THERMODYNAMICS: Given the following hypothetical thermochemical equations 2QJ3 (g) + 3Q2X (g) → 4Q2 (g) + 3J2X (1) AH = -1,070 kJ 4QJ3 (g) + 3X2 (g) –→ 202 (g) + 6J2X (1) AH = -1,580.2 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + ½ X2 (g) → Q2X (g (Round off the final answer to ONE decimal place. Do not include unit.)

THERMODYNAMICS: Given the following hypothetical thermochemical equations 2QJ3 (g) + 3Q2X (g) → 4Q2 (g) + 3J2X (1) AH = -1,070 kJ 4QJ3 (g) + 3X2 (g) –→ 202 (g) + 6J2X (1) AH = -1,580.2 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + ½ X2 (g) → Q2X (g (Round off the final answer to ONE decimal place. Do not include unit.)

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

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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Given the following thermochemical equations, 2Cu (s) + S (s) · → Cu₂S (s) AH° = -79.5 kJ S (s) + O₂ (g) → SO₂ (g) AH° = −297 kJ Cu₂ S (s) +20₂ (g) → 2CuO (s) + SO₂ (g) AH° = -527.5 kJ calculate the standard enthalpy of formation (in kJ mole¹¹) of CuO (s).
THERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) → 4Q2 (g) + 3J2X (I) AH = -1,096.4 kJ 4QJ3 (g) + 3X2 (g) → 2Q2 (g) + 6J2X (1) AH = -1,548.8 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + ½ X2 (g) → Q2X (g)
THERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) → 4Q2 (g) + 3J2X (I) AH = -1,089.5 kJ 4QJ3 (g) + 3X2 (g) → 2Q2 (g) + 6J2X (I) AH = -1,502 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + ½ X2 (g) → Q2X (g) (Round off the final answer to ONE decimal place. Do not include unit.)
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THERMODYNAMICS: Calculate the value of AH (in kJ) for the reaction, 2AR3 (g) + Q (s) → A2R4 (g) + QR2 (s) given the following hypothetical thermochemical equations: AH = -84 kJ AH = -151.3 kJ AH = -537.5 kJ 2AR3 (g) + 2AX (g) → A2R4 (g) + 2XAR (g) AX (g) + ½R2 (g) → XAR (g) Q (s) + R2 (g) → QR2 (s) (Round off final answer to ONE decimal place) Round your answer to 1 decimal place.
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