Question

How would you calculate the enthalpy of formation of the reaction:
$4{C_3}{H_5}{N_3}{O_9}(l) \to 2C{O_2} + 10{H_2}O + 6N$, where $\Delta H = - 5678J$.

Answer 1

Hint: We need to understand the concept of enthalpy change with the combination of Hess’ law. When a substance changes at constant pressure, enthalpy, $\Delta H$ determines the amount of heat and work which was added or removed from the substance. According to Hess’s Law of Constant Heat summation, if the reaction takes place in several steps then its Total heat change is the sum of the heat changes of the individual reactions.

Complete step by step answer:
To find the enthalpy of formation of nitroglycerin ${C_3}{H_5}{N_3}{O_9}$ we use Hess’s Law. This implies that we can express the standard enthalpy change $\left( {\Delta H^\circ } \right)$ of reaction by using the standard enthalpy changes of formation of the reactant and of the products.
\[\Delta {H^ \circ }_{reaction} = \sum (n \times \Delta H{^\circ _{products}}) - \sum (m \times \Delta H{^\circ _{reac\tan ts}})\], where
n, m - the stoichiometric coefficients of the species that take part in the reaction.
The standard enthalpy changes of formation for one mole of carbon dioxide, water, and nitrogen gas are constant and is given below
For $C{O_2} = - 393.51kJ/mol$
For ${H_2}O = - 241.82kJ/mol$
${N_2} = 0kJ/mol$ 4 moles of nitroglycerin.
Let us first calculate for the products:
$\left[ {12moles\left( { - 393.51kJ/mol} \right) + 10moles( - 241.82kJ/mol + 6moles(0kJ/mol)} \right] = - 7.140kJ$
In the reaction, there are 12 moles of carbon dioxide, 10 moles of water, 6 moles of nitrogen gas and
We are to calculate the $\Delta H^\circ $ for nitroglycerin ${C_3}{H_5}{N_3}{O_9}$.
Using Hess’ Law,
\[\Delta {H^ \circ }_{reaction} = \sum (n \times \Delta {H^ \circ }_{products}) - \sum (m \times \Delta {H^ \circ }_{reac\tan ts})\].
$ - 5.678kJ = - 7140.32kJ - (4 \times \Delta {H^ \circ }_{nitroglycerin})$
On rearranging, we get $\Delta {H^ \circ }_{nitroglycerin} = \dfrac{{ - 7134.642kJ}}{4} = - 1784kJ/mol$
Hence, the enthalpy of formation of the reaction:
$4{C_3}{H_5}{N_3}{O_9}(l) \to 2C{O_2} + 10{H_2}O + 6N$, where $\Delta H = - 5678J$ is $ - 1784kJ/mol$.

Note:
It must be noted that Heat formation or heat change is also known as standard enthalpy change and is independent of the path between initial state (reactants) and final state (products). The combination of Kirchhoff’s Law and Hess’s Law can be used to calculate the formation of heat at different temperatures. A negative value of $\Delta H$ indicates that the reaction is exothermic and a positive value of $\Delta H$ means the reaction is endothermic.