Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ equilibrium constants $\left(K_{\mathrm{p}} \text { values ) as a function of temperature }\right.\( are\)\begin{array}{ll}{300^{\circ} \mathrm{C},} & {4.34 \times 10^{-3}} \\ {500^{\circ} \mathrm{C},} & {1.45 \times 10^{-5}} \\\ {600^{\circ} \mathrm{C},} & {2.25 \times 10^{-6}}\end{array}$ Is the reaction exothermic or endothermic?

First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature. For the given equilibrium constants, we have: \[\begin{array}{ll}{T_1 = 300^{\circ} \mathrm{C} + 273.15 = 573.15 \mathrm{K},} & {K_{p1} = 4.34 \times 10^{-3}} \\{T_2 = 500^{\circ} \mathrm{C} + 273.15 = 773.15 \mathrm{K},} & {K_{p2} = 1.45 \times 10^{-5}} \\{T_3 = 600^{\circ} \mathrm{C} + 273.15 = 873.15 \mathrm{K},} & {K_{p3} = 2.25 \times 10^{-6}}\end{array}\] Now, we will use the Van't Hoff equation to analyze the relationship between \(K_p\) and temperature.

We want to see how the equilibrium constant \(K_p\) changes with temperature. If the reaction were exothermic, increasing temperature would result in a decrease in \(K_p\). Conversely, if the reaction were endothermic, increasing temperature would result in an increase in \(K_p\). By looking at the given equilibrium constants and the associated temperatures, we see that as temperature increases, the equilibrium constant \(K_p\) decreases: - At \(T_1 = 573.15\,\mathrm{K}\), \(K_{p1} = 4.34 \times 10^{-3}\) - At \(T_2 = 773.15\,\mathrm{K}\), \(K_{p2} = 1.45 \times 10^{-5}\) - At \(T_3 = 873.15\,\mathrm{K}\), \(K_{p3} = 2.25 \times 10^{-6}\) Since \(K_p\) decreases as temperature increases, this means \(\frac{d(\ln K_p)}{dT} < 0\). We can now look at the Van't Hoff equation and determine the sign of \(\Delta H^{\circ}\).

Recall the Van't Hoff equation: \[\frac{d(\ln K_p)}{dT} = \frac{\Delta H^{\circ}}{RT^2}\] We found that \(K_p\) decreases as temperature increases (\(\frac{d(\ln K_p)}{dT} < 0\)). Therefore, the right side of the equation must also be negative: \[\frac{\Delta H^{\circ}}{RT^2} < 0\] Since \(R\) and \(T^2\) are always positive, we can conclude that \(\Delta H^{\circ} < 0\). Now that we know the sign of \(\Delta H^{\circ}\) is negative, we can classify the reaction.

Since \(\Delta H^{\circ} < 0\), the reaction is exothermic. In other words, the reaction releases heat as it proceeds.