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Given the thermochemical equations X₂ + 3 Y₂ → 2XY3 X₂ +2Z₂ → 2XZ₂ 2Y₂+Z₂2Y₂Z AH3= -270 kJ Calculate the change in enthalpy for the reaction. 4XY3 +7Z₂6Y₂Z + 4XZ₂ AH = AH₁ = - 320 kJ AH₂ = -140 kJ kJ

Given the thermochemical equations X₂ + 3 Y₂ → 2XY3 X₂ +2Z₂ → 2XZ₂ 2Y₂+Z₂2Y₂Z AH3= -270 kJ Calculate the change in enthalpy for the reaction. 4XY3 +7Z₂6Y₂Z + 4XZ₂ AH = AH₁ = - 320 kJ AH₂ = -140 kJ kJ

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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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O Macmillan Learning Given the thermochemical equations X₂ + 3 Y₂ → 2XY3 X₂ +2Z₂ → 2XZ₂ -> AH₁ = -320 kJ ΔΗ, = -140 kJ 2Y₂+Z₂2Y₂Z AH3 = -270 kJ Calculate the change in enthalpy for the reaction. 4XY3 +7Z2₂6Y₂Z + 4 XZ₂ AH = kJ
Transcribed Image Text:O Macmillan Learning Given the thermochemical equations X₂ + 3 Y₂ → 2XY3 X₂ +2Z₂ → 2XZ₂ -> AH₁ = -320 kJ ΔΗ, = -140 kJ 2Y₂+Z₂2Y₂Z AH3 = -270 kJ Calculate the change in enthalpy for the reaction. 4XY3 +7Z2₂6Y₂Z + 4 XZ₂ AH = kJ
Expert Solution
Step 1: Lis the Given Data

The thermochemical equations are:

straight X subscript 2 plus 3 straight Y subscript 2 rightwards arrow 2 XY subscript 3 space space space space space space increment straight H subscript 1 equals negative 320 space kJ straight X subscript 2 plus 2 straight Z subscript 2 rightwards arrow 2 XZ subscript 2 space space space space space space increment straight H subscript 2 equals negative 140 space kJ 2 straight Y subscript 2 plus straight Z subscript 2 rightwards arrow 2 straight Y subscript 2 straight Z space space space space space space increment straight H subscript 3 equals negative 270 space kJ

The target equation is:

4 XY subscript 3 plus 7 straight Z subscript 2 rightwards arrow 6 straight Y subscript 2 straight Z plus 4 XZ subscript 2

To find, the value of increment straight H.

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