Answer: The pressures of [tex]NO[/tex], [tex]Cl_2[/tex], and [tex]NOCl[/tex] in an equilibrium mixture are 0.4 atm , 0.2 atm and 6.6 atm respectively.
Explanation:
Initial pressure of [tex]NO[/tex] = 7.0 atm
Initial pressure of [tex]Cl_2[/tex] = 3.5 atm
The given balanced equilibrium reaction is,
[tex]2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)[/tex]
Initial pressure 7.0 atm 3.5 atm 0 atm
At eqm. conc. (7-2x)atm (3.5-x)atm (2x) Matm
The expression for equilibrium constant for this reaction will be,
[tex]K_p=\frac{[NOCl]^2}{[NO]^2[Cl_2]}[/tex]
[tex]K_p=\frac{(2x)^2}{(7-2x)^2\times (3.5-x)}[/tex]
we are given : [tex]K_p=2.9\times 10^3[/tex]
Now put all the given values in this expression, we get :
[tex]2.9\times 10^3=\frac{(2x)^2}{(7-2x)^2\times (3.5-x)}[/tex]
[tex]x=3.3atm[/tex]
Pressure of [tex]NO[/tex] at equilibrium = (7-2x) = [tex](7-2\times 3.3)=0.4 atm[/tex]
pressure of [tex]Cl_2[/tex] at equilibrium = (3.5-x) = [tex](3.5-3.3)=0.2 atm[/tex]
pressure of [tex]NOCl[/tex] at equilibrium = (2x) = [tex](2\times 3.3)=6.6 atm[/tex]
