PCl5(g)→PCl3(g)+Cl2(g) in the reversible reaction, the moles of PCl5,PCl3andCl2 are a, b and c respectively and total pressure at equilibrium is P, then value of Kp is:
A
bca.RT
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B
b(a+b+c).P
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C
bcPa(a+b+c)
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D
ca(a+b+c).P
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Solution
The correct option is CbcPa(a+b+c) PCl5→PCl3 + Cl3 moles a b c mole fraction a(a+b+c)b(a+b+c)c(a+b+c) Partial Pressure: a(a+b+c).Pb(a+b+c).Pc(a+b+c).P Kp=b(a+b+c).Pc(a+b+c).Pa(a+b+c).P =bca(a+b+c)P