At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO 3 : 2 KClO 3 (s) → 2 KCl(s) + 3 O 2 (g) ΔH = -89.4 kJ c. Now consider the reverse reaction, in which KClO 3 is formed from KCl and O 2 . What is Δ𝐻 for the formation of 19.1 g KClO 3 from KCl and O 2 ?

Hey everyone. So here it says, considered the dissolution of calcium chloride determine the entropy that change in entropy for the reverse reaction. Alright, so here we have calcium chloride solid breaking down to give us calcium ion plus two chloride ions. Here, the entropy of reaction for this is negative 81.5 kg per mole. Here we have to do the reverse of it. So now our products will become our reactant. So calcium ion plus two chloride ions well combined together to give me my one mole of calcium chloride solid. Now remember that when you reverse the reaction, you also are going to have to reverse the sign of Delta H. Here it started off as negative 81.5 Na becomes positive 81.5 and it'll still be killer joules per mole. So this would be my final answer for the reverse reaction for the dissolution of calcium chloride.

Ch.5 - Thermochemistry

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Brown 15th Edition

Ch.5 - Thermochemistry

Problem 46c

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Chapter 5, Problem 46c

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ c. Now consider the reverse reaction, in which KClO₃ is formed from KCl and O₂. What is Δ𝐻 for the formation of 19.1 g KClO₃ from KCl and O₂?

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Identify the given reaction: 2 KClO_3(s) \rightarrow 2 KCl(s) + 3 O_2(g) with \Delta H = -89.4 \text{ kJ}.

Recognize that the reverse reaction is being considered: 2 KCl(s) + 3 O_2(g) \rightarrow 2 KClO_3(s).

For the reverse reaction, the enthalpy change (\Delta H) will be the opposite sign of the forward reaction, so \Delta H = +89.4 \text{ kJ} for the formation of 2 moles of KClO_3.

Calculate the molar mass of KClO_3: K (39.10 g/mol) + Cl (35.45 g/mol) + 3 \times O (16.00 g/mol) = 122.55 g/mol.

Determine the \Delta H for the formation of 19.1 g of KClO_3 by using the ratio: \Delta H = \left(\frac{19.1 \text{ g}}{2 \times 122.55 \text{ g/mol}}\right) \times 89.4 \text{ kJ}.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Thermochemistry

Thermochemistry is the study of the heat energy associated with chemical reactions and changes of state. It involves understanding how energy is absorbed or released during reactions, which is quantified as enthalpy change (ΔH). In this context, the ΔH value indicates whether a reaction is exothermic (releases heat) or endothermic (absorbs heat), which is crucial for calculating the energy changes in the formation of KClO3.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows chemists to calculate the amounts of substances consumed and produced in a reaction based on balanced chemical equations. In this question, stoichiometry is essential for determining how much KClO3 can be formed from a given mass of KCl and O2, using the coefficients from the balanced equation.

Reverse Reactions and Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the number of steps taken to complete the reaction. This principle is particularly useful when considering reverse reactions, as the ΔH for the reverse process is equal in magnitude but opposite in sign to that of the forward reaction. In this case, calculating the ΔH for the formation of KClO3 from KCl and O2 requires applying Hess's Law to the given reaction.

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Hess's Law

Related Practice

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates Ag⁺(aq) + Cl^-(aq) → AgCl(s) H = -65.5 kJ (a) Calculate H for the production of 0.450 mol of AgCl by this reaction. (b) Calculate H for the production of 9.00 g of AgCl. (c) Calculate H when 9.25⨉10^-4 mol of AgCl dissolves in water.

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Textbook Question

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (a) 1.36 mol of O₂

Consider the combustion of isopropanol, C₃H₇OH(l), which is the primary component of rubbing alcohol: C₃H₇OH(l) + 9/2 O₂(g) → 3 CO₂(g) + 4 H₂O(l) ΔH = -2248 kJ a. What is the enthalpy change for the reverse reaction?

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Textbook Question

Consider the combustion of isopropanol, C₃H₇OH(l), which is the primary component of rubbing alcohol: C₃H₇OH(l) + 9/2 O₂(g) → 3 CO₂(g) + 4 H₂O(l) ΔH = -2248 kJ (b) Balance the forward reaction with whole-number coefficients. What is ΔH for the reaction represented by this equation?

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Textbook Question

Consider the decomposition of liquid benzene, C₆H₆(l), to gaseous acetylene, C₂H₂(g): C₆H₆(l) → 3 C2H₂(g) ΔH = +630 kJ (a) What is the enthalpy change for the reverse reaction?

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