SELECT * FROM question_mgmt as q WHERE id=3016 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=48 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 11th chemistry Equilibrium find out the value of kc for each of the following

Question:

Find out the value of Kc for each of the following equilibria from the value of Kp:

(i) 2NOCl (g)  ↔  2NO (g) + Cl2 (g);                   Kp  =  1.8 × 10–2 at 500 K

(ii) CaCO3 (s)  ↔  CaO(s) + CO2(g);                  Kp  = 167 at 1073 K

Answer:

(a) The relation between Kp and Kc is given as:

Kp = Kc (RT)Δn

(a) Here, Δn = 3 - 2 = 1

R = 0.0831 barLmol-1K-1

T = 500 K

Kp = 1.8 x 10-2

Now, Kp = Kc (RT)Δn

⇒ Kc  = 1.8 x 10-2  /  (0.0831x500)

         = 4.33 x 10-4 (approx.)

 

(b) Here, Δn = 2 - 1 = 1

R = 0.0831 barLmol-1K-1

T = 1073 K

Kp = 167

Now, Kp = Kc (RT)Δn

⇒ Kc  = 167 / (0.0831x1073)

         = 1.87 (approx.)


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