Consider the reaction below: 2<(g) = 2 1(g) The Kc of this reaction is 3.820 × 10-5 at 791.0°C. Calculate Kp at th same temperature for the reverse reaction. 2 1(g) = l2(g). R = 0.08314 L·bar/mol·K.
Consider the reaction below: 2<(g) = 2 1(g) The Kc of this reaction is 3.820 × 10-5 at 791.0°C. Calculate Kp at th same temperature for the reverse reaction. 2 1(g) = l2(g). R = 0.08314 L·bar/mol·K.
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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
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![Consider the reaction below:
2<(g) = 2 1(g)
The Kc of this reaction is 3.820 × 10-5 at 791.0°C. Calculate Kp at the
same temperature for the reverse reaction.
2 I(g) = l2(g).
R = 0.08314 L·bar/mol·K.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcbe580c0-9ec2-448a-b9ab-e4b595776b9b%2F7a08ad99-4d8c-461d-aeb8-07225e588bdc%2Fee6f18_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the reaction below:
2<(g) = 2 1(g)
The Kc of this reaction is 3.820 × 10-5 at 791.0°C. Calculate Kp at the
same temperature for the reverse reaction.
2 I(g) = l2(g).
R = 0.08314 L·bar/mol·K.
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