At certain temperature (T) for the gas phase reaction
$2 H_{2} O (g) + 2 C l_{2} (g) ⇌_{}^{} 4 H C l (g) + O_{2} (g)$ $K_{p} = 12 \times 10^{8} a t m$
If $C l_{2}, H C l & O_{2}$ are mixed is such a manner that the partial pressure of each is 2 atm and the mixture is brough into contact with excess of liquid water. The approximate partial pressure of $C l_{2}$ when equilibrium is attained at temperature (T) in correct scientific notation is $A \times 10^{- B}$ atm. (A+B) value is
[Given: Vapour pressure of water is 380 mm Hg at temperature (T)]

Question

At certain temperature (T) for the gas phase reaction
$2 H_{2} O (g) + 2 C l_{2} (g) ⇌_{}^{} 4 H C l (g) + O_{2} (g)$ $K_{p} = 12 \times 10^{8} a t m$
If $C l_{2}, H C l & O_{2}$ are mixed is such a manner that the partial pressure of each is 2 atm and the mixture is brough into contact with excess of liquid water. The approximate partial pressure of $C l_{2}$ when equilibrium is attained at temperature (T) in correct scientific notation is $A \times 10^{- B}$ atm. (A+B) value is
[Given: Vapour pressure of water is 380 mm Hg at temperature (T)]

Accepted Answer

If KP is v.hight, any one of the reactant (limiting reagent) will be completely reacted at the time of reaching eq. state. Since liquid water is present in the vessel,  PH2O maintained constantly at 380 mm of Hg (0.5 atm)2H2O(g)+2Cl2(g)⇌ 4HCL(g)+O2(g)    t=0             0.5 atm      2 atm        2 atm     2 atmt = eq. state  0.5 atm   2-2= x     2+4 = 6 atm   2+1=3 atm12×108=64×30.52×x2⇒x2=64×10−8 ⇒x=36×10−4=3.6×10−3atm  ∴ A+B=3.6+3=6.6

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