Answer
See explanation
Work Step by Step
$\left(\right.$ a) The equilibrium constant for the reaction $\mathrm{CO}+\frac{1}{2} \mathrm{O}_2 \Leftrightarrow \mathrm{CO}_2$ can be expressed as
$$\begin{aligned}
K_p=\frac{N_{\mathrm{CO}_2}^{v_{\mathrm{CO}_2}}}{N_{\mathrm{CO}}^{v_{\mathrm{O}}} N_{\mathrm{O}_2}^{v_{\mathrm{O}_2}}}\left(\frac{P}{N_{\text {total }}}\right)^{\left(v_{\mathrm{CO}_2}-v_{\mathrm{CO}}-v_{\mathrm{O}_2}\right)}
\end{aligned}$$ Judging from the values in Table A-28, the $K_p$ value for this reaction decreases as temperature increases. That is, the indicated reaction will be less complete at higher temperatures. Therefore, the number of moles of $\mathrm{CO}_2$ will decrease and the number moles of $\mathrm{CO}$ and $\mathrm{O}_2$ will increase as the temperature increases.
(b) The value of the exponent in this case is $1-1-0.5=-0.5$, which is negative. Thus as the pressure increases, the term in the brackets will decrease. The value of $K_p$ depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products $\left(\mathrm{CO}_2\right)$ must increase, and the number of moles of the reactants $\left(\mathrm{CO}, \mathrm{O}_2\right)$ must decrease.
