Part 1:(a) Calculate K at 298 K for the following reaction:2C(graphite) + O₂(g) = 2CO(g)K = 1.25C(graphite)CO(g)0₂ (8)Part 2 out of 2AGⓇx 10||48(b) Use the equilibrium constant to calculate AG (in kJ) at 298 K for the following reaction:2HCl(g) + Br₂(1) ⇒ 2HBr(g) + Cl₂(g)K= 2.22 × 10-15 at 298 KAG KJkJCmolx 100-137.20kJ

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Part 1: (a) Calculate K at 298 K for the following reaction: 2C(graphite) + O₂(g) = 2CO(g) K = 1.25 C(graphite) CO(g) 0₂(g) Part 2 out of 2 AGⓇ x 10 || 48 kJ fmol (b) Use the equilibrium constant to calculate AGº (in kJ) at 298 K for the following reaction: 2HCl(g) + Br₂(1)→ 2HBr(g) + Cl₂(g) K = 2.22 × 10-¹5 at 298 K x 10 AG 0 -137.2 0 kJ

Part 1: (a) Calculate K at 298 K for the following reaction: 2C(graphite) + O₂(g) = 2CO(g) K = 1.25 C(graphite) CO(g) 0₂(g) Part 2 out of 2 AGⓇ x 10 || 48 kJ fmol (b) Use the equilibrium constant to calculate AGº (in kJ) at 298 K for the following reaction: 2HCl(g) + Br₂(1)→ 2HBr(g) + Cl₂(g) K = 2.22 × 10-¹5 at 298 K x 10 AG 0 -137.2 0 kJ

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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