Question

The value of ${\Delta _f}{H^ \circ }$for$N{H_3}$ is $ - 45.9KJmo{l^{ - 1}}$ . Calculate enthalpy change for the reaction:
$2N{H_3}(g) \to {N_2}(g) + 3{H_2}(g)$

Answer 1

Hint: In a chemical reaction, reactants are converted into products. Therefore the enthalpy change accompanying a reaction is called the reaction enthalpy. It is represented by the symbol ${\Delta _r}H$. It is required to maintain an industrial chemical reaction at constant temperature.

Complete answer:
Enthalpy of a reaction depends on the conditions under which a reaction is carried out. It is therefore necessary that we must specify out some standard conditions. The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states just like the reactants and products in the above equation.
The standard state of a substance at a specified temperature is its pure form at 1 bar. In the equation above we have been given standard enthalpy of formation for $N{H_3}$ is given that is the standard enthalpy change for the formation of one mole of a compound in this case $N{H_3}$ from its elements in their most stable states of aggregation also known as reference states is called standard molar enthalpy of formation.
Therefore enthalpy of reaction will be twice the enthalpy of formation as enthalpy of formation is for one mole and in the reaction we have two mole of $N{H_3}$
 $
  {\Delta _r}H = 2 \times {\Delta _f}{\rm H} \\
\Rightarrow {\Delta _r}H = 2 \times ( - 45.9KJmo{l^{ - 1}}) \\
\Rightarrow {\Delta _r}H = - 91.8KJmo{l^{ - 1}} \sim - 92KJmo{l^{ - 1}}
 $
Hence this is the enthalpy of the above reaction.

Note:
It is important to understand that a standard molar enthalpy of formation, ${\Delta _f}{H^ \circ }$ is just a special case of ${\Delta _r}H$, where one mole of a compound is formed from its constituent elements.