Question

Question: The triple bond in the ${\mathrm{N}}_2$molecule is very strong, but at high enough temperatures even it breaks down. At 5000 K, when the total pressure exerted by a sample of nitrogen is 1.00 atm,localid="1663413824968" ${\mathrm{N}}_2\left(\mathrm{g}\right)$islocalid="1663413838372" $0.65\%$dissociated at equilibrium:

${\mathrm{N}}_2\left(\mathrm{g}\right)\rightleftarrows 2\mathrm{N}\left(\mathrm{g}\right)$

At 6000 K with the same total pressure, the proportion of localid="1663414733966" ${\mathit{N}}_{\mathbf{2}\mathbf{}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$dissociated at equilibrium rises to$11.6\%$. Use the van’t Hoff equation to estimate thelocalid="1663413867221" $\mathrm{\Delta H}$of this reaction.

Short answer

The value of $\mathrm{\Delta H}$is 12252.08.

Step-by-step solution

We have to find K at 5000K which is$T_1$and at 6000K which is $T_2$

Step-by-step solution

At 5000K

When ${\mathrm{N}}_2\left(\mathrm{g}\right)$is$0.65\%$ dissociated at equilibrium the final equilibrium will be,

$$\begin{gathered} {\mathrm{N}}_2\left(\mathrm{g}\right)\rightleftarrows 2\mathrm{N}\left(\mathrm{g}\right) \\ \mathrm{I}1.00 \\ \mathrm{C}-0.0065+2\left(0.0065\right) \\ \mathrm{E}0.99350.0130 \end{gathered}$$

So,

$$\begin{gathered} {\mathrm{K}}_1=\frac{{\left({\mathrm{P}}_{\mathrm{N}}\right)}^2}{{\mathrm{P}}_{{\mathrm{N}}_2}} \\ =\frac{{\left(0.0130\right)}^2}{0.9935} \\ =1.70\times {10}^{-4} \end{gathered}$$

At 6000K

When${\mathrm{N}}_2\left(\mathrm{g}\right)$is $11.6\%$dissociated at equilibrium the final equilibrium will be,

$$\begin{gathered} {\mathrm{N}}_2\left(\mathrm{g}\right)\rightleftarrows 2\mathrm{N}\left(\mathrm{g}\right) \\ \mathrm{I}1.00 \\ \mathrm{C}-0.116+2\left(0.116\right) \\ \mathrm{E}0.8840.232 \end{gathered}$$

So,

$$\begin{gathered} {\mathrm{K}}_2=\frac{{\left({\mathrm{P}}_{\mathrm{N}}\right)}^2}{{\mathrm{P}}_{{\mathrm{N}}_2}} \\ =\frac{{\left(0.232\right)}^2}{0.884} \\ =6.09\times {10}^{-2} \end{gathered}$$

By using Van’t Hoff equation estimating the ΔH

According to Van’t Hoff equation,

$$\begin{gathered} ln\left(\frac{K_2}{K_1}\right)=-\frac{\Delta H°}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right] \\ \Rightarrow ln\left(\frac{6.09\times {10}^{-2}}{1.70\times {10}^{-4}}\right)=-\frac{\Delta H°}{0.083}\left[\frac{1}{6000}-\frac{1}{5000}\right] \\ \Rightarrow ln358.23=\frac{\Delta H°}{0.083}\times 0.00004 \\ \Rightarrow 5.881=\Delta H°\times 0.00048 \\ \Rightarrow \Delta H°=\frac{5.881}{0.00048} \\ =12252.08 \end{gathered}$$

Hence the value of $\mathrm{\Delta H}$is 12252.08.