Question

Iron(III) oxide and hydrogen react to form iron and water, like this: (s)(g)(s)(g) At a certain temperature, a chemist finds that a reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition: compound amount Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answer 1

The question is incomplete, here is the complete question:

Iron(III) oxide and hydrogen react to form iron and water, like this:

`Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)`

At a certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition:

Compound Amount

`Fe_2O_3` 3.54 g

`H_2` 3.63 g

Fe 2.37 g

`H_2O` 2.13 g

Calculate the value of the equilibrium constant for this reaction. Round your answer to 2 significant digits

Answer: The value of equilibrium constant for the given reaction is
`2.8* 10^(-4)`

Step-by-step explanation:

To calculate the molarity of solution, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}* \text{Volume of solution (in L)}}`

• For hydrogen gas:

Given mass of hydrogen gas = 3.63 g

Molar mass of hydrogen gas = 2 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

`\text{Molarity of hydrogen gas}=(3.63)/(2* 5.4)\\\\\text{Molarity of hydrogen gas}=0.336M`

• For water:

Given mass of water = 2.13 g

Molar mass of water = 18 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

`\text{Molarity of water}=(2.13)/(18* 5.4)\\\\\text{Molarity of water}=0.0219M`

The given chemical equation follows:

`Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)`

The expression of
`K_c` for above equation follows:

`K_c=([H_2O]^3)/([H_2]^3)`

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression. So, the concentration of iron and iron (III) oxide is not present in equilibrium constant expression.

Putting values in above equation, we get:

`K_c=((0.0219)^3)/((0.336)^3)\\\\K_c=2.77* 10^(-4)`

Hence, the value of equilibrium constant for the given reaction is
`2.8* 10^(-4)`