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A student ran the following reaction in the laboratory at 584 K: When he introduced cOCl(g) at a pressure of 0.738 atm into a 1.00 L evacuated container, he found the equilibrium partial pressure of COC1(g) to be 0.374 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction. Kp

A student ran the following reaction in the laboratory at 584 K: When he introduced cOCl(g) at a pressure of 0.738 atm into a 1.00 L evacuated container, he found the equilibrium partial pressure of COC1(g) to be 0.374 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction. Kp

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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A student ran the following reaction in the laboratory at 584 K: When he introduced COCI(g) at a pressure of 0.738 atm into a 1.00 L evacuated container, he found the equilibrium partial pressure of COCI(g) to be 0.374 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction. Kp
Transcribed Image Text:A student ran the following reaction in the laboratory at 584 K: When he introduced COCI(g) at a pressure of 0.738 atm into a 1.00 L evacuated container, he found the equilibrium partial pressure of COCI(g) to be 0.374 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction. Kp
Expert Solution
Step 1

Given reaction is

COCl2(0.738-x) = CO(x)+ Cl2(x)

Let us consider the disassociation constant to be x then at equilibrium pressure of CO= x and pressure of Cl2=x.and presssure of COCl2= 0.738-x 

But pressure of COCl2= 0.374 atm, so 0.738-x= 0.374, x=0.364 atm .

SO KP=( x×x)/0.738-x= (0.364×0.364)/0.374= 0.354

 

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