Question

Challenge ?H for the following reaction is -1789 kJ. Use this and Equation a to determine ?H for Equation b. \(4 {Al}({s})+3 {MnO}_{2}({s}) \rightarrow 2 {Al}_{2} {O}_{3}({s})+3 {Mn}({s}) \quad \Delta H=-1789 {kJ}\) a. \(4 \mathrm{Al}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}) \Delta H=-3352 \mathrm{kJ}\) b. \(\mathrm{Mn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{MnO}_{2}(\mathrm{s}) \Delta H=?\)

Short answer

The enthalpy change, ΔH, for the given Reaction b, \(Mn(s) + O_{2}(g) \rightarrow MnO_{2}(s)\), is 241.2 kJ.

Step-by-step solution

Write down the given reactions

The given reactions are as follows: Challenge: \(4 Al(s) + 3 MnO_{2}(s) \rightarrow 2 Al_{2}O_{3}(s) + 3 Mn(s); \ \Delta H = -1789 \ kJ\) Equation a: \(4 Al(s) + 3 O_{2}(g) \rightarrow 2 Al_{2}O_{3}(s); \ \Delta H = -3352 \ kJ\) Equation b (desired reaction): \(Mn(s) + O_{2}(g) \rightarrow MnO_{2}(s); \ \Delta H = ?\)

Manipulate the given reactions to match the desired reaction

First, we need to manipulate the Challenge reaction to isolate MnO_{2} on the right side. To do this, divide the entire reaction, including the enthalpy change, by 3. \(\frac{1}{3}\) Challenge: \( \frac{4}{3} Al(s) + MnO_{2}(s) \rightarrow \frac{2}{3} Al_{2}O_{3}(s) + Mn(s); \ \Delta H = -596.3 \ kJ\) Next, we need to manipulate Equation a to isolate Al_{2}O_{3} on the left side and O_{2}(g) on the right side. To do this, reverse the reaction and divide it by 2. \(-\frac{1}{2}\) Equation a: \( Al_{2}O_{3}(s) \rightarrow 2 Al(s) + \frac{3}{2} O_2(g); \ \Delta H = 837.5 \ kJ\)

Add the manipulated reactions and calculate the enthalpy change of the desired reaction

Now, we will add the manipulated reactions together to obtain the desired reaction (Equation b). \(\frac{1}{3}\) Challenge + \(-\frac{1}{2}\) Equation a: \(\frac{4}{3} Al(s) + MnO_{2}(s) + Al_{2}O_{3}(s) \rightarrow \frac{2}{3} Al_{2}O_{3}(s) + Mn(s) + 2 Al(s) + \frac{3}{2} O_{2}(g)\) This simplifies to: \(Mn(s) + O_{2}(g) \rightarrow MnO_{2}(s)\) The enthalpy change ΔH for this reaction: \(\Delta H_b = -596.3 \ kJ + 837.5 \ kJ = 241.2 \ kJ\)

Conclusion

The enthalpy change, ΔH, for the given Reaction b is 241.2 kJ.

Key concepts

Thermochemistry

Thermochemistry is a branch of chemistry that deals with the study of heat energy involved in chemical reactions and phase changes. Understanding the concept of enthalpy, symbolized as \( \Delta H \), is crucial in thermochemistry. Enthalpy is a measure of the total energy of a thermodynamic system, including both internal energy and the product of pressure and volume. The \( \Delta H \) associated with a chemical reaction indicates whether the reaction is exothermic (releasing heat to the surroundings, indicated by a negative \( \Delta H \) value) or endothermic (absorbing heat from the surroundings, indicated by a positive \( \Delta H \) value).

In thermochemical calculations, it is important to pay attention to the stoichiometry of the reactions since \( \Delta H \) is dependent on the amounts and states of substances involved. The provided exercise tasked us with calculating the enthalpy change for a specific reaction using given \( \Delta H \) values for other reactions, highlighting the practical application of thermochemistry in solving chemical problems.

Chemical Reactions

Chemical reactions involve the transformation of one set of chemical substances to another. During these reactions, bonds are broken and formed, which involves energy changes. The enthalpy change calculation in a chemical reaction is an expression of these energy changes. It's crucial to write down the chemical equations accurately, reflecting the correct stoichiometry, and state symbols (e.g., \(s\) for solid, \(l\) for liquid, \(g\) for gas, \(aq\) for aqueous solutions), as these details are significant for enthalpy calculations.

In the given exercise, we are asked to find the \( \Delta H \) for the oxidation of manganese. Writing down the given reactions helps us visualize the changes that occur and lays out a clear path for using these reactions to calculate the unknown enthalpy change for a new reaction.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass, which dictates that matter is neither created nor destroyed in a chemical reaction. Thus, the quantity of each element must remain the same before and after the reaction. Stoichiometry applies not only to the number of atoms but also to the energy changes in a reaction.

In our example, we manipulated the stoichiometry of the given reactions to mathematically isolate the desired product \(MnO_{2}\). This involved dividing the entirety of the challenge reaction by 3 and the equation \(a\) by 2. Stoichiometric coefficients play a crucial role in calculating enthalpy changes, emphasizing the need for students to understand this concept thoroughly.

Hess's Law

Hess's Law is a principle in chemistry that states the total enthalpy change during a chemical reaction is the same, regardless of the number of steps the reaction is carried out in. This principle allows for the calculation of enthalpy changes for reactions that are difficult to measure directly.

In the context of the exercise, we used Hess's Law to find the desired \( \Delta H \) for equation \(b\) by combining manipulated versions of the given reactions in such a way that unwanted substances cancel out. By applying Hess's Law, we can add the enthalpy changes of the modified reactions to obtain the enthalpy change for the target reaction. The law is immensely useful for indirectly calculating the enthalpy changes of complex chemical reactions. Using this exercise as an example, we illustrated how Hess's Law can aid in solving thermochemical equations.