Question

The minerals hematite (Fe2O3) and magnetite (Fe3O4) exist in equilibrium with atmospheric oxygen:4Fe3O4(s) + O2(g) ⇌ 6Fe2O3(s)Kp = 2.5 X1087 at 298 K(a) Determine PO2, at equilibrium

(b) Given that PO2 in air is 0.21 atm, in which direction will the reaction proceed to reach equilibrium?

(c) Calculate Kc at 298 K.

Answer 1

Answer:

For a: The partial pressure of oxygen gas at equilibrium is $4.0\times 10^{-88}atm$

For b: The reaction must proceed in the forward direction to reach equilibrium.

For c: The value of $K_c$ is $6.12\times 10^{88}$

Explanation:

For the given chemical equation:

$4Fe_3O_4(s)+O_2(g)\rightleftharpoons 6Fe_2O_3(s)$

• For a:

In the expression of $K_p$, the partial pressures of solids and liquids are taken as 1.

The expression of $K_p$ for above equation follows:

$K_p=\frac{1}{p_{O_2}}$

We are given:

$K_p=2.5\times 10^{87}$

Putting values in above equation, we get:

$2.5\times 10^{87}=\frac{1}{p_{O_2}}\\\\p_{O_2}=\frac{1}{2.5\times 10^{87}}=4.0\times 10^{-88}atm$

Hence, the partial pressure of oxygen gas at equilibrium is $4.0\times 10^{-88}atm$

• For b:

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{1}{p_{O_2}}$

We are given:

$p_{O_2}=0.21$

Putting values in above equation, we get:

$Q_p=\frac{1}{0.21}=4.76$

There are 3 conditions:

• When $K_{p}>Q_p$; the reaction is product favored.
• When $K_{p}<Q_p$; the reaction is reactant favored.
• When $K_{p}=Q_p$; the reaction is in equilibrium.

As, $K_p>Q_p$, the reaction will be favoring product side.

Hence, the reaction must proceed in the forward direction to reach equilibrium.

• For c:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure = $2.5\times 10^{87}$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 298 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=0-1=-1$

Putting values in above equation, we get:

$2.5\times 10^{87}=K_c\times (0.0821\times 298)^{-1}\\\\K_c=6.12\times 10^{88}$

Hence, the value of $K_c$ is $6.12\times 10^{88}$

Answer 2

The partial pressure of oxygen at equilibrium is $\rm 4.0\;\times\;10^{-88}\;atm$. With the partial pressure of oxygen in the air at 0.21 atm, the reaction process in forwarding direction, and the value of   $K_c=6.12\;\times\;10^{88}$.

What is the relation between equilibrium constant and pressure?

• The equilibrium constant for the partial pressure of the gas is given as $K_p$. The equilibrium constant for the reaction gives the pressure of oxygen at equilibrium as:

$K_p=\dfrac{1}{P_{\rm O_2}}$

The equilibrium constant for the given reaction is $K_P=2.5\;\times\;10^{87}$

Substituting the values for the partial pressure of oxygen:

$2.5\;\times\;10^{87}=\dfrac{1}{P{\text O_2}} \\\\P_{\text O_2}=\dfrac{1}{2.5\;\times\;10^{87}}\\\\ P_{\text O_2}=4.0\;\times\;10^{-88}\;\rm atm$

The partial pressure of oxygen at equilibrium is $\rm 4.0\;\times\;10^{-88}\;atm$.

• The oxygen in the air has a partial pressure of 0.21 atm. Thus, the air has high concentration of oxygen. The increased concentration of oxygen results in the reaction to process in the forward direction.

• The equilibrium constant for the reaction at 298 K is:

$K_p=K_c(RT)^{\Delta n}$

Where the equilibrium constant for the given reaction is $K_P=2.5\;\times\;10^{87}$

The gas constant for the reaction is, $R=0.0821\;\rm L.atm/mol.K$

The temperature of the reaction is, $T=298\;\rm K$

The change in moles of gas particles,

$\Delta n= \text {products - reatants of gas molecules}\\\Delta n=0-1\\\Delta n=-1$

Substituting the values for determining $K_c$:

$2.5\;\times\;10^{87}=K_c(0.0821\;\times\;298)^{-1}\\\\K_c=\dfrac{2.5\;\times\;10^{87}}{(0.0821\;\times\;298)^{-1}} \\\\K_c=6.12\;\times\;10^{88}$

The value of the equilibrium constant for the reaction at 298 K is $K_c=6.12\;\times\;10^{88}$.

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