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Standard XII

Chemistry

Derivation of Kp and Kc

A flask conta...

Question

A flask containing $0.5$ atm pressure of $A_{2} (g)$ , some solid AB added into flask which undergoes dissociation according to $2 A B (s) ⇌ A_{2} (g) + B_{2} (g)$ , $K_{p} = 0.06$ $a t m^{2}$
The total pressure (in atm) at equilibrium is:

0.70

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0.6

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0.10

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none of these

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Solution

The correct option is A $0.70$
According to the equation
$2 A B (s) ⇌ A_{2} (g) + B_{2} (g)$
Hence the pressure will be $0.5 + x$ and $x$ for $A_{2}$ and $b_{2}$ respectively.
Therefore,
$K_{p} = p_{A_{2}} . p_{B_{2}}$
$0.06 = (0.5 + x) (x)$
$x = 0.1$
Hence total pressure
$p (t o t a l) = p_{A_{2}} + p_{B_{2}}$
$= (0.5 + 0.1) + 0.1 = 0.70 a t m$

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A flask containing 0.5 atm pressure of A2(g), some solid AB added into flask which undergoes dissociation according to 2AB(s)⇌A2(g)+B2(g), Kp=0.06 atm2The total pressure (in atm) at equilibrium is:

The correct option is A 0.70According to the equation 2AB(s)⇌A2(g)+B2(g)Hence the pressure will be 0.5+x and x for A2 and b2 respectively.Therefore,Kp=pA2.pB20.06=(0.5+x)(x)x=0.1Hence total pressure p(total)=pA2+pB2=(0.5+0.1)+0.1=0.70atm

A flask containing $0.5$ atm pressure of $A_{2} (g)$ , some solid AB added into flask which undergoes dissociation according to $2 A B (s) ⇌ A_{2} (g) + B_{2} (g)$ , $K_{p} = 0.06$ $a t m^{2}$
The total pressure (in atm) at equilibrium is: