To determine the equilibrium constant [tex]\(\mathbf{K_p}\)[/tex] for the given reaction at the specified conditions, we start by writing down the balanced chemical equation and the expression for [tex]\(\mathbf{K_p}\)[/tex] in terms of the partial pressures of the reactants and products.
The balanced chemical equation is:
[tex]\[ 4 \text{HCl (g)} + \text{O}_2 \text{(g)} \rightarrow 2 \text{Cl}_2 \text{(g)} + 2 \text{H}_2\text{O (g)} \][/tex]
The expression for [tex]\(\mathbf{K_p}\)[/tex] based on this equation is:
[tex]\[ K_p = \frac{(P_{\text{Cl}_2})^2 (P_{\text{H}_2\text{O}})^2}{ (P_{\text{HCl}})^4 (P_{\text{O}_2}) } \][/tex]
Given the pressures at equilibrium, we have:
[tex]\[
\begin{aligned}
P_{\text{HCl}} &= 67.8 \text{ atm} \\
P_{\text{O}_2} &= 57.9 \text{ atm} \\
P_{\text{Cl}_2} &= 29.8 \text{ atm} \\
P_{\text{H}_2\text{O}} &= 50.4 \text{ atm}
\end{aligned}
\][/tex]
Now, substitute these values into the [tex]\(\mathbf{K_p}\)[/tex] expression:
[tex]\[
K_p = \frac{(29.8)^2 (50.4)^2}{(67.8)^4 (57.9)}
\][/tex]
First, we'll calculate the individual terms:
[tex]\[
(29.8)^2 = 888.04
\][/tex]
[tex]\[
(50.4)^2 = 2540.16
\][/tex]
[tex]\[
(67.8)^4 \approx 2118644.41
\][/tex]
Next, multiply the numerators and denominators:
[tex]\[
K_p = \frac{888.04 \times 2540.16}{2118644.41 \times 57.9}
\][/tex]
Calculate the multiplication:
[tex]\[
888.04 \times 2540.16 \approx 2255422.064
\][/tex]
And then the denominator:
[tex]\[
2118644.41 \times 57.9 \approx 122699309.379
\][/tex]
So, the expression for [tex]\(K_p\)[/tex] becomes:
[tex]\[
K_p = \frac{2255422.064}{122699309.379}
\][/tex]
Finally, performing the division we get:
[tex]\[
K_p \approx 0.0018437254979959442
\][/tex]
Rounding this final answer to 2 significant digits, we obtain:
[tex]\[
K_p \approx 0.0018
\][/tex]
Therefore, the equilibrium constant [tex]\( \boldsymbol{K_p} \)[/tex] for the reaction, rounded to two significant digits, is:
[tex]\[
\boxed{0.0018}
\][/tex]
