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Question

# Consider following equilibrium at 400 K and 30 atm 3A(g)⇌2B(g)+2C(g)+D(g) If the equilibrium pressure of A is 5 atm then Kc will be:

A
9.5 L2mol2
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B
0.39 L2mol2
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C
0.95 L2mol2
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D
3.9 L2mol2
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Solution

## The correct option is B 0.39 L−2mol23A⇌2B+2C+DInitialP∘000EquilibriumP∘−3x2x2xx Equilibrium pressure =P∘−3x+2x+2x+x ⇒30=P∘+2x...(1) From question 5=P∘−3x...(2) From equation (1) and (2) 25=5x⇒x=5Kp=P2BP2CPDP3A=(10)2(10)2553⇒Kp=400 atm2 We know, Kp=Kc(RT)Δn⇒Kc=Kp(RT)Δn=400(0.08×400)2 =0.39 L−2mol2

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