THERMODYNAMICS: Given the following hypothetical thermochemical equations: 20J3 (0) + 302X (e) - 402 (9) + 3J2X (1) AH = -1.017.3 kJ 40J3 (e) + 3X2 (0) – 202 (a) + OJ2X () AH = -1,548.8 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (0) + % X2 (a) – Q2X (@) (Round off the final answer to ONE decimal place. Do not include unit.)

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THERMODYNAMICS: Given the following hypothetical thermochemical equations:
20J3 (g) + 302x (g) – 402 (g) + 3J2X ()
AH = -1,017.3 kJ
4QJ3 (e) + 3X2 (@) – 202 (g) + 6J2X (1)
AH = -1,548.8 kJ
Calculate the value of AH (in kJ) for the reaction: a2 (g) + % X2 (a) - Q2X (g)
(Round off the final answer to ONE decimal place. Do not include unit.)
Transcribed Image Text:THERMODYNAMICS: Given the following hypothetical thermochemical equations: 20J3 (g) + 302x (g) – 402 (g) + 3J2X () AH = -1,017.3 kJ 4QJ3 (e) + 3X2 (@) – 202 (g) + 6J2X (1) AH = -1,548.8 kJ Calculate the value of AH (in kJ) for the reaction: a2 (g) + % X2 (a) - Q2X (g) (Round off the final answer to ONE decimal place. Do not include unit.)
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