Question

When one mole of nitroglycerine, \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l)\) decomposes, nitrogen, oxygen, carbon dioxide, and liquid water are formed. The decomposition liberates \(5725 \mathrm{~kJ}\) of heat. (a) Write a balanced thermochemical equation for the reaction. (b) Using Table \(8.3,\) calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for nitroglycerine.

Short answer

Answer: The standard enthalpy of formation for nitroglycerine is approximately -4151 kJ/mol.

Step-by-step solution

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the decomposition of nitroglycerine: C₃H₅(NO₃)₃(l) → N₂(g) + O₂(g) + CO₂(g) + H₂O(l) To balance the equation, we can start by counting the number of atoms of each element in the reactant and the products. The balanced equation becomes: 4 C₃H₅(NO₃)₃(l) → 6 N₂(g) + 10 O₂(g) + 12 CO₂(g) + 10 H₂O(l)

Include the heat liberated in the equation

Since we are given the heat liberated upon decomposition of one mole of nitroglycerine, we need to adjust the heat value for four moles of nitroglycerine in the balanced equation: 4 C₃H₅(NO₃)₃(l) → 6 N₂(g) + 10 O₂(g) + 12 CO₂(g) + 10 H₂O(l) + 22900 kJ

Apply Hess's law and use Table 8.3

Now, we will calculate ΔHf⁰ for nitroglycerine using Hess's law and the enthalpies of formation values from Table 8.3 for other substances in the equation: ΔH_{rxn} = Σ[ΔHf⁰(products)] - Σ[ΔHf⁰(reactants)] Where ΔH_{rxn} is given as 22,900 kJ for the balanced equation. Using the standard enthalpies of formation from Table 8.3, we have: ΔHf⁰(N₂) = 0 kJ/mol ΔHf⁰(O₂) = 0 kJ/mol ΔHf⁰(CO₂) = -393.5 kJ/mol ΔHf⁰(H₂O) = -285.8 kJ/mol 22,900 = [6 × 0 + 10 × 0 + 12 × (-393.5) + 10 × (-285.8)] - [4 × ΔHf⁰(C₃H₅(NO₃)₃)] Solving for ΔHf⁰(C₃H₅(NO₃)₃): ΔHf⁰(C₃H₅(NO₃)₃) = -4151 kJ/mol

Key concepts

Chemical Reaction Balancing

Understanding how to balance chemical reactions is crucial as it ensures the law of conservation of mass is respected, which states that matter cannot be created or destroyed in a closed system. When balancing a reaction, like the decomposition of nitroglycerine, it is important to count and equalize the number of atoms for each element on both sides of the equation. For instance, we start with a complex reactant \(\mathrm{C}_{3}\mathrm{H}_{5}(\mathrm{NO}_{3})_{3}\) and end with products like nitrogen, oxygen, carbon dioxide, and water. Each atom on the left must be accounted for on the right, leading to a balanced equation of:

\[4 \mathrm{C}_{3}\mathrm{H}_{5}(\mathrm{NO}_{3})_{3}(\ell) \rightarrow 6 \mathrm{N}_{2}(g) + 10 \mathrm{O}_{2}(g) + 12 \mathrm{CO}_{2}(g) + 10 \mathrm{H}_{2}\mathrm{O}(\ell)\]
Such an approach ensures that the mass remains constant before and after the reaction, enabling accurate quantitative descriptions of chemical processes.

Hess's Law

Hess's law offers a method to calculate the overall standard enthalpy change for a reaction. It states that the total enthalpy change during a chemical reaction is the same whether the reaction takes place in one step or through multiple steps. In our exercise, Hess's law allows us to determine the enthalpy of formation for nitroglycerine by summing the standard enthalpies of formation of the products and then subtracting the standard enthalpies of formation of the reactants. Here's a simplified expression for Hess's law:

\(\Delta H_{\mathrm{rxn}} = \Sigma[\Delta H_{\mathrm{f}}^\circ(\text{products})] - \Sigma[\Delta H_{\mathrm{f}}^\circ(\text{reactants})]\)
Using this relationship, the large calculation involving multiple substances is broken down into a series of simpler steps, each represented by its own enthalpy change.

Enthalpy of Formation

The enthalpy of formation, designated as \(\Delta H_{\mathrm{f}}^\circ\), is the heat change that results when one mole of a compound is formed from its elements in their standard states. Standard states reference the most stable form of an element or compound at 1 atm and 25°C (298.15 K). A critical aspect of \(\Delta H_{\mathrm{f}}^\circ\) values is that they are determined under these standard conditions and that the enthalpy of formation for any element in its standard state is zero. Therefore, in the given equation, the standard enthalpy of formation for nitrogen \(\mathrm{N}_{2}(g)\) and oxygen \(\mathrm{O}_{2}(g)\) is zero. When we calculate the \(\Delta H_{\mathrm{f}}^\circ\) for nitroglycerine, it involves the application of Hess's law to relate the known enthalpies of the other substances in the equation to the unknown enthalpy of formation of nitroglycerine.

Standard Enthalpy Change

Standard enthalpy change refers to the enthalpy change that occurs in a system when one mole of matter is transformed under standard conditions. For reactions like the decomposition of nitroglycerine, the standard enthalpy change is the heat absorbed or released. It is important to note that if the sign is negative, the reaction is exothermic; heat is released to the surroundings. Conversely, a positive sign indicates that the reaction is endothermic; heat is absorbed from the surroundings. In our exercise, the reaction releases \(5725 \mathrm{~kJ}\), hence, for four moles of nitroglycerine, the total heat released is \(22900 \mathrm{~kJ}\), indicating an exothermic reaction. The standard enthalpy change can be calculated using the balanced thermochemical equation and Hess's law, which allows the determination of energy changes associated with chemical reactions.