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(1) 2 H2 (g) + 02(g) → 2 H2O(g) AH = -484 kJ (2) N2 (g) + 3 H2(g) → 2 NH3 (g) ДН — —92 kJ (3) H20(1) → H20(g) AH +44 kJ (4) 2 N2(g) + 6 H20(1) → 4 NH3 (g) + 3 02(g) AH° = kJ

(1) 2 H2 (g) + 02(g) → 2 H2O(g) AH = -484 kJ (2) N2 (g) + 3 H2(g) → 2 NH3 (g) ДН — —92 kJ (3) H20(1) → H20(g) AH +44 kJ (4) 2 N2(g) + 6 H20(1) → 4 NH3 (g) + 3 02(g) AH° = kJ

Chemistry for Engineering Students
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter9: Energy And Chemistry
Section: Chapter Questions
Problem 9.41PAE: 9.41 Under what conditions does the enthalpy change equal the heat of a process?
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(1) 2 H2(g) + 02(g) → 2 H2O(g) AH -484 kJ (2) N2 (g) + 3 H2(g) → 2 NH3 (g) ΔΗ -92 kJ %3D (3) H20(1) → H2O(g) AH = +44 kJ (4) 2 N2(g) + 6 H20(1) → 4 NH3(g) + 3 02 (g) AH : kJ
Transcribed Image Text:(1) 2 H2(g) + 02(g) → 2 H2O(g) AH -484 kJ (2) N2 (g) + 3 H2(g) → 2 NH3 (g) ΔΗ -92 kJ %3D (3) H20(1) → H2O(g) AH = +44 kJ (4) 2 N2(g) + 6 H20(1) → 4 NH3(g) + 3 02 (g) AH : kJ
Given the thermochemical equations for reactions (1),(2) and (3) below, calculate the enthalpy change for reaction (4).
Transcribed Image Text:Given the thermochemical equations for reactions (1),(2) and (3) below, calculate the enthalpy change for reaction (4).
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