Respuesta :
Answer : The value of equilibrium constant [tex]K_p[/tex] for the reaction is, 3.82
Explanation :
First we have to calculate the pressure of phosgene gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = Pressure of phosgene gas = ?
V = Volume of phosgene gas = 1.50 L
n = number of moles phosgene = [tex]3.30\times 10^{-2}mol[/tex]
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of phosgene gas = 800 K
Putting values in above equation, we get:
[tex]P\times 1.50L=3.30\times 10^{-2}mol\times (0.0821L.atm/mol.K)\times 800K[/tex]
[tex]P=1.44atm[/tex]
Now we have to calculate the value of equilibrium constant.
The balanced equilibrium reaction is:
[tex]CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)[/tex]
Initial pressure 0 0 1.44
At eqm. P P (1.44-P)
The expression of equilibrium constant [tex]K_p[/tex] for the reaction will be:
[tex]K_p=\frac{(p_{COCl_2})}{(p_{CO})\times (p_{Cl_2})}[/tex]
As we are given that:
[tex]p_{CO}=0.497atm[/tex]
That means, P = 0.497 atm
[tex]K_p=\frac{(1.44-P)}{(P)\times (P)}[/tex]
Now put all the values in this expression, we get :
[tex]K_p=\frac{(1.44-0.497)}{(0.497)\times (0.497)}[/tex]
[tex]K_p=3.82[/tex]
Thus, the value of equilibrium constant [tex]K_p[/tex] for the reaction is, 3.82
