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  2 NO(g) + Cl2(g) ⇄ 2 NOCl(g)    Kp = 1.40 × 108 A reaction vessel initially contains 3.0 atm of NO and 2.0 atm of Cl2(g). What is the pressure of NO(g) when equilibrium is reached?
3.6 × 10-4 atm
3.8 × 10-8 atm
0.5 atm
2.0 atm
1.0 atm
1.1 × 10-7 atm

Respuesta :

nkirotedoreen46

Answer:

The partial pressure of NOCl is 5.02 x 10^4 atm.

Explanation:

  • Partial pressures in a reaction equation may be used to calculate the equilibrium constant, Kp.
  • In this case;  2 NO(g) + Cl2(g) ⇄ 2 NOCl(g)

[tex]Kp=\frac{(Partial pressureNOCl)^2}{(Partial pressureNO)^2(Partial pressureCl2)}[/tex]

[tex]Kp=\frac{P(NOCl)^2}{(P(NO)^2(P(Cl2)}[/tex]

[tex]1.40*10^8= \frac{P(NOCl)^2}{(3.0 atm)^2(2.0atm)}[/tex]

[tex]1.40*10^8= \frac{P(NOCl)^2}{18.0}[/tex]

Therefore;

[tex]P(NOCl)^2 = 25.2*10^8[/tex]

[tex]P(NOCl) =5.02*10^4[/tex]

Therefore;

The partial pressure of NOCl is 5.02 x 10^4 atm.

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