Complete Question
The complete question is shown on the first uploaded image
Answer:
The equilibrium constant is [tex]K_c= 2.8*10^{-4}[/tex]
Explanation:
From the question we are told that
The chemical reaction equation is
[tex]Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}[/tex]
The voume of the misture is [tex]V_m = 5.4L[/tex]
The molar mass of [tex]Fe_{2} O_{3}_{(s)}[/tex] is a constant with value of [tex]M_{Fe_{2} O_{3}_{(s)} } = 160g/mol[/tex]
The molar mass of [tex]H_{2}_{(g)}[/tex] is a constant with value of [tex]H_2 = 2g/mol[/tex]
The molar mass of [tex]H_{2}O[/tex] is a constant with value of [tex]H_2O = 18g/mol[/tex]
Generally the number of moles is mathematically given as
[tex]No \ of \ moles \ = \frac{mass}{molar\ mass}[/tex]
For [tex]Fe_{2} O_{3}_{(s)}[/tex]
[tex]No \ of\ moles = \frac{3.54}{160}[/tex]
[tex]= 0.022125 \ mols[/tex]
For [tex]H_{2}[/tex]
[tex]No \ of\ moles = \frac{3.63}{2}[/tex]
[tex]= 1.815 \ mols[/tex]
For [tex]H_{2}O[/tex]
[tex]No \ of\ moles = \frac{2.13}{18}[/tex]
[tex]= 0.12 \ mols[/tex]
Generally the concentration of a compound is mathematicallyrepresented as
[tex]Concentration = \frac{No \ of \ moles }{Volume }[/tex]
For [tex]Fe_{2} O_{3}_{(s)}[/tex]
[tex]Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}[/tex]
[tex]= 4.10*10^{-3}M[/tex]
For [tex]H_{2}[/tex]
[tex]Concentration[H_2] = \frac{1.815}{5.4}[/tex]
[tex]= 0.336M[/tex]
For [tex]H_{2}O[/tex]
[tex]Concentration [H_2O] = \frac{0.12}{5.4}[/tex]
[tex]= 0.022M[/tex]
The equilibrium constant is mathematically represented as
[tex]K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}[/tex]
Considering [tex]H_2O \ for \ product[/tex]
And [tex]H_2 \ for \ reactant[/tex]
At equilibrium the
[tex]K_c = \frac{0.022}{0.336}[/tex]
[tex]K_c= 2.8*10^{-4}[/tex]
