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Reaction Quotient

The value of ...

Question

The value of $K_{p}$ is $1 \times 10^{- 3} {a t m}^{- 1}$ at $25$ for the reaction : $2 N O + C l_{2} ⇌ 2 N O C l$ . A flask contains $N O$ at $0.02 a t m$ at $25^{0} C$ . $x \times 10^{- 2}$ mole of $C l_{2}$ must be added if 1% of the $N O$ is to be converted to $N O C l$ at equilibrium. The volume of the flask is such that $0.2$ mole of gas produce 1 atm pressure at $25$ . Also $y \times 10^{- 5}$ mole of $C l_{2}$ used. (Ignore probable association of $N O$ to $N_{2} O_{2}$ .)
Then calculate the value of $x$ and $y$ .

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Solution

The equilibrium expression, the initial pressure and the equilibrium pressure are given elow.
$2 N O + C l_{2} ⇌ 2 N O C l$
$P$ at $t = 0$ $0.02 P 0$
$P$ at equilibrium $\frac{0.02 \times 99}{100} [P - \frac{0.02 \times 1}{100}] \frac{0.02 \times 1}{100}$
$= 0.02 \times 0.99 = (P - 0.0002) = 0.02 \times 0.01$
The expression for the equilibrium constant is as given below.
$K_{p} = 1 \times 10^{- 3}$
$= \frac{{(P_{N O C l})}^{2}}{{(P_{N O})}^{2} (P_{{C l}_{2}})}$
$= \frac{{(0.02 \times 0.01)}^{2}}{{(0.02 \times 0.99)}^{2} \times (P - 0.0002)}$
$∴ P = 0.102 a t m$
The equilibrium pressure of chlorine is
$P_{C l_{2}}$ at equilibrium $= 0.102 - 0.0002 = 0.1018 a t m$
For $C l_{2}$ in vessel, using the ideal gas equation $P V = n R T$
$0.1018 \times V = n_{C l_{2}} R T$ ...(i)
For a gas in vessel $1 \times V = 0.2 \times R T$ ...(ii)
By equations (i) and (ii), the number of moles of $C l_{2}$ at equilibrium
$(n) = 0.0204$
If $V, T$ are constant $C l_{2}$
Also, $0.102 \times V = n_{C l_{2} T o t a l} \times R T$
...(iii)
By equations (ii) and (iii), $n_{C l_{2} T o t a l} = 0.02042 = 2.042 \times 10^{- 2}$
$∴ n_{C l_{2} U s e d} = 0.02042 - 0.0204$
$= 0.00002$
i.e. chlorine used= $2 \times 10^{- 5}$

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