Question

Nitroglycerine, \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l)\), is a powerful explosive used in rock blasting when roads are created. When ignited, it produces water, nitrogen, carbon dioxide, and oxygen. Detonation of one mole of nitroglycerine liberates \(5725 \mathrm{~kJ}\) of heat. (a) Write a balanced thermochemical equation for the reaction for the detonation of four moles of nitroglycerine. (b) What is \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l) ?\)

Short answer

a) The balanced thermochemical equation for the detonation of nitroglycerin is: $$4\,\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow 6\,\text{CO}_{2}(g) + 10\,\text{H}_{2}\text{O}(l) + \frac{1}{2}\,\text{O}_{2}(g) + 6\,\text{N}_{2}(g) \hspace{.5cm} \Delta H=-22900\,\text{kJ}$$ b) The standard enthalpy of formation (\(\Delta H_{\mathrm{f}}^{\circ}\)) for nitroglycerine is \(-5725\,\text{kJ/mol}\).

Step-by-step solution

Writing the Unbalanced Equation

First, let's write the unbalanced equation for the reaction of nitroglycerine: $$\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow \text{CO}_{2}(g) + \text{H}_{2}\text{O}(l) + \text{O}_{2}(g) + \text{N}_{2}(g)$$

Balancing the Equation

Now let's balance the equation. The balanced chemical equation is: $$4\,\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow 6\,\text{CO}_{2}(g) + 10\,\text{H}_{2}\text{O}(l) + \frac{1}{2}\,\text{O}_{2}(g) + 6\,\text{N}_{2}(g)$$ Here we have balanced the number of carbon, hydrogen, nitrogen, and oxygen atoms on both sides of the equation.

from a) Balanced Thermochemical Equation for the Detonation with 4 moles of nitroglycerine

Now, as we're asked to write a balanced thermochemical equation for the reaction, let's include the heat liberated for the 4 moles of nitroglycerine detonation. $$4\,\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow 6\,\text{CO}_{2}(g) + 10\,\text{H}_{2}\text{O}(l) + \frac{1}{2}\,\text{O}_{2}(g) + 6\,\text{N}_{2}(g) \hspace{.5cm} \Delta H=-4\cdot5725\,\text{kJ}$$ So that the answer for part (a) will be: $$4\,\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow 6\,\text{CO}_{2}(g) + 10\,\text{H}_{2}\text{O}(l) + \frac{1}{2}\,\text{O}_{2}(g) + 6\,\text{N}_{2}(g) \hspace{.5cm} \Delta H=-22900\,\text{kJ}$$

Calculating the \(\Delta H_{\mathrm{f}}^{\circ}\) for nitroglycerine

Now let's calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for 1 mole of nitroglycerine, given the released heat for 4 moles of nitroglycerin detonation as \(-22900\,\text{kJ}\). $$\Delta H_{\mathrm{f}}^{\circ} (\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}) = \frac{-22900\,\text{kJ}}{4\,\text{moles}} = -5725\,\text{kJ/mol}$$ The answer for part (b) is \(\Delta H_{\mathrm{f}}^{\circ}(\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9})=-5725\,\text{kJ/mol}\).

Key concepts

Enthalpy

Enthalpy is a measure of the total energy in a thermodynamic system. It includes internal energy along with the energy required to make room for it by displacing its environment. In chemical reactions, enthalpy change (\( \Delta H \)) is critical because it indicates whether a reaction absorbs or releases heat.

• If \( \Delta H < 0 \), the reaction releases heat (exothermic).
• If \( \Delta H > 0 \), the reaction absorbs heat (endothermic).

For the reaction involving nitroglycerine, a powerful explosive, the enthalpy change is significant. When detonating one mole, the process liberates 5725 kJ of heat, making it highly exothermic. Incorporating this value into a thermochemical equation helps us understand the energy dynamics in such a reaction.

Nitroglycerine Reaction

Nitroglycerine, a molecule comprising carbon, hydrogen, nitrogen, and oxygen, is commonly used as an explosive due to its rapid and violent decomposition. When ignited, it decomposes into various gases: carbon dioxide (\( \text{CO}_2 \)), water vapor (\( \text{H}_2\text{O} \)), oxygen (\( \text{O}_2 \)), and nitrogen (\( \text{N}_2 \)).

• This decomposition releases a significant amount of energy, making nitroglycerine highly explosive.
• Given the reaction of four moles of nitroglycerine, it liberates 22900 kJ.

Understanding the products and energy released in this reaction is crucial for safety and practical applications where controlled blasts are needed, such as in construction for road blasting.

Balancing Chemical Equations

Balancing chemical equations ensures that the mass and charge are conserved throughout a chemical reaction. For the detonation of nitroglycerine, the equation must account for all atoms on both sides to reflect the actual reaction process:

• Start by listing all elements involved: C, H, N, O.
• Carefully balance each element one by one.
• We end up with the correct ratio of molecules, e.g., 4 moles of nitroglycerine yielding 6 moles of CO₂, 10 moles of H₂O, etc.

Balancing the equation ensures it follows the Law of Conservation of Mass and reflects accurately how substances interact at the atomic level. This is critical not just for theoretical calculations, but also for predicting quantities of products in real-world chemical processes.

Heat of Formation

The heat of formation (\( \Delta H_{\text{f}}^\circ \)) is the energy change when one mole of a compound forms from its elements in their standard states. It provides insights into the stability and energy content of substances.

• For nitroglycerine, the \( \Delta H_{\text{f}}^\circ \) can be calculated based on the energy released during detonation.
• Given that 4 moles release 22900 kJ, the heat of formation for one mole is −5725 kJ/mol.

This negative value indicates that nitroglycerine is less stable than its constituent elements in their reference states, signifying a higher potential energy, which assists its application as an explosive. Accurately determining \( \Delta H_{\text{f}}^\circ \) allows chemists to understand energy shifts during reactions, essential for chemical manufacturing and safety assessments.