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For gas phase reaction, the rate law can be expressed in terms of partial pressure of the reactants. For gas phase reaction below: 4 PH3(g) → P4(g) + 6 H₂(g) The rate law is given by Rate = k(PPH3). At 680°C, the half-life of the reaction is 35.0 s. Given initial pressures of PPH3 = 0.67 atm, PP4 = 0.25 atm, and PH2 = 0.75 atm, how many seconds will it take for the pressure of PPH3 to drop to 0.50 atm?

For gas phase reaction, the rate law can be expressed in terms of partial pressure of the reactants. For gas phase reaction below: 4 PH3(g) → P4(g) + 6 H₂(g) The rate law is given by Rate = k(PPH3). At 680°C, the half-life of the reaction is 35.0 s. Given initial pressures of PPH3 = 0.67 atm, PP4 = 0.25 atm, and PH2 = 0.75 atm, how many seconds will it take for the pressure of PPH3 to drop to 0.50 atm?

Principles of Modern Chemistry
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
Chapter18: Chemical Kinetics
Section: Chapter Questions
Problem 54AP
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For gas phase reaction, the rate law can be expressed in terms of partial pressure of the reactants. For = gas phase reaction below: 4 PH3(g) → P4(g) + 6 H₂(g) The rate law is given by Rate = k(PPH3). At 680°C, the half-life of the reaction is 35.0 s. Given initial pressures of PPH3 = 0.67 atm, PP4 = 0.25 atm, and PH₂ = 0.75 atm, how many seconds will it take for the pressure of PPH3 to drop to 0.50 atm?
Transcribed Image Text:For gas phase reaction, the rate law can be expressed in terms of partial pressure of the reactants. For = gas phase reaction below: 4 PH3(g) → P4(g) + 6 H₂(g) The rate law is given by Rate = k(PPH3). At 680°C, the half-life of the reaction is 35.0 s. Given initial pressures of PPH3 = 0.67 atm, PP4 = 0.25 atm, and PH₂ = 0.75 atm, how many seconds will it take for the pressure of PPH3 to drop to 0.50 atm?
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