Respuesta :
Answer:
Equilibrium concentration of [tex][HBr]=0.2080 M[/tex]
Equilibrium concentration of [tex][H_2]=[Br_2]=0.0001408 M[/tex]
Explanation:
[tex]H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)[/tex]..[1]
The equilibrium constant of reaction [1] =[tex]K_c=2.18\times 10^6[/tex]
Initial moles of HBr = 2.50 mol
Volume of the container = 12. 0 L
Concentration of a substance is calculated by:
[tex]\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}[/tex]
So, concentration of methane = [tex]\frac{0.0410}{1.00}=0.0410M[/tex]
Concentration of HBr = [tex]\frac{2.50 mol}{12.0 L}=0.2083 M[/tex]
The given chemical equation follows:
[tex]2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)[/tex]
Initial: 0.2083 M
At eqllm: (0.2083 - 2x)M x x
The equilibrium constant of reaction [2] :
=[tex]K_c'=\frac{1}{K_c}=\frac{1}{2.18\times 10^6}=4.587\times 10^{-7}[/tex]
Evaluating the value of 'x', we get:
[tex]K_c'=\frac{[H_2][Br_2]}{[HBr]^2}[/tex]
[tex]4.587\times 10^{-7}=\frac{x\times x}{(0.2083 - 2x)^2}[/tex]
Solving for x:
x = 0.0001408
Equilibrium concentration of [tex][HBr]=(0.2083-2x)M =0.2080 M[/tex]
Equilibrium concentration of [tex][H_2]=[Br_2]=x =0.0001408 M[/tex]
