Question

Calculate the pressures of NO, Cl2, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 8.2 atm NO and 4.1 atm Cl2. (Hint: Kp is relatively large; assume the reaction goes to completion then comes back to equilibrium.)

Answer 1

The given question is incomplete. The complete question is as follows.

Calculate the pressures of NO, $Cl_{2}$, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 8.2 atm NO and 4.1 atm $Cl_{2}$. (Hint: KP is small; assume the reverse reaction goes to completion then comes back to equilibrium.)

      $2NO(g) + Cl_{2} (g) \rightleftharpoons 2NOCl(g)$

 $K_{P} = 2.5 \times 10^{3}$

Explanation:

According to the ICE table,

               $2NO(g) + Cl_{2} (g) \rightleftharpoons 2NOCl(g)$

Initial:      8.2              4.1              0

Change:    -4.1x           -x               +4.1x

Equilbm: (8.2 - 4.1x)   (4.1 - x)        +4.1x

Now, expression for $K_{p}$ of the reaction is as follows.

           $K_{P} = \frac{[NOCl]^{2}}{[NO]^{2}[Cl_{2}]}$

    $2.5 \times 10^{3} = \frac{(2x)^{2}}{(8.2 - 4.1x)(4.1 - x)}$

                   x = 1.9

Therefore, at equilibrium

    [NOCl] = $2 \times 1.9$ = 3.8

    [NO] = (8.2 - 7.79) = 0.41

    $Cl_{2}$ = 2 - 1.9 = 0.1