Respuesta :
The given question is incomplete. The complete question is as follows.
Calculate the pressures of NO, [tex]Cl_{2}[/tex], and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 8.2 atm NO and 4.1 atm [tex]Cl_{2}[/tex]. (Hint: KP is small; assume the reverse reaction goes to completion then comes back to equilibrium.)
[tex]2NO(g) + Cl_{2} (g) \rightleftharpoons 2NOCl(g)[/tex]
[tex]K_{P} = 2.5 \times 10^{3}[/tex]
Explanation:
According to the ICE table,
[tex]2NO(g) + Cl_{2} (g) \rightleftharpoons 2NOCl(g)[/tex]
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 [tex]K_{p}[/tex] of the reaction is as follows.
[tex]K_{P} = \frac{[NOCl]^{2}}{[NO]^{2}[Cl_{2}]}[/tex]
[tex]2.5 \times 10^{3} = \frac{(2x)^{2}}{(8.2 - 4.1x)(4.1 - x)}[/tex]
x = 1.9
Therefore, at equilibrium
[NOCl] = [tex]2 \times 1.9[/tex] = 3.8
[NO] = (8.2 - 7.79) = 0.41
[tex]Cl_{2}[/tex] = 2 - 1.9 = 0.1
