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**Chapter 5 Homework: Reaction Enthalpy Calculation** **Consider these reactions, where M represents a generic metal:** 1. \( 2 \text{M(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{MCl}_3\text{(aq)} + 3 \text{H}_2\text{(g)} \) - \( \Delta H_1 = -931.0 \text{ kJ} \) 2. \( \text{HCl(g)} \rightarrow \text{HCl(aq)} \) - \( \Delta H_2 = -74.8 \text{ kJ} \) 3. \( \text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2 \text{HCl(g)} \) - \( \Delta H_3 = -1845.0 \text{ kJ} \) 4. \( \text{MCl}_3\text{(s)} \rightarrow \text{MCl}_3\text{(aq)} \) - \( \Delta H_4 = -118.0 \text{ kJ} \) **Task:** Use the given information to determine the enthalpy of the reaction: \[ 2 \text{M(s)} + 3 \text{Cl}_2\text{(g)} \rightarrow 2 \text{MCl}_3\text{(s)} \] **Note:** There are no graphs or diagrams included in this problem. To solve the task, apply Hess's Law to manipulate and combine the provided reactions and their enthalpies.

**Chapter 5 Homework: Reaction Enthalpy Calculation** **Consider these reactions, where M represents a generic metal:** 1. \( 2 \text{M(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{MCl}_3\text{(aq)} + 3 \text{H}_2\text{(g)} \) - \( \Delta H_1 = -931.0 \text{ kJ} \) 2. \( \text{HCl(g)} \rightarrow \text{HCl(aq)} \) - \( \Delta H_2 = -74.8 \text{ kJ} \) 3. \( \text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2 \text{HCl(g)} \) - \( \Delta H_3 = -1845.0 \text{ kJ} \) 4. \( \text{MCl}_3\text{(s)} \rightarrow \text{MCl}_3\text{(aq)} \) - \( \Delta H_4 = -118.0 \text{ kJ} \) **Task:** Use the given information to determine the enthalpy of the reaction: \[ 2 \text{M(s)} + 3 \text{Cl}_2\text{(g)} \rightarrow 2 \text{MCl}_3\text{(s)} \] **Note:** There are no graphs or diagrams included in this problem. To solve the task, apply Hess's Law to manipulate and combine the provided reactions and their enthalpies.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Chapter 5 Homework: Reaction Enthalpy Calculation** **Consider these reactions, where M represents a generic metal:** 1. \( 2 \text{M(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{MCl}_3\text{(aq)} + 3 \text{H}_2\text{(g)} \) - \( \Delta H_1 = -931.0 \text{ kJ} \) 2. \( \text{HCl(g)} \rightarrow \text{HCl(aq)} \) - \( \Delta H_2 = -74.8 \text{ kJ} \) 3. \( \text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2 \text{HCl(g)} \) - \( \Delta H_3 = -1845.0 \text{ kJ} \) 4. \( \text{MCl}_3\text{(s)} \rightarrow \text{MCl}_3\text{(aq)} \) - \( \Delta H_4 = -118.0 \text{ kJ} \) **Task:** Use the given information to determine the enthalpy of the reaction: \[ 2 \text{M(s)} + 3 \text{Cl}_2\text{(g)} \rightarrow 2 \text{MCl}_3\text{(s)} \] **Note:** There are no graphs or diagrams included in this problem. To solve the task, apply Hess's Law to manipulate and combine the provided reactions and their enthalpies.
Transcribed Image Text:**Chapter 5 Homework: Reaction Enthalpy Calculation** **Consider these reactions, where M represents a generic metal:** 1. \( 2 \text{M(s)} + 6 \text{HCl(aq)} \rightarrow 2 \text{MCl}_3\text{(aq)} + 3 \text{H}_2\text{(g)} \) - \( \Delta H_1 = -931.0 \text{ kJ} \) 2. \( \text{HCl(g)} \rightarrow \text{HCl(aq)} \) - \( \Delta H_2 = -74.8 \text{ kJ} \) 3. \( \text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2 \text{HCl(g)} \) - \( \Delta H_3 = -1845.0 \text{ kJ} \) 4. \( \text{MCl}_3\text{(s)} \rightarrow \text{MCl}_3\text{(aq)} \) - \( \Delta H_4 = -118.0 \text{ kJ} \) **Task:** Use the given information to determine the enthalpy of the reaction: \[ 2 \text{M(s)} + 3 \text{Cl}_2\text{(g)} \rightarrow 2 \text{MCl}_3\text{(s)} \] **Note:** There are no graphs or diagrams included in this problem. To solve the task, apply Hess's Law to manipulate and combine the provided reactions and their enthalpies.
Expert Solution
Step 1

Given:

The reactions are as follows,

2M(s) + 6HCl(aq) → 2MCl3(aq) + 3H2(g)     ∆H1 = -931.0 kJHCl(g) → HCl(aq)                                                 ∆H2 = -74.8 kJH2(g) + Cl2(g) → 2HCl(g)                                   ∆H3 = -1845.0 kJMCl3(s) → MCl3(aq)                                              ∆H4 = -118.0 kJ

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