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The pressure at which equilibrium consta...

The pressure at which equilibrium constant in terms of pressures is found to be equal to that in terms of mole fraction for the equilibrium,
` PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g) `

A

10 atm

B

1 atm

C

`0*1 `atm

D

2 atm

Text Solution

Verified by Experts

The correct Answer is:
B
`K_(p) = K_(x) (P)^(Delta n _(g))`
` "Here " , Delta n _(g) = 1. " Hence " , K_(p) = K_(x)(P).`
` " Thus ", K_(p)= K_(x) " only when " P=1 atm`
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For the reaction, PCl_(5)(g)to PCl_(3)(g)+Cl_(2)(g)

1 mol of Cl_(2) and 3 mol of PCl_(5) are placed in a 100 L vessel heated to 227^(@)C . The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation for PCl_(5) and K_(p) for the reaction. PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)

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