Answer
See explanation
Work Step by Step
The equilibrium constant for the reaction $\mathrm{CO}+\frac{1}{2} \mathrm{O}_2 \Leftrightarrow \mathrm{CO}_2$ can be expressed as
$$
K_p=\frac{N_{\mathrm{CO}_2}^{v_{\mathrm{O}_2}}}{N_{\mathrm{CO}}^{v_{\infty}} N_{\mathrm{O}_2}^{v_{\mathrm{O}_2}}}\left(\frac{P}{N_{\text {total }}}\right)^{\left(v_{\mathrm{CO}_2}-v_{\infty \mathrm{O}}-v_{\mathrm{O}_2}\right)}
$$ Adding more $\mathrm{N}_2$ (an inert gas) at constant temperature and pressure will increase $N_{\text {total }}$ but will have no direct effect on other terms. 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.
