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Given the thermochemical equations X₂ + 3 Y₂ → 2XY3 X₂ +2Z₂ → 2XZ₂ AH₁ = -380 kJ AH₂ = -120 kJ 2Y₂+Z₂2Y₂Z AH3 = -270 kJ Calculate the change in enthalpy for the reaction. 4XY3 +7Z₂ →→ 6Y₂Z + 4 XZ₂

Given the thermochemical equations X₂ + 3 Y₂ → 2XY3 X₂ +2Z₂ → 2XZ₂ AH₁ = -380 kJ AH₂ = -120 kJ 2Y₂+Z₂2Y₂Z AH3 = -270 kJ Calculate the change in enthalpy for the reaction. 4XY3 +7Z₂ →→ 6Y₂Z + 4 XZ₂

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter6: Thermochemisty
Section: Chapter Questions
Problem 6.109QP: A 21.3-mL sample of 0.977 M NaOH is mixed with 29.5 mL of 0.918 M HCl in a coffee-cup calorimeter...
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**Thermochemical Equation and Enthalpy Calculation** Given the following thermochemical equations: 1. \( X_2 + 3Y_2 \rightarrow 2XY_3 \) \( \Delta H_1 = -380 \text{ kJ} \) 2. \( X_2 + 2Z_2 \rightarrow 2XZ_2 \) \( \Delta H_2 = -120 \text{ kJ} \) 3. \( 2Y_2 + Z_2 \rightarrow 2Y_2Z \) \( \Delta H_3 = -270 \text{ kJ} \) **Objective:** Calculate the change in enthalpy for the reaction: \[ 4XY_3 + 7Z_2 \rightarrow 6Y_2Z + 4XZ_2 \] ### Explanation: #### Step-by-Step Approach: 1. **Identify the Target Reaction:** \[ 4XY_3 + 7Z_2 \rightarrow 6Y_2Z + 4XZ_2 \] 2. **Express the Target Reaction using Given Equations:** By manipulating the given equations, try to arrive at the target reaction. Consider the following transformations and additions: Multiply the first equation by 2: \[ 2(X_2 + 3Y_2 \rightarrow 2XY_3) \] Simplifies to: \[ 2X_2 + 6Y_2 \rightarrow 4XY_3 \] \( \Delta H_1' = 2 \times -380 \text{ kJ} = -760 \text{ kJ} \) Using the third equation as it is: \[ 2Y_2 + Z_2 \rightarrow 2Y_2Z \] \( \Delta H_3 = -270 \text{ kJ} \) To match the required coefficients: By multiplying the third equation by 3: \[ 3(2Y_2 + Z_2 \rightarrow 2Y_2Z) \] Simplifies to: \[ 6Y_2 + 3Z_2 \rightarrow 6Y_2Z \] \( \Delta H_3' = 3 \times -270 \text{ kJ} = -810 \text{
Transcribed Image Text:**Thermochemical Equation and Enthalpy Calculation** Given the following thermochemical equations: 1. \( X_2 + 3Y_2 \rightarrow 2XY_3 \) \( \Delta H_1 = -380 \text{ kJ} \) 2. \( X_2 + 2Z_2 \rightarrow 2XZ_2 \) \( \Delta H_2 = -120 \text{ kJ} \) 3. \( 2Y_2 + Z_2 \rightarrow 2Y_2Z \) \( \Delta H_3 = -270 \text{ kJ} \) **Objective:** Calculate the change in enthalpy for the reaction: \[ 4XY_3 + 7Z_2 \rightarrow 6Y_2Z + 4XZ_2 \] ### Explanation: #### Step-by-Step Approach: 1. **Identify the Target Reaction:** \[ 4XY_3 + 7Z_2 \rightarrow 6Y_2Z + 4XZ_2 \] 2. **Express the Target Reaction using Given Equations:** By manipulating the given equations, try to arrive at the target reaction. Consider the following transformations and additions: Multiply the first equation by 2: \[ 2(X_2 + 3Y_2 \rightarrow 2XY_3) \] Simplifies to: \[ 2X_2 + 6Y_2 \rightarrow 4XY_3 \] \( \Delta H_1' = 2 \times -380 \text{ kJ} = -760 \text{ kJ} \) Using the third equation as it is: \[ 2Y_2 + Z_2 \rightarrow 2Y_2Z \] \( \Delta H_3 = -270 \text{ kJ} \) To match the required coefficients: By multiplying the third equation by 3: \[ 3(2Y_2 + Z_2 \rightarrow 2Y_2Z) \] Simplifies to: \[ 6Y_2 + 3Z_2 \rightarrow 6Y_2Z \] \( \Delta H_3' = 3 \times -270 \text{ kJ} = -810 \text{
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