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Please help...Chemical equilibrium...Both 66 and 67...

Please help...Chemical equilibrium...Both 66 and 67...

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Grade:11

3 Answers

Adarsh
768 Points
6 years ago
in question 66 ,
kp= (partial pressure of CO)2/partial pressure of CO2
=> 16/8 = 2 hence 2 will be the correct answer for question no 66
Arun
25750 Points
6 years ago
Question 66
Kp = (partial pressure of CO)² / partial pressure of CO2
= 8²/4
= 64/4
= 16
Regards
Arun(askIITians forum expert)
Akshat Gupta
21 Points
5 years ago
Q.66 Only CO and CO2 are in gaseous form so
Kp
=(Partial pressure of CO)2/Partial pressure of CO2
=82/4=16
 
Q67- Let Initial concentration of A and B=P
Qc=0(c)
So Reaction will shift towards Right
Let final Concentration of  A and B= P-x
  So final Concentration of C and D = x
Now-
Kc=[C][D]/[A][B]
9=x2/(P-x)2
3=x/P-x
 
x=3P/4
Fraction of Moles converted in C and D=3/4=0.75

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