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Question

Find out the value of Kc for the given equilibria (i) and (ii) respectively: 
(i)2NOCl (g)2NO (g)+Cl2 (g); KP=1.8×102  at 500 K
(ii)CaCO3 (s)CaO (s)+CO2 (g); Kp=167  at 1073 K

(R=0.0831 bar L mol1K1) 
  1. 8.8×104 and 3.8
  2. 8.8×104 and 3.8
  3. 4.38×104 and 1.87
  4. 4.38×104 and 1.87


Solution

The correct option is C 4.38×104 and 1.87
(i)2NOCl (g)2NO (g)+Cl2 (g);KP=1.8×102  at 500 K
 KP=KC (RT)ng
ng=32=1
KC=KPRT=1.8×1020.0831×500=4.38×104

(ii)CaCO3 (s)CaO (s)+CO2 (g);   Kp=167  at 1073 K
 ng=10=1
KC=KPRT=1670.0831×1073=1.87



Theory:

Relation between Kp and Kc :
aA (g)+bB (g)cC (g)+dD (g)
For an ideal gas :
PV=nRT
P=(nV)RT
Where nV=C (molar concentration)
C=PRT
Kc=[C]ceq[D]deq[A]aeq[B]beq

Kc=(PC)ceq(PD)deq(RT)c(RT)d(PA)aeq(PB)beq(RT)a(RT)b

Kc=[PC]ceq×[PD]deq(RT)(a+b)(c+d)[PA]aeq×[PB]beq


By using the property 1ax×1ay=1ax+y

KC=KP(RT)ΔngKP=KC(RT)Δng
Δng = (number of moles of gaseous products) - (number of moles of gaseous reactants)
If Δng=0 then Kc=Kp

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