Question

in equilibrium the initial concentration of hydrogen and bromine is 0.50mol/L what is the equilibrium concentration of hydrogen bromide ( H2 + Br2 <> 2HBr ) kc= 50.0

Answer 1

`[HBr]_(eq)=0.78M`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`H_2 + Br_2 \rightleftharpoons 2HBr`

In that way, the Law of mass action results:

`Kc=([HBr]^2)/([H_2][Br_2])`

Hence, in terms of the change
`x` due to the reaction extent, we obtain:

`50.0=((2x)^2)/((0.5M-x)(0.5M-x))`

In such a way, by solving via the quadratic equation (two results), the results are:

`x_1=0.390M\\x_2=0.697M`

But the correct one is 0.390M as the other one results in negative concentrations of hydrogen and bromine at equilibrium. Therefore, the concentration of hydrogen bromide results:

`[HBr]_(eq)=2x=2*0.390M`

`[HBr]_(eq)=0.78M`

Best regards.

Answer 2

The concentration of HBr at the equilibrium is 0.780 M

Step-by-step explanation:

Step 1: Data given

The initial concentration of H2 = 0.50 mol/L

The initial concentration of Br2 = 0.50 mol/L

Kc = 50.0

Step 2: The balanced equation

H2 + Br2 ⇆ 2HBr

Step 3: The concentration at equilibrium

[H2] = 0.50 - X M

[Br2] = 0.50 - X M

[HBr] = 2X

Step 4: Calculate concentration of HBr

Kc = [HBr]² [H2][Br2]

50.0 = 4X² / (0.50 - X)(0.50 -X)

X = 0.390

[H2] = 0.50 - 0.390 = 0.110 M

[Br2] = 0.50 - 0.390 = 0.110 M

[HBr] = 2*0.390 = 0.780 M