Question

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ}\) For this reaction, calculate \(\Delta H\) for the formation of (a) \(0.632 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) and (b) \(8.57 \mathrm{~g}\) of \(\mathrm{KCl}\). (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\), is likely to be feasible under ordinary conditions? Explain your answer.

Short answer

(a) For the formation of 0.632 mol of O2, the enthalpy change (ΔH) is -29.8 kJ. (b) For the formation of 8.57 g of KCl, the enthalpy change (ΔH) is -10.3 kJ. (c) The reverse reaction (formation of KClO3 from KCl and O2) is unlikely to be feasible under ordinary conditions, as it would be endothermic and non-spontaneous.

Step-by-step solution

(a) Calculate ΔH for the formation of 0.632 mol of O2

First, we will use the stoichiometry of the balanced chemical equation to determine the amount of KClO3 that would be needed to form 0.632 mol of O2. The balanced equation states that 2 mol of KClO3 decompose to form 3 mol of O2. Therefore, we can set up a proportion: \(\frac{2 \, \text{mol KClO}_{3}}{3 \, \text{mol O}_{2}} = \frac{\text{x mol KClO}_{3}}{0.632 \, \text{mol O}_{2}}\) To solve for x, we can cross-multiply and divide: \(\text{x mol KClO}_{3}=\frac{(2 \, \text{mol KClO}_{3})(0.632 \, \text{mol O}_{2})}{3 \, \text{mol O}_{2}}\) Now, let's plug in the given ΔH value (-89.4 kJ) for the decomposition of 2 mol of KClO3 to determine the enthalpy change for the formation of 0.632 mol of O2. \(\Delta H_{\text{formation}} = \frac{-89.4 \, \text{kJ}}{2 \, \text{mol KClO}_{3}} \times \text{x mol KClO}_{3}\) Now we can plug in the value of x we found earlier and calculate the enthalpy change.

(b) Calculate ΔH for the formation of 8.57 g of KCl

First, we need to convert the mass of KCl to moles using its molar mass (74.55 g/mol): \(\text{moles of KCl} = \frac{8.57 \, \text{g}}{74.55 \, \text{g/mol}}\) Next, let's use the stoichiometry of the balanced chemical equation to determine the amount of KClO3 that would be needed to produce this amount of KCl. The balanced equation states that 2 mol of KClO3 produce 2 mol of KCl. Therefore, we can set up a proportion: \(\frac{2 \, \text{mol KClO}_{3}}{2 \, \text{mol KCl}} = \frac{\text{y mol KClO}_{3}}{\text{moles of KCl}}\) To solve for y, we can cross-multiply and divide: \(\text{y mol KClO}_{3}=\frac{(2 \, \text{mol KClO}_{3})(\text{moles of KCl})}{2 \, \text{mol KCl}}\) Now, let's plug in the given ΔH value (-89.4 kJ) for the decomposition of 2 mol of KClO3 to determine the enthalpy change for the formation of 8.57 g of KCl in the same way as we did for the formation of O2: \(\Delta H_{\text{formation}} = \frac{-89.4 \, \text{kJ}}{2 \, \text{mol KClO}_{3}} \times \text{y mol KClO}_{3}\) Now we can plug in the value of y we found earlier and calculate the enthalpy change.

(c) Feasibility of the reverse reaction

The reverse reaction involves the formation of KClO3 from KCl and O2. The forward reaction (decomposition of KClO3) is spontaneous when heated and has a negative enthalpy change, indicating that it is an exothermic reaction. For the reverse reaction to be feasible under ordinary conditions, it would need to be thermodynamically favored, meaning it should have a negative Gibbs free energy change (ΔG). However, since the forward reaction is exothermic (negative ΔH) and spontaneous at high temperatures, the reverse reaction is likely to be endothermic (positive ΔH) and non-spontaneous at normal conditions. Therefore, the reverse reaction is unlikely to be feasible under ordinary conditions.

Key concepts

Enthalpy Change

In thermochemistry, enthalpy change ( Delta H) measures the heat absorbed or released during a chemical reaction. For the given reaction, which decomposes potassium chlorate ( KClO_3), the enthalpy change is negative, at -89.4 kJ. This indicates that the reaction is exothermic, meaning it releases heat to the surroundings.
An enthalpy change of this nature suggests a few things:

• The reactants have higher energy than the products, leading to energy release.
• The reaction is generally more stable if it tends towards products with lower energy.

In this specific problem, you are asked to calculate the enthalpy change for different amounts of product, using stoichiometric conversions based on the provided balanced chemical equation.
The balanced equation helps determine how much heat is involved for a given amount of product, by establishing a ratio between reactants and products. So, whether you are calculating for 0.632 mol of O_2 or for grams of KCl, understanding enthalpy change is crucial in anticipating how energy dynamics work in this reaction.

Stoichiometry

Stoichiometry is a fundamental tool in chemistry that allows us to calculate the amounts of reactants and products involved in a chemical reaction using the balanced chemical equation.
In the given reaction where KClO_3 decomposes, stoichiometry enables us to set up ratios that relate to the balanced equation: 2 KClO_3 decomposing yields 2 KCl and 3 O_2.
Through stoichiometry, you can answer questions like:

• How much KClO_3 is needed to produce a certain amount of O_2?
• How can we convert grams of KCl into the corresponding moles using its molar mass?

Using these relationships and given amounts, such as 0.632 mol of O_2 or 8.57 g of KCl, you can determine how much KClO_3 reacted. By setting up proportions, you can convert between grams and moles and find the connection between reactants and products efficiently.

Exothermic Reactions

Exothermic reactions are chemical processes that release energy in the form of heat. In these reactions, the products are at a lower energy level than the reactants, resulting in a net release of energy. That's why the enthalpy change ( Delta H) is negative.
In the context of KClO_3 decomposition, the reaction is notably exothermic, as seen by the Delta H value of -89.4 kJ. This is significant because:

• It influences the feasibility of the reaction, making it spontaneous under heating.
• Products formed (like KCl and O_2) are in a more stable, low-energy configuration compared to the reactants.

When considering the reverse reaction, forming KClO_3 from KCl and O_2, we suspect it would be endothermic due to the need to input energy to convert low-energy products back to high-energy reactants, making it non-spontaneous under typical conditions.