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Iron (III) oxide and hydrogen react to form iron and water, like this:
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.
<u>Answer:</u> The value of equilibrium constant for given equation is
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
- <u>For hydrogen gas:</u>
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:
- <u>For water:</u>
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:
For the given chemical equation:
The expression of equilibrium constant for above equation follows:
Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.
Putting values in above expression, we get:
Hence, the value of equilibrium constant for given equation is