Question

Nitroglycerin, \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l),\) is an explosive most often used in mine or quarry blasting. It is a powerful explosive because four gases \(\left(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2},\right.\) and steam \()\) are formed when nitroglycerin is detonated. In addition, \(6.26 \mathrm{~kJ}\) of heat are given off per gram of nitroglycerin detonated. (a) Write a balanced thermochemical equation for the reaction. (b) What is \(\Delta H\) when \(4.65 \mathrm{~mol}\) of products are formed?

Short answer

Answer: The enthalpy change, ΔH, for the formation of 4.65 mol of products is -412.98 kJ.

Step-by-step solution

Write the unbalanced chemical equation

The unbalanced chemical equation for the detonation of nitroglycerin can be written as: C3H5(NO3)3(l) ⟶ N2(g) + O2(g) + CO2(g) + H2O(g)

Balance the chemical equation

To balance the chemical equation, we will adjust the coefficients for each reactant and product. We get: 2 C3H5(NO3)3(l) ⟶ 6 N2(g) + 10 O2(g) + 6 CO2(g) + 10 H2O(g)

Convert the heat given off per gram to the heat given off per mole for the ΔH value

Given that 6.26 kJ of heat are given off per gram of nitroglycerin, we first need to find the molar mass of nitroglycerin: Molar mass of C3H5(NO3)3 = 3(12.01) + 5(1.01) + 3(14.01+3(16.00)) = 227.09 g/mol Now, we can find the heat given off per mole of nitroglycerin: 6.26 kJ/g × 227.09 g/mol = 1421.42 kJ/mol We can now write the balanced thermochemical equation: 2 C3H5(NO3)3(l) ⟶ 6 N2(g) + 10 O2(g) + 6 CO2(g) + 10 H2O(g) ΔH = -1421.42 kJ The negative sign indicates that the reaction is exothermic.

Calculate the ΔH value for 4.65 mol of products formed

From the balanced thermochemical equation, we can see that 6 moles of N2, 10 moles of O2, 6 moles of CO2, and 10 moles of H2O are formed per 2 moles of nitroglycerin reacted. Therefore, for every mole of nitroglycerin reacted, 4.65 mol of products are formed: 4.65 mol of products / 32 moles of products (total in balanced equation) = x moles of nitroglycerin / 2 moles of nitroglycerin x = 0.290625 moles of nitroglycerin Now, we can calculate the ΔH value for 4.65 mol of products formed: ΔH = -1421.42 kJ/mol × 0.290625 mol = -412.98 kJ The enthalpy change, ΔH, for forming 4.65 mol of products is -412.98 kJ.

Key concepts

Nitroglycerin Detonation

When we talk about nitroglycerin detonation, we're discussing an explosive chemical reaction that releases a vast amount of energy in a very short period. Nitroglycerin, or \( \mathrm{C}_{3}\mathrm{H}_{5}(\mathrm{NO}_{3})_{3} \), undergoes a rapid decay resulting in a series of gases such as nitrogen, oxygen, carbon dioxide, and steam.

A notable aspect of such detonation is the transformation of solid or liquid reactants into gaseous products, which leads to a dramatic increase in volume, and this is what causes the explosion. Detonations are highly exothermic, meaning they release heat. In practical applications, this reaction is controlled to be used for constructive purposes like mining and demolition.

Understanding the detonation of nitroglycerin provides an excellent opportunity to see the laws of thermodynamics in action, particularly the rapid release of energy from high-energy bonds found in the nitroglycerin's molecular structure.

Chemical Equation Balancing

Balancing chemical equations is an essential step in chemistry, as it reflects the conservation of mass and atoms. To balance a chemical equation, we adjust the coefficients (the numbers placed before the chemical formulas) to ensure that the number of atoms for each element is the same on both sides of the equation.

In the case of nitroglycerin detonation, we started with an unbalanced equation and proceeded step by step, modifying coefficients to keep track of each atom present in the reactants and products. Through this meticulous process, we achieved the balanced equation: \[2 \mathrm{C}_{3}\mathrm{H}_{5}(\mathrm{NO}_{3})_{3}(l) ⟶ 6 \mathrm{N}_{2}(g) + 10 \mathrm{O}_{2}(g) + 6 \mathrm{CO}_{2}(g) + 10 \mathrm{H}_{2}\mathrm{O}(g)\]

Each element's atom count on both sides of the equation must be consistent. This skill is fundamental as it lays the groundwork for further quantitative analyses, such as stoichiometry, which are used to predict the outcomes of chemical reactions.

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a measure of heat change during a chemical reaction at constant pressure. It is a central concept in chemical thermodynamics, indicating whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).

The enthalpy change for the detonation of nitroglycerin was determined by first finding the exothermic heat given off per mole of reactant and then calculating it for a given amount of products. Given the exothermic nature of this explosion, the \( \Delta H \) is negative, showing that energy is released to the surroundings, which is consistent with our understanding of an explosion.

In our exercise, the exact \( \Delta H \) for 4.65 moles of products was calculated to be -412.98 kJ. Such calculations are critical in designing and managing industrial processes to ensure safety and efficiency and in understanding the energy profiles of chemical reactions in academic and research settings.