Стандартна ентальпія утворення

Standard Enthalpy of Formation ($\Delta H_{f} ^{o}$):

the enthalpy change associated with the formation of one mole of a substance from its constitute elements, in their standard states. A pure element that is in its standard state has a standard enthalpy of formation equal to zero.

To calculate $\Delta H_{f} ^{o}$ of a reaction, the following formula is used:

$\Delta$$_{f}$$H^{o}$= $\Sigma$v$\Delta H_{f}^{o}$(products) -$\Sigma$v$\Delta H_{f}^{o}$(reactants)

• $\Delta$$_{f}$$H^{o}$= standard enthalpy of formation for a chemical reaction at standard conditions. The $H$ symbolizes the change in enthalpy and the $f$ stands for formation. The superscript, $o$ refers to the fact that everything is under standard state conditions

• $v$= coefficients of each reactant or product in a balanced chemical reaction

• $\Sigma$v$\Delta H_{f} ^{o}$(products)= the sum of the enthalpy of each individual product in the balanced chemical reaction

• $\Sigma$v$\Delta H_{f} ^{o}$(reactants): sum of the enthalpies of every reactant in the balanced chemical reaction

• $\Delta$$_{f}$$H^{o}$ is measured in kilojoules (kJ)

This equation essentially states that the standard enthalpy change of formation is equal to the sum of the standard enthalpies of the products minus the sum of the standard enthalpies of the reactants.

The use of the standard enthalpy of formation is an application of Hess's law. Simply put, we look up the values, add them up for all the reactants and the products, then subtract!

Let's look at the combustion of methane as an example:

$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2}(g) + 2H_{2}O_{(l)}$

• $CH_{4(g)} \rightarrow C(s) + 2H_{2(g)}$, $\Delta H_{f}=74.8\frac{kJ}{mol}$
• Oxygen is a diatomic element. It's already in its elemental state. So, its $\Delta H_{f}=0$.
• $C_{s} + O_{2(g)} \rightarrow CO_{2(g)}$, $\Delta H_{f}=393.2\frac{kJ}{mol}$
• $H_{2(g)} + \frac{1}{2}O_{2(g)}\rightarrow 1H_{2}O_{l}$, $\Delta H_{f}=285.8\frac{KJ}{mol}$

Notice that our reaction forms two moles of liquid water. So, we'll need to multiple it's $\Delta H_{f}$ value by 2.

$\Delta H= \Sigma \Delta H_{f(reactants)}-\Sigma \Delta H_{f(products)}= (74.8KJ+0kJ)-(393.2kJ+(2\times 285.8kJ))= -890.0 kJ$.

(Video) Chemistry 10.7a Standard Heat of Formation

by IsaacsTEACH

In this video, Isaac explains the standard heat of formation. He starts by defining the standard conditions. Then he defines formation reactions and gives you some examples. After watching this video you'll understand the standard enthalpy of formation.