Answer
$T=3133 \mathrm{~K}$
Work Step by Step
The stoichiometric and actual reactions can be written as
Stoichiometric $\mathrm{O}_2 \Leftrightarrow 2 \mathrm{O}$ (thus $v_{\mathrm{O}_2}=1$ and $v_{\mathrm{O}}=2$ )
Actual: $\quad \mathrm{O}_2 \Leftrightarrow \underbrace{0.95 \mathrm{O}_2}_{\text {react. }}+\underbrace{0.10}_{\text {prod. }}$
The equilibrium constant $K_p$ can be determined from
$$
K_p=\frac{N_{\mathrm{O}}^{v_0}}{N_{\mathrm{O}_2}^{v_{\mathrm{O}_2}}}\left(\frac{P}{N_{\text {total }}}\right)^{v_0-v_{\mathrm{O}_2}}=\frac{0.1^2}{0.95}\left(\frac{3}{0.95+0.1}\right)^{2-1}=0.03008
$$ From Table A-28, the temperature corresponding to this $K_p$ value is $$
T=3133 \mathrm{~K}
$$
