Respuesta :
Answer:
[Cl2] equilibrium = 0.0089 M
Explanation:
Given:
[SbCl5] = 0 M
[SbCl3] = [Cl2] = 0.0546 M
Kc = 1.7*10^-3
To determine:
The equilibrium concentration of Cl2
Calculation:
Set-up an ICE table for the given reaction:
[tex]SbCl5(g)\rightleftharpoons SbCl3(g)+Cl2(g)[/tex]
I 0 0.0546 0.0546
C +x -x -x
E x (0.0546-x) (0.0546-x)
[tex]Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M[/tex]
The equilibrium concentration of Cl2 is:
= 0.0546-x = 0.0546-0.0457 = 0.0089 M
Answer : The concentration of [tex]Cl_2[/tex] at equilibrium will be, 0.0086 M
Explanation : Given,
Equilibrium constant = [tex]1.7\times 10^{-3}[/tex]
The balanced equilibrium reaction is,
[tex]SbCl_3(g)+Cl_2(g)\rightleftharpoons SbCl_5(g)[/tex]
Initial conc. 0.0546 0.0546 0
At eqm. (0.0.546-x) (0.0.546-x) x
The expression of equilibrium constant for the reaction will be:
[tex]K_c=\frac{[SbCl_5]}{[SbCl_3][Cl_2]}[/tex]
For the given the value of [tex]K_c[/tex] will be, [tex]\frac{1}{1.7\times 10^{-3}}[/tex]
Now put all the values in this expression, we get :
[tex]\frac{1}{1.7\times 10^{-3}}=\frac{(x)}{(0.0546-x)\times (0.0546-x)}[/tex]
By solving the term 'x', we get:
[tex]x=0.065M\text{ and }0.046M[/tex]
From the values of 'x' we conclude that, x = 0.065 can not be more than initial concentration. So, the value of 'x' which is equal to 0.065 is not consider.
So, x = 0.045 M
Thus, the concentration of [tex]Cl_2[/tex] at equilibrium = [tex](0.0546-x)M=[0.0546-2(0.045)]M=0.0086M[/tex]
