{kJ/mole}}$