Answer
See explanation
Work Step by Step
(a) The equilibrium constant for the reaction $\mathrm{N}_2 \Leftrightarrow 2 \mathrm{~N}$ can be expressed as
$$
K_p=\frac{N_{\mathrm{N}}^{v_{\mathrm{N}}}}{N_{\mathrm{N}_2}^{v_{\mathrm{N}_2}}}\left(\frac{P}{N_{\text {total }}}\right)^{\left(v_{\mathrm{N}}-v_{\mathrm{N}_2}\right)}
$$ Judging from the values in Table A-28, the $K_p$ value for this reaction increases as the temperature increases. That is, the indicated reaction will be more complete at higher temperatures. Therefore, the number of moles of $\mathrm{N}$ will increase and the number moles of $\mathrm{N}_2$ will decrease as the temperature increases.
(b) The value of the exponent in this case is 2-1 $=1$, which is positive. Thus as the pressure increases, the term in the brackets also increases. The value of $K_p$ depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products $(\mathrm{N})$ must decrease, and the number of moles of the reactants $\left(\mathrm{N}_2\right)$ must increase.
