H2(9) + 02(g) = H2O2cs) (Kc = 7.109 note:this is NOT Kp) If 0.420 atm of H2O2(g) is initially present, calculate the equilibrium partial pressures of H2(g), 02(g), and H2O2(g). (Hint: You need a value from the important information section to solve this).

Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 15P
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Consider the following equilibrium reaction at 160.00 °C.
H2(9) + 02(g)
= H202(g)
(K. = 7.109 note:this is NOT Kp)
If 0.420 atm of H2O2(g) is initially present, calculate the equilibrium partial pressures of H2(g), 02(g), and
H2O2(g). (Hint: You need a value from the important information section to solve this).
Transcribed Image Text:Consider the following equilibrium reaction at 160.00 °C. H2(9) + 02(g) = H202(g) (K. = 7.109 note:this is NOT Kp) If 0.420 atm of H2O2(g) is initially present, calculate the equilibrium partial pressures of H2(g), 02(g), and H2O2(g). (Hint: You need a value from the important information section to solve this).
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