A homogeneous ideal gaseous reaction AB₂(g) $\stackrel{ }{\rightleftharpoons }$A(g) + 2B(g) is carried out in a 25 litre flask at 27⁰C. The initial amount of AB₂ was 1 mole and the equilibrium pressure was 1.9 atm. The value of K_P is x × 10^–2. The value of is (Interger answer)

Step 1: Given: R=0. dm³atmK^–1mol^–1, equilibrium pressure = 1.9atm, V=25liter,T=27°C

Step 2: Formula used: Kp=(P_A)×(P_B)²/P_AB,PV=nRT

Kp=(X_APT)×(X_BPT)2/X_ABPT

Step 3:

                             AB₂(g)⇌A(g)+2B(g)

Initial                          1-x      x           2x

Atequillibrium                  1/1+2x      1/1.9

PV=nRT

1.9×25=n_T×0.082×300

n_T=1.931+2x=1.932x=0.93x=0.465

Kp=(X_APT)×(X_BPT)2/X_ABPT

Kp=(0.465×1.9/1.93)×(0.93×1.9/1.93)2/(0.535×1.9/1.93)

Kp=0.072atm²

Kp=72×10^-2atm²

Hence, a homogeneous ideal gaseous reaction, AB₂(g)⇌A(g)+2B(g) is carried out in a 25liter flask at 27°C. The initial amount of AB2was1moleand the equilibrium pressure was 1.9atm. The value of Kp is x×10^–2. The value of x here is 72.

 Or 

$K_p =\frac{\left({\displaystyle \frac{0.465}{1.93}\times 19}\right){\left({\displaystyle \frac{0.93}{1.93}\times 1.9}\right)}^2}{\left({\displaystyle \frac{0.535}{1.93}\times 1.9}\right)}=7.3\times {10}^{-2} at{m}^2$

Hence, the correct answer is $7.3\times {10}^{-2} at{m}^2$.