Bromine is one of the few elements that is a liquid at room temperature. It vaporizes easily and has a beautiful red color in both the liquid and gaseous states. Bromine reacts readily with hydrogen gas to make hydrogen bromide. One way to represent this equilibrium is: 2 HBr(g)H2(g) + Br2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) HBr(g) 1/2 H2(g) + 1/2 Br2(g) K1 = 2) 1/2 H2(g) + 1/2 Br2(g) HBr(g) K2 = 3) H2(g) + Br2(g) 2HBr(g) K3 = Drag and drop your selection from the following list to complete the answer: K1/2(1/K)1/21/K
Bromine is one of the few elements that is a liquid at room temperature. It vaporizes easily and has a beautiful red color in both the liquid and gaseous states. Bromine reacts readily with hydrogen gas to make hydrogen bromide. One way to represent this equilibrium is: 2 HBr(g)H2(g) + Br2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) HBr(g) 1/2 H2(g) + 1/2 Br2(g) K1 = 2) 1/2 H2(g) + 1/2 Br2(g) HBr(g) K2 = 3) H2(g) + Br2(g) 2HBr(g) K3 = Drag and drop your selection from the following list to complete the answer: K1/2(1/K)1/21/K
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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| Bromine is one of the few elements that is a liquid at room temperature. It vaporizes easily and has a beautiful red color in both the liquid and gaseous states. Bromine reacts readily with hydrogen gas to make hydrogen bromide. One way to represent this equilibrium is: 2 HBr(g)H2(g) + Br2(g) |
We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.
| 1) | HBr(g) | 1/2 H2(g) + 1/2 Br2(g) | K1 = | |||
| 2) | 1/2 H2(g) + 1/2 Br2(g) | HBr(g) | K2 = | |||
| 3) | H2(g) + Br2(g) | 2HBr(g) | K3 = |
Drag and drop your selection from the following list to complete the answer:
K1/2(1/K)1/21/K
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