In this example, n0 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 α=δeq/n0 , where is 2δeq the number of moles of Cl(g)

Question

In this example, [latex]n_{0}[/latex] 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 [latex]\alpha=\delta_{eq}/n_{0}[/latex] , where is [latex]2\delta_{eq}[/latex] the number of moles of [latex] Cl\left( g \right)[/latex] present at equilibrium, and [latex]n_{0}[/latex] represents the number of moles of [latex]Cl_{2}\left( g \right)[/latex] that would be present in the system if [latex]n_{0},[/latex] dissociation occurred. Derive an expression for [latex]K_{P}[/latex] in terms of [latex]n_{0},\delta_{eq}[/latex], and P.
b. Derive an expression for [latex]\alpha[/latex] as a function of [latex]K_{P}[/latex] and P.

Accepted Answer

a. We set up the following table:

	[latex]Cl_{2}\left( g \right)[/latex]	[latex]\rightleftharpoons [/latex]	[latex]2Cl\left( g \right)[/latex]
Initial number of moles	[latex]n_{0}[/latex]		0
Moles present at equilibrium	[latex]n_{0}-\delta_{eq}[/latex]		[latex]2\delta_{eq}[/latex]
Mole fraction present at equilibrium,[latex]x_{i}[/latex]	[latex]\frac{n_{0}-\delta_{eq}}{n_{0}+\delta_{eq}}[/latex]		[latex]\frac{2\delta_{eq}}{n_{0}+\delta_{eq}}[/latex]
Partial pressure at equilibrium,[latex]P_{i}=x_{i}P[/latex]	[latex]\left( \frac{n_{0}-\delta_{eq}}{n_{0}+\delta_{eq}} \right)P[/latex]		[latex]\left( \frac{2\delta_{eq}}{n_{0}+\delta_{eq}}\right)P[/latex]

We next express [latex]K_{P}[/latex] in terms of [latex]n_{0},\delta_{eq}[/latex] and P:

[latex]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°}}[/latex]

[latex]=\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}[/latex]

This expression is converted into one in terms of [latex]\alpha[/latex] :

[latex]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°}[/latex]

b. [latex]\left( K_{P}\left( T \right)+4\frac{P}{P°} \right)\alpha^{2}=K_{P}\left( T \right)[/latex]

[latex]\alpha=\sqrt{\frac{K_{P}\left( T \right)}{K_{P}\left( T \right)+4\frac{P}{P°}}}[/latex]

Because [latex]\left( K_{P}\left( T \right)\right)[/latex] depends strongly on temperature, will also be a strong function of temperature. Note that also depends on both a [latex]K_{P}[/latex] and P for this reaction.

Question 6.10: In this example, n0 moles of chlorine gas are placed in a re...

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