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For the equilibrium 21Br(g) = 1₂(g) + Br₂(g), Kc = 8.50 × 10 at 150°C. What is the equilibrium constant for the reverse reaction?

For the equilibrium 21Br(g) = 1₂(g) + Br₂(g), Kc = 8.50 × 10 at 150°C. What is the equilibrium constant for the reverse reaction?

Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter15: Principles Of Chemical Reactivity: Equilibria
Section15.4: Using Equilibrium Constants In Calculations
Problem 15.5CYU: The decomposition of PCl5(g) to form PCl3(g) and Cl2(g) has Kc = 33.3 at a high temperature. If the...
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For the equilibrium 21Br(g) = 1₂(g) + Br₂(g), Kc = 8.50 × 10 at 150°C. What is the equilibrium constant for the reverse reaction?
Transcribed Image Text:For the equilibrium 21Br(g) = 1₂(g) + Br₂(g), Kc = 8.50 × 10 at 150°C. What is the equilibrium constant for the reverse reaction?
For the reaction: 2 H₂S (g) + 3 02₂ (g) = 2 H₂O (g) + 2 SO₂ (g) K 7.71 x 10-9 If a reaction is done in a 0.500 L flask with 0.0609 moles H₂S, 0.152 moles O2, what will be the equilibrium concentration of SO₂?
Transcribed Image Text:For the reaction: 2 H₂S (g) + 3 02₂ (g) = 2 H₂O (g) + 2 SO₂ (g) K 7.71 x 10-9 If a reaction is done in a 0.500 L flask with 0.0609 moles H₂S, 0.152 moles O2, what will be the equilibrium concentration of SO₂?
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