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

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The relation between Kp and Kc is given as: 

Kp = Kc (RT)∆n 

(a) Here, 

∆n = 3 – 2 = 1 

R = 0.0831 barLmol–1K–1 

T = 500 K Kp = 1.8 × 10–2 

Now, 

Kp = Kc (RT) ∆n 

(b) Here, 

∆n = 2 – 1 = 1 

R = 0.0831 barLmol–1K–1 

T = 1073 K 

Kp= 167 

Now, 

Kp = Kc (RT) ∆n

 

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