Respuesta :
The question is incomplete, here is the complete question:
Consider the following chemical reaction: H₂ (g) + I₂ (g) ⇔ 2HI (g) At equilibrium in a particular experiment, the concentrations of H₂, I₂, and HI were 0.15 M, 0.033 M and 0.55 M respectively. The value of Keq for this reaction is
Answer: The value of [tex]K_{eq}[/tex] for the given reaction is 61.11
Explanation:
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_[eq}[/tex]
For a general chemical reaction:
[tex]aA+bB\rightarrow cC+dD[/tex]
The expression for [tex]K_{eq}[/tex] is written as:
[tex]K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]
For the given chemical equation:
[tex]H_2(g)+I_2(g)\rightleftharpoons 2HI(g)[/tex]
The expression of [tex]K_{eq}[/tex] for above equation follows:
[tex]K_{eq}=\frac{[HI]^2}{[H_2][I_2]}[/tex]
We are given:
[tex][HI]_{eq}=0.55M[/tex]
[tex][H_2]_{eq}=0.15M[/tex]
[tex][I_2]_{eq}=0.033M[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(0.55)^2}{0.15\times 0.033}\\\\K_{eq}=61.11[/tex]
Hence, the value of [tex]K_{eq}[/tex] for the given reaction is 61.11
