Answer
$6.4\%$
Work Step by Step
The stoichiometric and actual reactions in this case are
Stoichiometric: $\quad \mathrm{CO}_2 \Leftrightarrow \mathrm{CO}+\frac{1}{2} \mathrm{O}_2$ (thus $v_{\mathrm{CO}_2}=1, v_{\mathrm{CO}}=1$, and $v_{\mathrm{O}_2}=\frac{1}{2}$ )
Actual: $$
\mathrm{CO}_2 \longrightarrow \underbrace{x \mathrm{CO}_2}_{\text {neact. }}+\underbrace{y \mathrm{CO}+z \mathrm{O}_2}_{\text {products }}
$$ C balance: $$
1=x+y \longrightarrow y=1-x
$$ O balance: $$
2=2 \mathrm{x}+\mathrm{y}+2 \mathrm{z} \longrightarrow \mathrm{z}=0.5-0.5 \mathrm{x}
$$ Total number of moles:
$$
N_{\text {total }}=x+y+z=1.5-0.5 x
$$ The equilibrium constant relation can be expressed as $$
K_p=\frac{N_{\mathrm{CO}}^{v_{\mathrm{CO}}} N_{\mathrm{O}_2}^{v_{\mathrm{O}_2}}}{N_{\mathrm{CO}_2}^{v_{\mathrm{CO}_2}}}\left(\frac{P}{N_{\text {total }}}\right)^{\left(v_{\mathrm{CO}}+v_{\mathrm{O}_2}-v_{\mathrm{CO}_2}\right)}
$$ From Table A-28, $\ln K_p=-3.860$ at $2400 \mathrm{~K}$. Thus $K_p=0.02107$ Substituting,
$$
0.02107=\frac{(1-x)(0.5-0.5 x)^{1 / 2}}{x}\left(\frac{3}{1.5-0.5 x}\right)^{1.5-1}
$$ Solving for $x$, $$
x=0.936
$$ Thus the percentage of $\mathrm{CO}_2$ which dissociates into $\mathrm{CO}$ and $\mathrm{O}_2$ is
$$
1-0.936=0.064 \text { or } 6.4 \%
$$
