Respuesta :
Answer : The value of [tex]K_c[/tex] 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,
[tex]Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)[/tex]
The expression of [tex]K_c[/tex] will be,
[tex]K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}[/tex]
First we have to calculate the concentration of [tex]Br_2,Cl_2\text{ and }BrCl[/tex].
[tex]\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M[/tex]
[tex]\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M[/tex]
[tex]\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M[/tex]
Now we have to calculate the value of [tex]K_c[/tex] for the given reaction.
[tex]K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}[/tex]
[tex]K_c=\frac{(0.6)^2}{(1)\times (1)}[/tex]
[tex]K_c=0.36[/tex]
Therefore, the value of [tex]K_c[/tex] for the given reaction is, 0.36
Answer:
Kc for the reaction is 0.735
Explanation:
Step 1: Data given
Moles of Br2 = 0.500 moles
Moles of Cl2 = 0.500 moles
Volumes = 0.500 L
At the equilibrium we have 0.300 moles of BrCl
Step 2: The balanced equation
Br2(sol) + Cl2(sol) ⇌ 2 BrCl (sol)
Step 3: The concentration at the start
[Br2] = 0.500 moles / 0.500 L = 1 M
[Cl2] = 0.500 moles / 0.500 L = 1M
[BrCl] = 0 M
Step 4: Concentration at the equilibrium
For Br2 and Cl2 there will react X
For BrCl there will be consumed 2X
[Br2] = (1-X)M
[Cl2] = (1-X)M
[BrCl] = 2X = 0.300/0.500 = 0.600 M
X = 0.300
[Br2] = (1-X)M = 0.700
[Cl2] = (1-X)M = 0.700
Step 5: Calculate Kc
Kc = [BrCl]²/[Br2][Cl2]
Kc = 0.600² / (0.700*0.700)
Kc = 0.735
Kc for the reaction is 0.735
