A student ran the following reaction in the laboratory at 495 K: PCI3(g) + Cl₂(g) =PC15(g) When he introduced PCI3(g) and Cl₂(g) into a 1.00 L evacuated container, so that the initial partial pressure of PCI3 was 1.58 atm and the initial partial pressure of Cl₂ was 1.09 atm, he found that the equilibrium partial pressure of Cl₂ was 0.352 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction. Kp = ||
A student ran the following reaction in the laboratory at 495 K: PCI3(g) + Cl₂(g) =PC15(g) When he introduced PCI3(g) and Cl₂(g) into a 1.00 L evacuated container, so that the initial partial pressure of PCI3 was 1.58 atm and the initial partial pressure of Cl₂ was 1.09 atm, he found that the equilibrium partial pressure of Cl₂ was 0.352 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction. Kp = ||
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![A student ran the following reaction in the laboratory at 495 K:
PCI3(g) + Cl₂(g)
PCI5(g)
When he introduced PCI3(g) and Cl₂(g) into a 1.00 L evacuated container, so that the initial partial pressure of
PCI3 was 1.58 atm and the initial partial pressure of Cl₂ was 1.09 atm, he found that the equilibrium partial
pressure of Cl₂ was 0.352 atm.
Calculate the equilibrium constant, Kp, she obtained for this reaction.
Kp = ||](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0a8a901-2497-4ce7-b3d7-44550ff0e1ad%2F320e4493-3730-405f-92eb-67290d148161%2Fprz4mut_processed.png&w=3840&q=75)
Transcribed Image Text:A student ran the following reaction in the laboratory at 495 K:
PCI3(g) + Cl₂(g)
PCI5(g)
When he introduced PCI3(g) and Cl₂(g) into a 1.00 L evacuated container, so that the initial partial pressure of
PCI3 was 1.58 atm and the initial partial pressure of Cl₂ was 1.09 atm, he found that the equilibrium partial
pressure of Cl₂ was 0.352 atm.
Calculate the equilibrium constant, Kp, she obtained for this reaction.
Kp = ||
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