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Question

At 527 oC, the reaction given below has Kc=4
NH3(g)12N2(g)+32H2(g)

What is the value of Kp for the reaction given below?
N2(g)+3H2(g)2NH3(g)
  1. 16×(800 R)2
  2. (800 R4)2
  3. (14×800 R)2
  4. None of these


Solution

The correct option is C (14×800 R)2
Given: 527 oC=800 K and
NH3(g)12N2(g)+32H2(g);     Kc=4

Reversing the reaction and multipling by 2, we get
N2(g)+3H2(g)2NH3(g);     Kc=(14)2
Here, ng=mole of product (g) - moles of reactant (g)=2
 
Since,
Kp=Kc(RT)ng

Kp=(14)2(800 R)2

Kp=(14×800 R)2


 

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