Problem 45 - Chemistry: The Central Science

Question

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36\) \(\mathrm{mol}\) of \(\mathrm{O}_{2}\) and (b) \(10.4 \mathrm{~g}\) 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.

Step-by-Step Solution

Answer

The enthalpy change for forming (a) 1.36 mol of O₂ is -44.7 kJ and (b) 10.4 g of KCl is -6.23 kJ. (c) The reverse reaction, the formation of KClO₃ from KCl and O₂, is not likely to be feasible under ordinary conditions, because it is endothermic and requires energy input to proceed.

Step 1: Write down the given information

We have a balanced chemical equation for the reaction $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ The standard enthalpy change, \(\Delta H = -89.4\) kJ for this reaction. We should first calculate the enthalpy change when 1.36 mol of O2 is formed.

Step 2: Use stoichiometry to find moles of KClO3 reacting

From the balanced chemical equation, we see that 3 mol of O2 is produced from 2 mol of KClO3. To find out the amount of KClO3 needed to produce 1.36 mol of O2, we can set up a proportion: $$ \frac{3 \,\text{mol O}_2}{2\, \text{mol KClO}_3} = \frac{1.36\, \text{mol O}_2}{x \,\text{mol KClO}_3} $$ Solve for x to get the moles of KClO3: $$ x = \frac{1.36 \,\text{mol O}_2 \times 2 \,\text{mol KClO}_3}{3 \,\text{mol O}_2} = 0.9067\, \text{mol KClO}_3 $$

Step 3: Calculate ∆H for the given amount of O2

Now that we know how many moles of KClO3 are needed to produce 1.36 mol of O2, we can determine the enthalpy change for this amount using the given standard enthalpy change, ∆H, for the reaction. $$ \Delta H_{1.36\,\text{mol O}_2} = \frac{-89.4 \,\text{kJ}}{2\rightarrow 0.9067 \,\text{mol KClO}_3} = -44.7 \,\text{kJ} $$ Therefore, the enthalpy change for forming 1.36 mol of O2 is -44.7 kJ. #b - Calculate ∆H for forming 10.4 g of KCl#

Step 4: Calculate the moles of KCl produced

To find the enthalpy change for forming 10.4 g of KCl, we must first convert grams of KCl to moles: $$ \text{Moles of KCl} = \frac{10.4\, \text{g}}{74.55\, \text{g/mol}} = 0.1395\, \text{mol KCl} $$

Step 5: Use stoichiometry to find moles of KClO3 reacting

From the balanced chemical equation, we see that 2 mol of KCl is produced from 2 mol of KClO3. To find out the amount of KClO3 needed to produce 0.1395 mol of KCl, we can set up a proportion: $$ \frac{2 \,\text{mol KCl}}{2\, \text{mol KClO}_3} = \frac{0.1395\, \text{mol KCl}}{x \,\text{mol KClO}_3} $$ Solve for x to get the moles of KClO3: $$ x = \frac{0.1395 \,\text{mol KCl}}{1} = 0.1395\, \text{mol KClO}_3 $$

Step 6: Calculate ∆H for the given amount of KCl

Now that we know how many moles of KClO3 are needed to produce 0.1395 mol of KCl, we can determine the enthalpy change for this amount using the given standard enthalpy change ∆H for the reaction: $$ \Delta H_{10.4\,\text{g KCl}} = \frac{-89.4 \,\text{kJ}}{2\,\text{mol KClO}_3\rightarrow 0.1395 \,\text{mol KClO}_3} = -6.23 \,\text{kJ} $$ Therefore, the enthalpy change for forming 10.4 g of KCl is -6.23 kJ. #c - Analyze the feasibility of the reverse reaction#

Step 7: Comment on the feasibility of the reverse reaction

The reaction has a negative enthalpy change, meaning that it is exothermic. This indicates that energy is released when the KClO3 decomposes, making the forward reaction spontaneous. The reverse reaction, the formation of KClO3 from KCl and O2, would have a positive enthalpy change (since the enthalpy change would have the opposite sign). A positive enthalpy change means the reaction is endothermic and requires energy input to proceed. Therefore, the reverse reaction is not likely to be feasible under ordinary conditions, as it would require an external energy source to drive the reaction.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. When analyzing a chemical equation, stoichiometry helps us understand the proportional relationships.
For example, in the reaction:

• 2 \(\text{KClO}_3\)(s) \(\rightarrow\) 2 \(\text{KCl}\)(s) + 3 \(\text{O}_2\)(g)

This balanced equation tells us that 2 moles of potassium chlorate (\(\text{KClO}_3\)) decomposes to form 2 moles of potassium chloride (\(\text{KCl}\)) and 3 moles of oxygen gas (\(\text{O}_2\)).
The ratio given by stoichiometry is utilized to predict how much of one substance will react with a set quantity of another, or how much product will be produced. For instance:

• From the equation, 3 moles of \(\text{O}_2\) are produced by 2 moles of \(\text{KClO}_3\). This is crucial when using stoichiometry for calculations, such as determining the amount of \(\text{KClO}_3\) needed to produce a certain quantity of \(\text{O}_2\).

Overall, stoichiometry acts like a recipe in chemistry, enabling chemists to scale up reactions and predict outcomes with accurate, quantified information.

Endothermic Reaction

Endothermic reactions are processes that absorb energy from their surroundings. They require an input of energy to proceed. This energy absorption usually occurs in the form of heat, making the surroundings feel colder.
In the context of the reverse reaction given in the exercise — forming \(\text{KClO}_3\) from \(\text{KCl}\) and \(\text{O}_2\) — would involve an endothermic process since energy is required to make the reaction occur. This energy need signifies a positive enthalpy change (\(\Delta H > 0\)).
Some characteristics of endothermic reactions include:

• They require a continuous energy supply to maintain the reaction.
• Common examples include photosynthesis in plants and the melting of ice.

In chemical terms, an endothermic reaction on a molecular level means that the bonds being formed in the products are weaker than the bonds being broken in the reactants. Thus, more energy is taken in than is released, necessitating external energy to occur.

Exothermic Reaction

Exothermic reactions release energy into their surrounding environment, generally as heat. This release makes the surroundings feel warmer as heat dissipates.
The decomposition of \(\text{KClO}_3\) into \(\text{KCl}\) and \(\text{O}_2\) in the provided exercise is an exothermic reaction. It is identified by its negative enthalpy change (\(\Delta H = -89.4 \, \text{kJ}\)).
Features of exothermic reactions include:

• They release energy spontaneously which sometimes can be used to do work, such as power a heat engine or provide warmth.
• Examples include combustion reactions, such as burning wood or gasoline.
• Exothermic reactions often require an initial energy input to start, called activation energy, but then proceed by releasing more energy than what was consumed.

In summary, exothermic reactions are typically associated with processes that occur naturally and spontaneously due to their ability to lower the energy of the system by releasing energy.