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iPad ??3:43 cvg.cengagenow.com ???? Bing Google Yahoo ???? ?+ Write The Cell Notat ????-???? ???????? CengageBrain - My.. oWLv2 | Online t... REVIEW SPRING BREAK The partial pressures of an equilibrium mixture of N204(9) and NO2(9) are P,o, -0.35 atm and Po,1.19 atm at a certain temperature The volume of the container is doubled. Find the partial pressures of the two gases when a new equilibrium is established. Question 1 1 pt Question 2 Question 3 Question 4 Question 5 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt Partial pressure of NO2 atm Partial pressure of N204 Question 6 Submit Answer Try Another Version Question Question 8 Question 9 10 item attempts remaining Question 10 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Progress: 0/16 items Due Apr 8 at 11:55 PM Previous Next Finish Assignment Email Instructor Save and Exit Cengage Learning |Cengage Technical SupportExplanation / Answer
N2O4 ? 2 NO2
If volume increases , according to Le Chatlier's Principle , the equilibrium shifts towards the side with more no. of particles . i.e. NO2
now when volume is doubled , Lets say change in N2O4 is -x , while the change in NO2 is 2x . Initial partial pressures being 0.35 and 1.19 atm .
Kp= (P(NO2 ))2/ P(N2O4)
Kp= (1.19)2/0.35 = 4.046
N2O4 ? 2 NO2
0.35-x 2x
Since the reaction is again in equilibrium , Kp = 4.046 i.e. the same doesn't change (Le Chartlier's principle equilibrium shifts but the Equilibrium constant is the same)
Thus ,
Kp= (P(NO2 ))2/ P(N2O4)
= (2x)/(0.35-x)
x= 0.234221
Thus the new partial pressures are , P(N2O4) = 0.115779 , (P(NO2 )=0.468442
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