Question

Use the result of the previous problem to estimate the equilibrium constant of the reaction${\mathrm{N}}_2+3{\mathrm{H}}_2\leftrightarrow 2{\mathrm{NH}}_3$at 500° C, using only the room- temperature data at the back of this book. Compare your result to the actual value of K at 500° C quoted in the text.

Short answer

Therefore, the equilibrium constant at temperature 500°C is$6.528\times {10}^{-5}$

Step-by-step solution

Given information

Reaction

${\mathrm{N}}_2+3{\mathrm{H}}_2\leftrightarrow 2{\mathrm{NH}}_3$

The equilibrium constant from previous problem

$\ln \left(K\left(T_2\right)\right)=\ln \left(K\left(T_1\right)\right)+\frac{\Delta H°}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$

Explanation

We have the reaction

${\mathrm{N}}_2+3{\mathrm{H}}_2\leftrightarrow 2{\mathrm{NH}}_3$

The equilibrium constant from the previous problem is:

role="math" localid="1647210871040" $\ln \left(K\left(T_2\right)\right)=\ln \left(K\left(T_1\right)\right)+\frac{\Delta H°}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$

Where,

$K\left(T_1\right)$is the equilibrium constant at room temperature and is given by

$K\left(T_1\right)=5.9\times {10}^5$

$\Delta H°$is the enthalpy at standard condition and is given by

$\begin{array}{cc}{H}_{{\mathrm{N}}_2}& 0\mathrm{kJ}\\ H_{{\mathrm{H}}_2}& 0\mathrm{kJ}\\ H_{{\mathrm{NH}}_3}& -46.11\mathrm{kJ}\end{array}$

From this table the change in the enthalpy at the standard condition is:

$$\begin{gathered} \Delta H°=2H_{{\mathrm{NH}}_3}-H_{{\mathrm{N}}_2}-H_{{\mathrm{H}}_2} \\ =2(-46.11\mathrm{kJ})-(0\mathrm{kJ})-(0\mathrm{kJ}) \\ =-92.22\mathrm{kJ} \end{gathered}$$

Substitute the values in (1), we get
role="math" localid="1647212511484" $$\begin{gathered} \ln \left(K\left(T_2\right)\right)=\ln \left(5.6\times {10}^5\right)-\frac{92.22\times {10}^3\mathrm{J}}{8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}}\left[\frac{1}{298\mathrm{K}}-\frac{1}{773\mathrm{K}}\right] \\ \ln \left(K\left(T_2\right)\right)=-9.6368 \end{gathered}$$

Take the natural logarithm,

role="math" localid="1647212684960" $$\begin{gathered} K\left(T_2\right)=e^{-9.6368} \\ K\left(T_2\right)=6.528\times {10}^{-5} \end{gathered}$$