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At 500 K the reaction 2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) has Kp = 51 In an equilibrium mixture at 500 K, the partial pressure of NO is 0.125 atm and Cl2 is 0.165 atm. What is the partial pressure of NOCl in the equilibrium mixture?

Respuesta :

Answer:

p3=0.36atm (partial pressure of NOCl)

Explanation:

2 NO(g) + Cl2(g) ⇌ 2 NOCl(g)  Kp = 51

lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively

[tex]Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }[/tex]

[tex]Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }[/tex]

p1=0.125atm;

p2=0.165atm;

p3=?

Kp=51;

On solving;

p3=0.36atm (partial pressure of NOCl)

Ver imagen ruchi86
Cricetus

The partial pressure of NOCl will be "0.36 atm".

The reaction,

  • [tex]2 NO(g)+Cl_2(g) \rightleftharpoons 2 NOCl (g)[/tex]

Given values,

Pressure,

  • [tex]P_1 = 0.125 \ atm[/tex]
  • [tex]P_2 = 0.165 \ atm[/tex]

Value of Kp,

  • [tex]51[/tex]

Now,

→ [tex]K_p = \frac{[NOCl]^2}{[NO]^2[Cl_2]^2}[/tex]

or,

→ [tex]K_p = \frac{[P_3]^2}{[P_1]^2[P_2]}[/tex]

By substituting the values,

   [tex]51 = \frac{[P_3]^2}{[0.125]^2[0.165]}[/tex]

   [tex]P_3 = 0.36 \ atm[/tex]

Thus the response above is appropriate.  

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