Problem 49 - Chemistry: Principles and Reactions

Question

An explosion used five tons \((1\) ton \(=2000 \mathrm{lb}\) ) of ammonium nitrate \((\Delta E=-37.0 \mathrm{~kJ} / \mathrm{mol})\). (a) How much energy was released by the explosion? (b) How many grams of TNT \((\Delta E=-2.76 \mathrm{~kJ} / \mathrm{g})\) are needed to release the energy calculated in (a)? (c) How many grams of U-235 are needed to obtain the same amount of energy calculated in (a)? (See the equation in Problem 45.)

Step-by-Step Solution

Answer

How many grams of TNT and U-235 are needed to release the same amount of energy? Answer: (a) The explosion released approximately 2,096,226.5 kJ of energy. (b) To release the same amount of energy, about 759,508.2 grams of TNT are needed. (c) To obtain the same amount of energy, about 0.0008385 grams of U-235 are needed.

Step 1: Convert mass of ammonium nitrate to moles

We are given that 1 ton is equal to 2000 lb. We have 5 tons of ammonium nitrate, so we need to convert this to mass in pounds: \(5~tons \times \frac{2000 ~lb}{1 ~ton} =10000 ~lb\) Now, we need to convert this mass to grams. We know that 1 lb is equal to 453.592 grams: \(10000 ~lb \times \frac{453.592 ~grams}{1 ~lb}= 4535920 ~grams\) Next, we will use the molar mass of ammonium nitrate (NH4NO3) to convert grams to moles. The molar mass is approximately: \(2 \times 1.007 + 4 \times 1.007 + 14.007 + 3 \times 16.00 = 80.044 ~grams/mol\) Now we can find the number of moles of ammonium nitrate: \(\frac{4535920 ~grams}{80.044 ~grams/mol} \approx 56654.5 ~moles\)

Step 2: Calculate the energy released by the explosion

We are given that the energy released per mole of ammonium nitrate is \(-37.0 ~kJ/mol\). We can now calculate the total energy released by multiplying the energy released per mole with the number of moles: \(-37.0 ~kJ/mol \times 56654.5 ~moles \approx -2096226.5 ~kJ\)

Step 3: Calculate the mass of TNT needed to release the same amount of energy

We are given that TNT releases an energy of \(-2.76 ~kJ/g\). We can now find how many grams of TNT are needed to release \(-2096226.5 ~kJ\): \(\frac{-2096226.5 ~kJ}{-2.76 ~kJ/g} \approx 759508.2 ~grams\)

Step 4: Calculate the mass of U-235 needed to obtain the same amount of energy

From Problem 45, we know that 0.2021 g of U-235 releases 2.5e12 J of energy. First, we need to convert this energy into kJ: \(2.5e12 ~J \times \frac{1 ~kJ}{1000 ~J} = 2.5e9 ~kJ\) Now, we can calculate the mass of U-235 required to release \(-2096226.5 ~kJ\) using the ratio of energy released per gram: \(\frac{-2096226.5 ~kJ}{2.5e9 ~kJ/g} \approx -0.0008385 ~grams\) The negative signs indicate energy being released, but we are interested in the absolute values. So, we have: (a) The explosion released approximately \(2096226.5 ~kJ\) of energy. (b) To release the same amount of energy, about \(759508.2 ~grams\) of TNT are needed. (c) To obtain the same amount of energy, about \(0.0008385 ~grams\) of U-235 are needed.

Key Concepts

Enthalpy of Reaction

Understanding the enthalpy of reaction is crucial when studying energy changes in chemical reactions. Enthalpy, represented by the symbol \( H \), is a measure of the total heat content of a system at constant pressure. The enthalpy of reaction, or \( \Delta H \), refers to the heat change that occurs during a chemical reaction.

For exothermic reactions, such as the explosion of ammonium nitrate, the enthalpy of reaction is negative, indicating that energy is being released to the surroundings. In contrast, endothermic reactions absorb energy, resulting in a positive \( \Delta H \). The value for the enthalpy change in the given problem is \( -37.0 \mathrm{~kJ/mol} \), describing the amount of energy each mole of ammonium nitrate releases upon decomposition.

To calculate the total energy released in the explosion, one would multiply the enthalpy change per mole by the total number of moles of reactants consumed. Here, improved comprehension can be achieved by reassuring a firm grasp of the concept of moles and molar enthalpy change as indicators of energy per unit of a substance.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the relationship between heat changes and chemical reactions. It encompasses the study of the energy and entropy of a system related to the spontaneity of a reaction.

The first law of thermodynamics states that energy cannot be created or destroyed, only transformed. This is directly applicable when calculating the energy released by the explosion of ammonium nitrate and equivalent energy from TNT or U-235, as shown in the exercise. The amounts of ammonium nitrate, TNT, and U-235 calculated reflect the energy that they can release, which is conserved and follows the law of conservation of energy.

As students work through chemical thermodynamics problems, they should be mindful that variables such as temperature, pressure, and the nature of reactants and products significantly impact the thermodynamic quantities of a system. The challenge often lies in translating these abstract concepts into actual quantitative values, as in the given exercise.

Stoichiometry

Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of the mole.

When utilizing stoichiometry to solve problems like the one given, one must first convert quantities into moles, as it allows for the comparison between different substances based on their molar ratios dictated by the balanced chemical equation. In this case, we began by converting the mass of ammonium nitrate into moles.

After determining the number of moles, the next step usually involves using the mole ratio to find the amount of another substance involved in the reaction, in terms of moles, grams, or energy released or absorbed. Students can improve problem-solving by clearly understanding and applying stoichiometric coefficients and the molar mass of each substance, essentially the heart of stoichiometry, to ensure accurate calculations and a strong conceptual foundation.