Answer
(a) $T=3628 \mathrm{~K}$
(b) $T=3909 \mathrm{~K}$
Work Step by Step
(a) The stoichiometric and actual reactions can be written as
Stoichiometric $\mathrm{N}_2 \Leftrightarrow 2 \mathrm{N}$ (thus $v_{\mathrm{N}_2}=1$ and $v_{\mathrm{N}}=2$ )
Actual: $\quad \mathrm{N}_2 \Leftrightarrow \underbrace{0.998 \mathrm{N}_2}_{\text {react. }}+\underbrace{0.004N}_{\text {prod. }}$
The equilibrium constant $K_p$ can be determined from $$
K_p=\frac{N_{\mathrm{N}}^{v_N}}{N_{\mathrm{N}_2}^{v_{\mathrm{N}_2}}}\left(\frac{P}{N_{\text {total }}}\right)^{v_N-v_{\mathrm{N}_2}}=\frac{0.004^2}{0.998}\left(\frac{1/101.325}{0.998+0.004}\right)^{2-1}=1.579\times 10^{-7}
$$ and $$\ln K_p=-15.66.$$ From Table A-28, the temperature corresponding to this $K_p$ value is $$
T=3628 \mathrm{~K}
$$ (b) At $10\text{ kPa}$, $$
K_p=\frac{N_{\mathrm{N}}^{v_N}}{N_{\mathrm{N}_2}^{v_{\mathrm{N}_2}}}\left(\frac{P}{N_{\text {total }}}\right)^{v_N-v_{\mathrm{N}_2}}=\frac{0.004^2}{0.998}\left(\frac{10/101.325}{0.998+0.004}\right)^{2-1}=1.579\times 10^{-6}
$$ and $$\ln K_p=-13.36.$$ From Table A-28, the temperature corresponding to this $K_p$ value is $$
T=3909 \mathrm{~K}
$$
