Respuesta :
The question is incomplete, here is the complete question:
Iron (III) oxide and hydrogen react to form iron and water, like this:
[tex]Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)[/tex]
At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.
Compound Amount
Fe₂O₃ 3.95 g
H₂ 4.77 g
Fe 4.38 g
H₂O 2.00 g
Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.
Answer: The value of equilibrium constant for given equation is [tex]1.0\times 10^{-4}[/tex]
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
- For hydrogen gas:
Given mass of hydrogen gas = 4.77 g
Molar mass of hydrogen gas = 2 g/mol
Volume of the solution = 8.9 L
Putting values in above expression, we get:
[tex]\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M[/tex]
- For water:
Given mass of water = 2.00 g
Molar mass of water = 18 g/mol
Volume of the solution = 8.9 L
Putting values in above expression, we get:
[tex]\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M[/tex]
For the given chemical equation:
[tex]Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)[/tex]
The expression of equilibrium constant for above equation follows:
[tex]K_{eq}=\frac{[H_2O]^3}{[H_2]^3}[/tex]
Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.
Putting values in above expression, we get:
[tex]K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}[/tex]
Hence, the value of equilibrium constant for given equation is [tex]1.0\times 10^{-4}[/tex]
