The value of KP is 1×10−3 atm−1 at 25 ∘C for the reaction 2NO(g)+Cl2(g)⇌2NOCl(g). A flask contains NO at 0.02 atm and at 25 ∘C. The mole of Cl2 that must be added if 1% of the NO is to be converted to NOCl at equilibrium is x×10−5. The volume of the flask is such that 0.2 mole of gas produces 1 atm pressure at 25 ∘C (Ignore probable association of NO to N2O2). Find the value of x.(Given: 1(99)2≈1.02×10−4)

The value of KP is 1×10−3 atm−1 at 25 ∘C for the reaction 2NO(g

1.	The value of KP is 1×10−3 atm−1 at 25 ∘C for the reaction 2NO(g)+Cl2(g)⇌2NOCl(g). A flask contains NO at 0.02 atm and at 25 ∘C. The mole of Cl2 that must be added if 1% of the NO is to be converted to NOCl at equilibrium is x×10−5. The volume of the flask is such that 0.2 mole of gas produces 1 atm pressure at 25 ∘C (Ignore probable association of NO to N2O2). Find the value of x.(Given: 1(99)2≈1.02×10−4)
Answer» The value of $K_{P}$ is $1 \times 10^{- 3} {atm}^{- 1}$ at $25^{\circ} C$ for the reaction $2 N O (g) + C l_{2} (g) ⇌ 2 N O C l (g) .$ A flask contains $N O$ at $0.02 atm$ and at $25^{\circ} C$ . The mole of $C l_{2}$ that must be added if $1 %$ of the $N O$ is to be converted to $N O C l$ at equilibrium is $x \times 10^{- 5}$ . The volume of the flask is such that $0.2$ mole of gas produces $1 atm$ pressure at $25^{\circ} C$ (Ignore probable association of $N O$ to $N_{2} O_{2}$ ). Find the value of $x$ . (Given: $\frac{1}{(99)^{2}} \approx 1.02 \times 10^{- 4}$ )