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

Short Answer

Expert verified
Answer: The enthalpy change when 4.65 mol of products are formed is approximately 6610 kJ.

Step by step solution

01

Write the unbalanced equation.

The unbalanced chemical equation for the reaction involving nitroglycerine is: C3H5(NO3)3(l) -> N2(g) + O2(g) + CO2(g) + H2O(g)
02

Balance the equation.

There are 3 carbons, 6 oxygens (from the nitrate), and 3 nitrogens in nitroglycerine and are shown below in the balanced equation. C3H5(NO3)3(l) -> 3CO2(g) + H2O(g) + 3N2(g) + 2.5O2(g) This equation is not yet thermochemical; we need to add heat given off per gram of nitroglycerine, which is 6.26 kJ/g.
03

Add heat to the balanced equation.

To have the complete balanced thermochemical equation, we need to add the heat given off by the reaction. The net heat produced (per mole) is: Heat (kJ/mol) = Grams of nitroglycerine × Heat per gram Heat (kJ/mol) = (227 g/mol ) (6.26 kJ/g) Heat (kJ/mol) = 1421 kJ Now, we can write the balanced thermochemical equation as follows: C3H5(NO3)3(l) -> 3CO2(g) + H2O(g) + 3N2(g) + 2.5O2(g) + 1421 kJ b) What is ∆H when 4.65 mol of products is formed?
04

Calculate the moles of nitroglycerine consumed

Calculate how many moles of nitroglycerine are consumed when 4.65 mol of products is formed. Since the products formed are not specified, use the product with the least number of moles (H2O) in the equation. Moles of Nitroglycerin consumed = Moles of H2O formed / Moles of H2O in balanced equation Moles of Nitroglycerin consumed = 4.65 mol / 1 Moles of Nitroglycerin consumed = 4.65 mol
05

Calculate the enthalpy change (∆H)

Use the enthalpy change for the balanced thermochemical equation (1421 kJ/mol) to calculate ∆H when 4.65 mol of products is formed. ΔH = moles of Nitroglycerine consumed × Enthalpy change per mole of Nitroglycerine ΔH = 4.65 mol × 1421 kJ/mol ΔH ≈ 6610 kJ So, the enthalpy change (∆H) when 4.65 mol of the product is formed is approximately 6610 kJ.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction Balancing
The process of balancing chemical equations is akin to solving a puzzle where each side of the equation must have the same number of atoms for each element. In the detonation of nitroglycerine, we deal with a complex molecule that breaks down into simpler gaseous products. To balance the equation, we tally the atoms of each element on both sides, adjusting coefficients—the numbers in front of molecules—accordingly.

It's crucial that no atoms are lost or gained in the reaction; this consistency upholds the law of conservation of mass. After balancing the equation, we can see that for every mole of nitroglycerine decomposed, we get 3 moles of carbon dioxide, a mole of water, 3 moles of nitrogen gas, and 2.5 moles of oxygen gas. These proportions are essential for stoichiometric calculations and enthalpy change computations, providing a roadmap to understanding the reaction's energetics.
Enthalpy Change Calculations
Enthalpy change, denoted as \( \Delta H \), measures the heat exchange in a chemical reaction at constant pressure. For exothermic reactions like the detonation of nitroglycerine, \( \Delta H \) is negative, signifying heat release. The calculation of enthalpy change requires knowledge of the reaction's stoichiometry and the amount of heat released or absorbed per mole of reactant or product.

In our example, each mole of nitroglycerine releases 1421 kJ of energy upon detonation—this figure is determined by multiplying the molecular weight of nitroglycerine by the heat given off per gram. When calculating \( \Delta H \) for a given amount of product, it's essential to first establish the stoichiometric relationship between reactants and products—a key concept intertwining stoichiometry and thermochemistry.
Stoichiometry
Stoichiometry bridges the quantitative relationship between reactants and products in a chemical reaction. It's fundamental in ensuring that calculations for mass, moles, and energy are consistent and accurate. For a given reaction, stoichiometry dictates the exact proportions of each substance involved.

To elucidate this, consider the reaction we are studying. We must first identify the 'limiting reactant' or in this case, utilize the stoichiometric ratios to find out how much nitroglycerine gives rise to 4.65 moles of the product—water, in this scenario. We calculate how much energy is exchanged (\( \Delta H \) value) by applying the stoichiometric principles to relate moles of products back to moles of reactants, essentially showcasing the interconnectedness of reactants, products, and their respective energy changes in a coherent stoichiometric framework.

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Most popular questions from this chapter

Mercury has a specific heat of \(0.140 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). Assume that a thermometer has 20 (2 significant figures) grams of mercury. How much heat is absorbed by the mercury when the temperature in the thermometer increases from \(98.6^{\circ} \mathrm{F}\) to \(103.2^{\circ} \mathrm{F} ?\) (Assume no heat loss to the glass of the thermometer.)

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