Answer : The value of
for the given reaction is, 0.36
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given equilibrium reaction is,

The expression of
will be,
![K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BBrCl%5D%5E2%7D%7B%5BBr_2%5D%5BCl_2%5D%7D)
First we have to calculate the concentration of
.



Now we have to calculate the value of
for the given reaction.
![K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BBrCl%5D%5E2%7D%7B%5BBr_2%5D%5BCl_2%5D%7D)


Therefore, the value of
for the given reaction is, 0.36
What's wrong with this setup is the substrate on which you have positioned
the drop is "dirty and unclean" meaning it is not being dampened by
the solution. This action can be corrected by comprehensively cleaning the
substrate where the drop will be positioned.
"CH3CH2CH2CH2OH " is known by the name of "n-butanol" and "CH3CH(OH)CH3" is known by the name of "<span>Isopropyl alcohol". These two given products are basically alcohols. I hope that this is the answer that you were looking for and the answer has actually come to your desired help. Thanks for joining brainly and getting your questions solved.</span>