Consider the combustion of isopropanol, C 3 H 7 OH(l), which is the primary component of rubbing alcohol: C 3 H 7 OH(l) + 9/2 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(l) ΔH = -2248 kJ a. What is the enthalpy change for the reverse reaction?

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 47a

Chapter 5, Problem 47a

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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Identify the given reaction: C_3H_7OH(l) + \frac{9}{2} O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(l) with \Delta H = -2248 \text{ kJ}.

Understand that the enthalpy change (\Delta H) for a reaction is the energy change when the reaction occurs as written.

For the reverse reaction, the products and reactants are swapped: 3 CO_2(g) + 4 H_2O(l) \rightarrow C_3H_7OH(l) + \frac{9}{2} O_2(g).

The enthalpy change for the reverse reaction is the negative of the enthalpy change for the forward reaction.

Calculate the enthalpy change for the reverse reaction by changing the sign of the given \Delta H: \Delta H_{reverse} = -(-2248 \text{ kJ}).

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Key Concepts

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

Enthalpy Change

Enthalpy change (ΔH) is a measure of the heat absorbed or released during a chemical reaction at constant pressure. In exothermic reactions, like the combustion of isopropanol, ΔH is negative, indicating that heat is released. Understanding enthalpy is crucial for predicting the energy changes associated with chemical reactions, including both forward and reverse processes.

Combustion Reactions

Combustion reactions involve the reaction of a substance with oxygen, producing heat and light. In the case of isopropanol, the combustion results in carbon dioxide and water as products. Recognizing the characteristics of combustion reactions helps in understanding the energy dynamics and the stoichiometry involved in the reaction, which is essential for calculating enthalpy changes.

Reverse Reaction

The reverse reaction of a chemical process is simply the reaction proceeding in the opposite direction. For the combustion of isopropanol, the reverse reaction would involve the formation of isopropanol from carbon dioxide and water. The enthalpy change for the reverse reaction is equal in magnitude but opposite in sign to that of the forward reaction, which is a fundamental principle in thermodynamics.

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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

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 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?

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?

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?

Textbook Question

Consider the decomposition of liquid benzene, C₆H₆(l), to gaseous acetylene, C₂H₂(g): C₆H₆(l) → 3 C2H₂(g) ΔH = +630 kJ (b) What is H for the formation of 1 mol of acetylene?

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?

Verified video answer for a similar problem:

Key Concepts

Enthalpy Change

Combustion Reactions

Reverse Reaction

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?