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6. HI was heated in a sealed tube at 440°C till the equilibrium was reached. HI was found to be 22% decomposed. The equilibrium constant for dissociation is (A) 0.282 (C) 0.0155 (B) 0.0769. (D) 0.0199 The question is from chemical equilibrium lecture’s practice paper, in the video where the solutions were given and the problem was solved, they gave the answer to this question as option C. However my answer was option B. I want to know exactly how you got the answer to be option C? Thank you

6. HI was heated in a sealed tube at 440°C till the equilibrium was reached. HI was found to be 22% decomposed. The equilibrium constant for dissociation is 

(A) 0.282        (C) 0.0155

(B) 0.0769.       (D) 0.0199    

The question is from chemical equilibrium lecture’s practice paper, in the video where the solutions were given and the problem was solved, they gave the answer to this question as option C. However my answer was option B. I want to know exactly how you got the answer to be option C? Thank you

 

Grade:10

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
There r 2of eq. consistent 
. Kc and Kp 
rxn 2HI - > H2 +I2 
moles initally 2 0 
moles @eqlb. 2-2x x 
Kc = x*x/{2-2x}^2 = (0.22)^2/4*(0.78)^2 = 0.19888 
conc.(C)= P/RT for gas stage rxn 
RT *(x)= PH2 = PI2 =0.19888 (same as Kc bcz control o RT counteract ) .
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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