Q17.56P - Chemistry: Molecular Nature Of Matter And Change

Question

 A key step in the extraction of iron from its or 

$\mathbf{FeO}\mathbf{ }\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{ }\mathbf{+}\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathbf{\square }\mathbf{ }\mathbf{Fe}\mathbf{ }\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{ }\mathbf{+}{\mathbf{CO}}_{\mathbf{2}\mathbf{ }}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }{\mathbf{K}}_{\mathbf{P}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{403}\mathbf{at}{\mathbf{1000}}^{\mathbf{\circ }}\mathbf{C}$

This step occurs in the ${\mathbf{700}}^{\mathbf{\circ }}$C to ${\mathbf{1200}}^{\mathbf{\circ }}\mathbf{C}$  zone within a blast furnace. What are the equilibrium partial pressures of CO(g) and ${\mathbf{CO}}_{\mathbf{2}}$(g) when 1.00 atm of CO(g) and excess FeO(s) react in a sealed container at ${\mathbf{1000}}^{\mathbf{\circ }}\mathbf{C}$?

Step-by-Step Solution

Answer

Partial pressure of carbon dioxide, ${\mathrm{P}}_{{\mathrm{CO}}_2}=0.287\mathrm{atm}$  

Partial pressure of carbon monoxide, ${\mathrm{P}}_{\mathrm{CO}}=0.713\mathrm{atm}$ 

Step 1: Reaction equilibrium

The reaction equilibrium is a point in a reaction system at which there is no further change in the concentration of reactants and products in the system. 

This occurs because the rate of formation of products and reactants becomes equal.

At a particular temperature, the ratio of the concentration of products to reactants at 

the equilibrium of a particular reaction is independent of the initial reactant 

concentration. This ratio is called equilibrium constant and compares the rates of 

forward and backward reactions.

Step 2: Equilibrium constant

The equilibrium constant is the reaction quotient at equilibrium and can be expressed in terms of concentration or partial pressure. The equilibrium constant in partial pressure is used for reactions involving gases. The value of the equilibrium constant can be used to determine the direction of the reaction.

A higher value of the equilibrium constant implies that the concentration of products is greater than reactants at equilibrium. Therefore, the forward reaction is favored. The equilibrium constant in terms of partial pressures is the ratio of the pressure of gaseous product to gaseous reactant at equilibrium.

Step 3: Finding the partial pressures

The reaction between the ferrous oxide and carbon monoxide can be given by the following equation

$\mathrm{FeO} \left(\mathrm{s}\right) +\mathrm{CO}\left(\mathrm{g}\right) \square \mathrm{Fe} \left(\mathrm{s}\right) +{\mathrm{CO}}_{2 }\left(\mathrm{g}\right)$

The equilibrium constant in terms of partial pressure is 0.403 at 1000 degrees Celsius and can be calculated using the equation:

${\mathrm{K}}_{\mathrm{P}}=\frac{{\mathrm{P}}_{{\mathrm{CO}}_2}}{{\mathrm{P}}_{\mathrm{CO}}}$

The partial pressure of gases is only considered.

The initial pressure of carbon monoxide is 1.00 atm. If the pressure of carbon dioxide produced at equilibrium is x atm, the partial pressure of carbon monoxide is:

${\mathrm{P}}_{\mathrm{CO}}=\left(1-\mathrm{x}\right)\mathrm{atm}$

This is because carbon monoxide and carbon dioxide are the only gases in the system.

Substituting the known values for the equilibrium constant,

$$\begin{gathered} 0.403=\frac{x}{1-x} \\ x=0.403 \left(1-x\right) \\ = 0.403-0.403x \end{gathered}$$

Therefore, x can be calculated as:

$$\begin{gathered} x+0.403x=0.403 \\ 1.403x =0.403 \\ x=\frac{0.403}{1.403} \\ =0.287 \end{gathered}$$

Therefore, the partial pressure of carbon dioxide is 0.287 atm.

${\mathrm{P}}_{{\mathrm{CO}}_2}=0.287\mathrm{atm}$

The partial pressure of carbon monoxide can be calculated as:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{CO}}=\left(1-\mathrm{x}\right)\mathrm{atm} \\ =\left(1-0.287\right)\mathrm{atm} \\ = 0.713\mathrm{atm} \end{gathered}$$