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Calculate AGO at 595 K for H₂O(g) + 1/2 O2(g) H₂O2(g) using the following data: H2(g) +O2(g) H₂O2(g) K= 3.7 x 10³7 at 595 K 2H2(g) + O2(g) 2H₂O(g) K= 2.6 x 106 at 595 K AGO = -319 kj/mol

Calculate AGO at 595 K for H₂O(g) + 1/2 O2(g) H₂O2(g) using the following data: H2(g) +O2(g) H₂O2(g) K= 3.7 x 10³7 at 595 K 2H2(g) + O2(g) 2H₂O(g) K= 2.6 x 106 at 595 K AGO = -319 kj/mol

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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says -319 is wrong an sources my sources S Calculate AGO at 595 K for using the following data: H₂O(g) + 1/2O2(g) H₂O2(g) AGO=-319 c cleats Monthly Incomes H2(g) +O2(g) H₂O2(g) K = 3.7 x 10³7 Submit Answer 2H2(g) + O2(g) 2H₂O(g) K = 2.6 x 106 at 595 K at 595 K kj/mol An error has been detected in your answer. Check for typos, miscalculations etc. before submitting your answer. Try Another Version 1 item
Transcribed Image Text:says -319 is wrong an sources my sources S Calculate AGO at 595 K for using the following data: H₂O(g) + 1/2O2(g) H₂O2(g) AGO=-319 c cleats Monthly Incomes H2(g) +O2(g) H₂O2(g) K = 3.7 x 10³7 Submit Answer 2H2(g) + O2(g) 2H₂O(g) K = 2.6 x 106 at 595 K at 595 K kj/mol An error has been detected in your answer. Check for typos, miscalculations etc. before submitting your answer. Try Another Version 1 item
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