Question

Small amounts of oxygen gas can be prepared in the laboratory by decomposing potassium chlorate with heat. A by-product of the decomposition is potassium chloride. When one mole of potassium chlorate decomposes, \(44.7 \mathrm{~kJ}\) are evolved. (a) Write a balanced thermochemical equation for the decomposition of one mole of potassium chlorate. (b) Is the reaction exothermic? (c) Draw an energy diagram showing the path of this reaction. (Figure 8.6 is an example of such an energy diagram.) (d) What is \(\Delta H\) when \(3.00 \mathrm{~g}\) of potassium chlorate decompose? (e) How many grams of potassium chlorate need to be decomposed to liberate fifteen kilojoules of heat?

Short answer

Question: Write a balanced thermochemical equation for the decomposition of potassium chlorate and determine if the reaction is exothermic. Calculate the change in enthalpy when 3.00 g decompose and how many grams are needed to liberate 15 kJ of heat. Answer: The balanced thermochemical equation is: KClO3 (s) → KCl (s) + 3/2 O2 (g); ΔH = -44.7 kJ/mol The reaction is exothermic as ΔH is negative. The change in enthalpy when 3.00 g of potassium chlorate decompose is -1.095 kJ. To liberate 15 kJ of heat, 41.12 g of potassium chlorate needs to be decomposed.

Step-by-step solution

(a) Balanced thermochemical equation

First, we write the balanced equation for the decomposition of potassium chlorate (KClO3) into oxygen gas (O2) and potassium chloride (KCl): $$2\mathrm{KClO_{3}(s)} \rightarrow 2\mathrm{KCl(s)} + 3\mathrm{O_{2}(g)}$$ Now, let's include the enthalpy change for the decomposition of one mole of potassium chlorate: $$\mathrm{KClO_{3}(s)} \rightarrow \mathrm{KCl(s)} + \frac{3}{2}\mathrm{O_{2}(g)} \quad; \quad \Delta H = -44.7 \mathrm{~kJ/mol}$$

(b) Is the reaction exothermic?

Since the value of \(\Delta H\) is negative, the reaction is exothermic as it releases heat during the decomposition process.

(c) Draw an energy diagram

The energy diagram for this exothermic reaction will show a decrease in energy as the reactants form the products: Reactants → Transition State → Products Energy Level: \(\mathrm{KClO_{3}}\) → TS → \(\mathrm{KCl + \frac{3}{2}O_{2}}\) (-44.7 kJ/mol) 【Energy Diagram】 I cannot draw an energy diagram here, but you can refer to Figure 8.6 in your textbook for an example of an energy diagram.

(d) Calculate \(\Delta H\) when 3.00 g of potassium chlorate decompose

Let's first find out how many moles of KClO3 are in 3.00 g: Molar mass of KClO3 = 39.10 (K) + 35.45 (Cl) + 3 × 16.00 (O) = 122.55 g/mol Moles in 3.00 g of KClO3: \(\frac{3.00\, \mathrm{g}}{122.55\, \mathrm{g/mol}} = 0.0245\, \mathrm{mol}\) Then, we can find the change in enthalpy by multiplying the moles by the given \(\Delta H\): \(\Delta H = -44.7\, \mathrm{kJ/mol} \times 0.0245 \mathrm{mol} = -1.095\, \mathrm{kJ}\)

(e) How many grams of potassium chlorate need to be decomposed to liberate fifteen kilojoules of heat?

Since \(\Delta H = -44.7\, \mathrm{kJ/mol}\) for the decomposition of potassium chlorate, we will first find the moles needed to produce 15 kJ of heat: $$\mathrm{moles} = \frac{15\, \mathrm{kJ}}{-44.7\, \mathrm{kJ/mol}} = -0.3356\, \mathrm{mol}$$ Notice that we have a negative value for the moles, this doesn't make sense in the context. So, we will make this value positive: 0.3356 mol. Now, we will convert moles to grams using the molar mass of KClO3 (122.55 g/mol): $$\mathrm{grams} = 0.3356\, \mathrm{mol} \times 122.55\, \mathrm{g/mol} = 41.12\, \mathrm{g}$$ To liberate fifteen kilojoules of heat, 41.12 grams of potassium chlorate need to be decomposed.