Question

consider the equilibrium system described by the chemical reaction below. at equilibrium, a 2.0 l reaction vessel contained a mixture of 1.2 mol fe, 1.0×10⁻³ mol o₂, and 2.0 mol of fe₂o₃ at 2000.0 °c. what are the values of kc and kp for this reaction? 4 fe(s) + 3 o₂(g) ⇌ 2 fe₂o₃(s) based on your value of kc, determine the value of kp under these conditions. kp =

Explanation:

Step1: Calculate molar - concentrations

The molar - concentration formula is $c=\frac{n}{V}$. For $O_2$, $n = 1.0\times10^{-3}\ mol$ and $V = 2.0\ L$. So, $[O_2]=\frac{1.0\times10^{-3}\ mol}{2.0\ L}=5.0\times10^{-4}\ M$. Solids ($Fe$ and $Fe_2O_3$) are not included in the $K_c$ expression. The $K_c$ expression for the reaction $4Fe(s)+3O_2(g)
ightleftharpoons2Fe_2O_3(s)$ is $K_c=\frac{1}{[O_2]^3}$.

Step2: Calculate $K_c$

Substitute $[O_2]=5.0\times10^{-4}\ M$ into the $K_c$ expression:
$K_c=\frac{1}{(5.0\times10^{-4})^3}=\frac{1}{1.25\times10^{-10}} = 8.0\times10^{9}$

Step3: Relate $K_p$ and $K_c$

The relationship between $K_p$ and $K_c$ is $K_p = K_c(RT)^{\Delta n}$, where $R = 0.0821\ L\cdot atm\cdot K^{-1}\cdot mol^{-1}$, $T=(2000.0 + 273.15)\ K=2273.15\ K$, and $\Delta n$ is the change in the number of moles of gaseous species. For the reaction $4Fe(s)+3O_2(g)
ightleftharpoons2Fe_2O_3(s)$, $\Delta n=0 - 3=-3$.

Step4: Calculate $K_p$

$K_p=K_c(RT)^{\Delta n}=8.0\times10^{9}\times(0.0821\times2273.15)^{-3}$
First, calculate $0.0821\times2273.15 = 186.625615$. Then, $(0.0821\times2273.15)^{-3}=\frac{1}{(186.625615)^3}=\frac{1}{6.479\times10^{6}}$.
$K_p=8.0\times10^{9}\times\frac{1}{6.479\times10^{6}}\approx1.24\times10^{3}$

Answer:

$K_c = 8.0\times10^{9}$, $K_p=1.24\times10^{3}$