A reação é a seguinte:
[tex3]\begin{aligned} & \text{CO}_{2(g)} \text{+} & \text{C}_{(s)} \rightleftharpoons 2\text{CO}_{(g)} \\ \end{aligned}[/tex3]
Sabemos que [tex3]K_p = \frac{(p{\text{CO}})^2}{(p{ \text{CO}_2})}.[/tex3]
a)
[tex3]\begin{array}{cccc}
⠀&\text{CO}_{2(g)} & + \,\,\, \text{C}_{(s)} & \rightleftharpoons \,\,\, 2\text{CO}_{(g)}\\
\text{Início (atm)}&x & 0 & 0\\
\text{ Reage e forma (atm)}& y & 0 & 2y \\
\text{ Equilíbrio (atm)}& x -y & 0 & 2y
\end{array} \\⠀\\
\text{Pressão total} = x -y + 2y = 4 \,\,\, \Rightarrow \,\,\,\, x + y= 4 \\⠀\\
\text{K}_p = \frac{\( 2y\) ^2}{x - y} \,\,\,\, \Rightarrow \,\,\,\, 10 = \frac{4y^2}{4 -2y} \,\,\,\, \therefore \,\,\,\, y \approx 1,53 \\⠀\\ (p{\text{CO}}) = 2y \approx 3,06 \, \text{atm} \, \wedge \, (p{ \text{CO}_2}) = 4 - 2y \approx 0,937 \, \text{atm}[/tex3]
b) dps respondo esse item.