Respuesta :
Answer:
Tthe value of [tex]K_{eq}=2.31\times 10^{-3}[/tex]
Explanation:
Initial moles of hydrogen peroxide = 1.75 mole
Volume of the container = 25 L
Concentration of a substance is calculated by:
[tex]\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}[/tex]
So, concentration of hydrogen peroxide = [tex]\frac{1.75}{25 L}=0.07 M[/tex]
The given chemical equation follows:
[tex]2H_2O_2(g)\rightleftharpoons 2H_2O(g)+O_2(g)[/tex]
Initially 0.07 M 0 0
At equilibrium
(0.07-2x) M 2x x
We are given:
Equilibrium moles of hydrogen peroxide = 1.20 mole
Equilibrium concentration of hydrogen peroxide = [tex]\frac{1.20}{25 L}=0.048 M[/tex]
Evaluating the value of 'x', we get:
(0.07-2x) M = 0.048 M
x = 0.011 M
So, the equilibrium concentrations of water vapors and oxygen gas are:
[tex][H_2O]=2x=2\times 0.011 M= 0.022 M[/tex]
[tex][O_2]=x=0.011 M[/tex]
The expression of [tex]K_{eq}[/tex] for the above reaction follows:
[tex]K_{eq}=\frac{[H_2O]^2\times [O_2]}{ [H_2O_2]^2}[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(0.022 M)^2\times (0.011 M)}{(0.048 M)^2}\\\\K_{eq}=2.31\times 10^{-3}[/tex]
Hence, the value of [tex]K_{eq}=2.31\times 10^{-3}[/tex]
