Answer
$24.0\%$
Work Step by Step
The stoichiometric and actual reactions can be written as
Stoichiometric: $\mathrm{H}_2 \Leftrightarrow 2 \mathrm{H}$ (thus $v_{\mathrm{H}_2}=1$ and $v_{\mathrm{H}}=2$ )
Actual: $$
\mathrm{H}_2 \longrightarrow \underbrace{x \mathrm{H}_2}_{\text {react. }}+\underbrace{y \mathrm{H}}_{\text {prod. }}
$$ H balance:
$$
2=2 x+y \text { or } y=2-2 x
$$ Total number of moles:
$$
N_{\text {total }}=x+y=x+2-2 x=2-x
$$ The equilibrium constant relation can be expressed as
$$
K_p=\frac{N_{\mathrm{H}}^{v_{\mathrm{H}}}}{N_{\mathrm{H}_2}^{v_{\mathrm{H}_2}}}\left(\frac{P}{N_{\text {total }}}\right)^{v_{\mathrm{H}}-v_{\mathrm{H}_2}}
$$ From Table A-28, $\ln K_p=0.202$ at $3800 \mathrm{~K}$. Thus $K_p=1.224$. Substituting,
$$
1.224=\frac{(2-2 x)^2}{x}\left(\frac{5}{2-x}\right)^{2-1}
$$ Solving for $x$, $$
x=0.7599
$$ Thus the percentage of $\mathrm{H}_2$ which dissociates to $\mathrm{H}$ at $3200 \mathrm{~K}$ and $5 \mathrm{~atm}$ is $$
1-0.7599=0.240 \text { or } \mathbf{2 4 . 0} \%
$$
