Atomic number
atomic mass
group|family
periods
and element symbols
Answer:
It kind of is logical so my answer is yes
Answer :
The equilibrium concentration of CO is, 0.016 M
The equilibrium concentration of Cl₂ is, 0.034 M
The equilibrium concentration of COCl₂ is, 0.139 M
Explanation :
The given chemical reaction is:
![CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)](https://tex.z-dn.net/?f=CO%28g%29%2BCl_2%28g%29%5Crightleftharpoons%20COCl_2%28g%29)
Initial conc. 0.1550 0.173 0
At eqm. (0.1550-x) (0.173-x) x
As we are given:
![K_c=255](https://tex.z-dn.net/?f=K_c%3D255)
The expression for equilibrium constant is:
![K_c=\frac{[COCl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCOCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
Now put all the given values in this expression, we get:
![255=\frac{(x)}{(0.1550-x)\times (0.173-x)}](https://tex.z-dn.net/?f=255%3D%5Cfrac%7B%28x%29%7D%7B%280.1550-x%29%5Ctimes%20%280.173-x%29%7D)
x = 0.139 and x = 0.193
We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.139
The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M
The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M
The equilibrium concentration of COCl₂ = x = 0.139 M