Question 6.10: In this example, n0 moles of chlorine gas are placed in a re...
In this example, n_{0} moles of chlorine gas are placed in a reaction vessel whose temperature can be varied over a wide range, so that molecular chlorine can partially dissociate to atomic chlorine.
a. Define the degree of dissociation as \alpha=\delta_{eq}/n_{0} , where is 2\delta_{eq} the number of moles of Cl\left( g \right) present at equilibrium, and n_{0} represents the number of moles of Cl_{2}\left( g \right) that would be present in the system if n_{0}, dissociation occurred. Derive an expression for K_{P} in terms of n_{0},\delta_{eq}, and P.
b. Derive an expression for \alpha as a function of K_{P} and P.
a. We set up the following table:
| Cl_{2}\left( g \right) | \rightleftharpoons | 2Cl\left( g \right) | |
| Initial number of moles | n_{0} | 0 | |
| Moles present at equilibrium | n_{0}-\delta_{eq} | 2\delta_{eq} | |
| Mole fraction present at equilibrium,x_{i} | \frac{n_{0}-\delta_{eq}}{n_{0}+\delta_{eq}} | \frac{2\delta_{eq}}{n_{0}+\delta_{eq}} | |
| Partial pressure at equilibrium,P_{i}=x_{i}P | \left( \frac{n_{0}-\delta_{eq}}{n_{0}+\delta_{eq}} \right)P | \left( \frac{2\delta_{eq}}{n_{0}+\delta_{eq}}\right)P |
We next express K_{P} in terms of n_{0},\delta_{eq} and P:
K_{p}\left( T \right)=\frac{\left( \frac{P_{Cl}^{eq}}{P°} \right)^{2}}{\left( \frac{P_{Cl_{2}}^{eq}}{P°} \right)}=\frac{\left[ \left( \frac{2\delta_{eq}}{n_{0}+\delta_{eq}} \right)\frac{P}{P°} \right]^{2}}{\left( \frac{n_{o}+\delta_{eq}}{n_{0}+\delta_{eq}} \right)\frac{P}{P°}}=\frac{4\delta^{2}_{ea}}{\left( n_{0}+\delta_{eq} \right)\left( n_{0}-)\delta_{eq}\right)}\frac{P}{P}=\frac{4\delta_{eq}^{2}}{\left( n_{0} \right)^{2}-\delta_{eq}^{2}}\frac{P}{P}
This expression is converted into one in terms of \alpha :
K_{P}\left( T \right)=\frac{4\delta_{eq}^{2}}{\left( n_{0} \right)^{2}-\delta^{2}_{eq}}\frac{P}{P°}=\frac{4\alpha^{2}}{1-\alpha^{2}}\frac{P}{P°}b. \left( K_{P}\left( T \right)+4\frac{P}{P°} \right)\alpha^{2}=K_{P}\left( T \right)
\alpha=\sqrt{\frac{K_{P}\left( T \right)}{K_{P}\left( T \right)+4\frac{P}{P°}}}
Because \left( K_{P}\left( T \right)\right) depends strongly on temperature, will also be a strong function of temperature. Note that also depends on both a K_{P} and P for this reaction.