Q20.110CP - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the formation of ammonia: ${\mathbf{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{2}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

(a) Assuming that $\mathbf{\Delta H}\mathbf{°}$ and $\mathbf{\Delta S}\mathbf{°}$are constant with temperature, find the temperature at which  ${\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}$ . 

(b) Find ${\mathbf{K}}_{\mathbf{p}}$at 400°C, a typical temperature for ${\mathbf{NH}}_{\mathbf{3}}$production. 

(c) Given the lower ${\mathbf{K}}_{\mathbf{p}}$ at the higher temperature, why are these conditions used industrially?

Step-by-Step Solution

Answer

a) At a temperature of 465.28 K, ${\mathrm{K}}_{\mathrm{p}}=1$.

b) At ${400}^{\mathrm{o}}\mathrm{C}$, the equilibrium constant is $6.592\times {10}^{-4}$.

c) They seek to manufacture less product in a shorter period of time in the sector

Step 1: Definition of Equilibrium

Equilibrium refers to a situation in which the concentrations of reactants and products do not change significantly. Although there appears to be no change in equilibrium, this does not imply that all chemical reactions have stopped.

Step 2: Calculating temperature at K p = 1 . 00

1. At first, glance ${\mathrm{K}}_{\mathrm{p}}=1.00$ implies that $\mathrm{\Delta G}=0$, because $\ln 1=0$.

$\mathrm{\Delta G}=-\mathrm{R}\times \mathrm{T}\times \mathrm{lnK}$

As a result, the reaction has reached equilibrium. Calculate the ${\mathrm{\Delta H}}^{\mathrm{o}}$ and ${\mathrm{\Delta S}}^{\mathrm{o}}$ for the reaction, assuming they are unaffected by temperature changes, to obtain the equilibrium temperature.

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

• Because entropy is a state function, it can be stated in the following way:

${\mathrm{\Delta S}}^{\mathrm{o}}=\sum _{}^{}{\mathrm{\Delta S}}_{\mathrm{products} }^{\mathrm{o}}-\sum _{}^{}{\mathrm{\Delta S}}_{\mathrm{reactants} }^{\mathrm{o}}$

• In the case of the stated reaction,

$$\begin{gathered} {\mathrm{\Delta S}}^{\mathrm{o}}=\left[{2\mathrm{mol}\times \mathrm{\Delta S}}_{{\mathrm{NH}}_3\left(\mathrm{g}\right)}^{\mathrm{o}}\right]-\left[{1\mathrm{mol}\times \mathrm{\Delta S}}_{{\mathrm{N}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}{+3\mathrm{mol}\times \mathrm{\Delta S}}_{{\mathrm{H}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}\right] \\ {\mathrm{\Delta S}}^{\mathrm{o}}=\left[2\mathrm{mol}\times 193.00\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\right]-\left[3\mathrm{mol}\times 130.60\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}+1\mathrm{mol}\times 191.50\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\right] \\ {\mathrm{\Delta S}}^{\mathrm{o}}=-197.30\frac{\mathrm{J}}{\mathrm{K}} \end{gathered}$$

Step 3: Calculating enthalpy

• Enthalpy can be calculated in the same way:

$$\begin{gathered} {\mathrm{\Delta H}}^{\mathrm{o}}=\left[{2\mathrm{mol}\times \mathrm{\Delta H}}_{{\mathrm{NH}}_3\left(\mathrm{g}\right)}^{\mathrm{o}}\right]-\left[{1\mathrm{mol}\times \mathrm{\Delta H}}_{{\mathrm{N}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}{+3\mathrm{mol}\times \mathrm{\Delta H}}_{{\mathrm{H}}_2\left(\mathrm{g}\right)}^{\mathrm{o}}\right] \\ {\mathrm{\Delta H}}^{\mathrm{o}}=[2\mathrm{mol}\times (-45.9\mathrm{kJ}/\mathrm{mol}\left)\right]-[1\mathrm{mol}\times (0.00\mathrm{kJ}/\mathrm{mol})+3\mathrm{mol}\times (0.00\mathrm{kJ}/\mathrm{mol}\left)\right] \\ {\mathrm{\Delta H}}^{\mathrm{o}}=-91.8\mathrm{kJ} \end{gathered}$$

• Lastly,

$$\begin{gathered} {\mathrm{\Delta G}}^{\mathrm{o}}={\mathrm{\Delta H}}^{\mathrm{o}}-\mathrm{T}\times {\mathrm{\Delta S}}^{\mathrm{o}}=0 \\ \mathrm{T}=\frac{{\mathrm{\Delta H}}^{\mathrm{o}}}{{\mathrm{\Delta S}}^{\mathrm{o}}} \\ \mathrm{T}=\frac{-91.8\times {10}^3\mathrm{J}}{-197.30\frac{\mathrm{J}}{\mathrm{K}}} \\ \mathrm{T}=465.28\mathrm{K} \end{gathered}$$

At a temperature of $465.28 \mathrm{K}$, ${\mathrm{K}}_{\mathrm{p}}=1$.

Step 4: Calculating K p at 400

(b) 

Assume $\mathrm{T}=400{ }^{\mathrm{o}}\mathrm{C}=673 \mathrm{K}$.

$$\begin{gathered} \mathrm{\Delta G}=\left[-91.8\times {10}^3\mathrm{J}\right]-673\mathrm{K}\times \left[-197.30\frac{\mathrm{J}}{\mathrm{K}}\right] \\ \mathrm{\Delta G}=40982.9\mathrm{J}=40.983\mathrm{kJ} \end{gathered}$$

• Then the ${\mathrm{K}}_{\mathrm{p}}$ is as follows:

$\begin{array}{rcl}\mathrm{\Delta G}& =& -\mathrm{R}\times \mathrm{T}\times {\mathrm{lnK}}_{\mathrm{p}}\\ {\mathrm{lnK}}_{\mathrm{p}}& =& -\frac{\mathrm{\Delta G}}{\mathrm{R}\times \mathrm{T}}\\ {\mathrm{lnK}}_{\mathrm{p}}& =& -\frac{40982.9\mathrm{J}}{8.314\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\times 673\mathrm{K}}\\ {\mathrm{lnK}}_{\mathrm{p}}& =& -7.32449\\ {\mathrm{K}}_{\mathrm{p}}& =& {\mathrm{e}}^{{\mathrm{lnK}}_{\mathrm{p}}}\\ {\mathrm{K}}_{\mathrm{p}}& =& 6.592\times {10}^{-4}\end{array}$

At ${400}^{\mathrm{o}}\mathrm{C}$, the equilibrium constant is $6.592\times {10}^{-4}$.

Step 5: Determining why these conditions are used industrially

(c)

 A lower ${\mathrm{K}}_{\mathrm{p}}$ indicates that product formation is less favorable, resulting in a lower yield.

The process is faster at higher temperatures because the kinetic energy of molecules and entropy are both larger.

They strive to manufacture fewer products in a shorter amount of time in the industry.