Question

.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was found to contain .300 mol of brcl. What is the value of Kc for this reaction?Br2(sol) + Cl2(sol) ⇌ 2 BrCl (sol)

Answer 1

Kc for the reaction is 0.735

Step-by-step 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

Answer 2

Answer : The value of
`K_c` 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,

`Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)`

The expression of
`K_c` will be,

`K_c=([BrCl]^2)/([Br_2][Cl_2])`

First we have to calculate the concentration of
`Br_2,Cl_2\text{ and }BrCl`.

`\text{Concentration of }Br_2=(Moles)/(Volume)=(0.500mol)/(0.500L)=1M`

`\text{Concentration of }Cl_2=(Moles)/(Volume)=(0.500mol)/(0.500L)=1M`

`\text{Concentration of }BrCl=(Moles)/(Volume)=(0.300mol)/(0.500L)=0.6M`

Now we have to calculate the value of
`K_c` for the given reaction.

`K_c=([BrCl]^2)/([Br_2][Cl_2])`

`K_c=((0.6)^2)/((1)* (1))`

`K_c=0.36`

Therefore, the value of
`K_c` for the given reaction is, 0.36