Chapter 14, Problem 14.112QA

Interpretation Introduction

To find:

The value of equilibrium constant for the given reaction at ${500}^0\mathrm{C}=773\mathrm{ }\mathrm{K}$.

Answer to Problem 14.112QA

Solution:

The value of the equilibrium constant for the given reaction is $1.16\mathrm{ }\times {10}^7$ at $500℃.$

Explanation of Solution

1. Concept:

We are asked to calculate Kp from the two $∆{\mathrm{G}}^0$ values.We can use the relation between $∆{\mathrm{G}}^0$ and K to calculate K. Next, we calculate the Kp values from the relation of Kp and K.

2. Formula

i) $∆{\mathrm{G}}^0=-\mathrm{R}\mathrm{T}\mathrm{l}\mathrm{n}\mathrm{ }{K}_p$

ii) $\mathrm{ ln}{K}_p=\frac{-∆{\mathrm{G}}^0}{\mathrm{R}\mathrm{T}}$

3. Given information:

i) $T= {500}^0\mathrm{C}=773\mathrm{ }\mathrm{K}$

ii) ${3\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{ }\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{ }\mathrm{N}\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}$

iii) ${2\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}\rightleftharpoons 2\mathrm{ }{\mathrm{N}\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}∆{\mathrm{G}}^0=-69.7\mathrm{ }\mathrm{k}\mathrm{J}$(equation1)

iv) ${2\mathrm{ }\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}\rightleftharpoons {2\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{N}}_{2\mathrm{ }\left(\mathrm{g}\right)}∆{\mathrm{G}}^0=-33.8\mathrm{ }\mathrm{k}\mathrm{J}$(equation2)

v) ${\mathrm{N}}_{2\mathrm{ }\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}\rightleftharpoons {2\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}∆{\mathrm{G}}^0=173.2\mathrm{ }\mathrm{k}\mathrm{J}$(equation3)

4. Calculations:

To calculate the $∆{\mathrm{G}}^0$ for the following reaction

${3\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{ }\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{ }\mathrm{N}\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}$

We manipulated the three reactions to get the final reaction. For this, we have to reverse equation (2) and (3)

${2\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{N}}_{2\mathrm{ }\left(\mathrm{g}\right)}\rightleftharpoons {2\mathrm{ }\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}∆{\mathrm{G}}^0=33.8\mathrm{ }\mathrm{k}\mathrm{J}$---------- (4)

${2\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{N}}_{2\mathrm{ }\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}∆{\mathrm{G}}^0=-173.2\mathrm{ }\mathrm{k}\mathrm{J}$ ------- (5)

Now we need to add equations (1), (4) and (5).

${2\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}\mathrm{ }\rightleftharpoons \mathrm{ }{2\mathrm{ }\mathrm{N}\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}∆{\mathrm{G}}^0=-69.7\mathrm{ }\mathrm{k}\mathrm{J}$------------ (1)

${2\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{N}}_{2\mathrm{ }\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{ }2\mathrm{ }\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}∆{\mathrm{G}}^0=33.8\mathrm{ }\mathrm{k}\mathrm{J}$---------- (4)

${2\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}\mathrm{ }\mathrm{ }\rightleftharpoons \mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{N}}_{2\mathrm{ }\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}∆{\mathrm{G}}^0=-173.2\mathrm{ }\mathrm{k}\mathrm{J}$ ------- (5)

${6\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}\rightleftharpoons {2\mathrm{ }\mathrm{ }\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{ }2\mathrm{ }\mathrm{N}\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}∆{\mathrm{G}}^0=-209.1\mathrm{ }\mathrm{k}\mathrm{J}$ ------ (6)

After dividing equation (6) by 2, we can get the main reaction equation

${3\mathrm{ }\mathrm{N}\mathrm{O}}_{\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{ }\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{ }\mathrm{N}\mathrm{O}}_{2\mathrm{ }\left(\mathrm{g}\right)}\mathrm{ }∆{\mathrm{G}}^0=-104.55\mathrm{ }\mathrm{k}\mathrm{J}$

To find equilibrium constant at $773\mathrm{ }\mathrm{K}$ we have,

$\ln K_p=\frac{-∆{\mathrm{G}}^0}{\mathrm{R}\mathrm{T}}$

$\ln K_p=\frac{-(-104.55\mathrm{ }\times {10}^3\mathrm{ }\mathrm{J})}{(8.3145\mathrm{ }\mathrm{J}/\mathrm{K}.\mathrm{m}\mathrm{o}\mathrm{l}\left)\right(773\mathrm{ }\mathrm{K})}$

$\ln K_p=16.27$

$K_p=1.16\times {10}^7$

Conclusion:

By using the formula for standard free energy, we calculated the value of the equilibrium constant for the given reaction.