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
Chemistry
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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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![**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{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbccee3c-a9d1-4424-96e1-9881a48faaa2%2F60ec16b6-7c0f-4da7-94c1-e38b8ad0f9a2%2Fyd257k_processed.png&w=3840&q=75)
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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