Respuesta :
Answer: Concentration of [tex]H_2[/tex] at equilibrium = 0.08 M
Concentration of [tex]I_2[/tex] at equilibrium =0.08 M
Concentration of [tex]Hl[/tex] = 0.64M
Explanation:
Moles of [tex]HI[/tex] = 4.00 mole
Volume of solution = 5.00 L
Initial concentration of [tex]HI[/tex] =[tex]\frac{moles}{Volume}=\frac{4.00mol}{5.00L}=0.8M[/tex]
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_c[/tex]
For the given chemical reaction:
[tex]2HI(g)\rightarrow H_2(g)+I_2(g) [/tex]
Initial conc. 0.8 M 0 M 0 M
At eqm. conc. (0.8-2x) M (x) M (x) M
The expression for [tex]K_c[/tex] is written as:
[tex]K_c=\frac{[H_2]^1[I_2]^1}{[HI]^2}[/tex]
[tex]0.016=\frac{(x)\times (x)}{(0.8-2x)}[/tex]
[tex]x=0.08M[/tex]
Concentration of [tex]H_2[/tex] at equilibrium = x M = 0.08 M
Concentration of [tex]I_2[/tex] at equilibrium = x M = 0.08 M
Concentration of [tex]Hl[/tex] = (0.8-2x) = [tex](0.8-2\times 0.08)M= 0.64M[/tex]
