Using the result of Example Problem 6.10 and the data tables, consider the dissociation equilibrium $Cl_{2}\rightleftharpoons 2Cl\left( g \right)$ .

a. Calculate $K_{P}$ at 800. K, 1500. K, and 2000. K for P=0.010 bar.

b. Calculate the degree of dissociation $\alpha$ at 300. K, 1500. K, and 2000. K.

Question

Using the result of Example Problem 6.10 and the data tables, consider the dissociation equilibrium [latex]Cl_{2}\rightleftharpoons 2Cl\left( g \right)[/latex].

a. Calculate [latex]K_{P}[/latex] at 800. K, 1500. K, and 2000. K for P=0.010 bar.

b. Calculate the degree of dissociation [latex]\alpha[/latex] at 300. K, 1500. K, and 2000. K.

Accepted Answer

a. [latex]\Delta G°_{R}=2\Delta G°_{f}\left( Cl,g \right)-\Delta G°_{f}\left( Cl_{2},g \right)=2×105.7 ×10^{3} J mol^{-1} -0[/latex]

[latex]=211.4 kJ mol^{-1}[/latex]

[latex]\Delta H°_{R}=2\Delta H°_{f}\left( Cl,g \right)-\Delta H°_{f}\left( Cl_{2},g \right)=2×121.3 ×10^{3} J mol^{-1}-0[/latex]

[latex]=242.6 kJ mol^{-1}[/latex]

[latex]\ln K_{P}\left( T_{f} \right)=-\frac{\Delta G°_{R}}{RT}-\frac{\Delta H°_{R}}{R}\left( \frac{1}{T_{f}} -\frac{1}{298.15 K}\right)[/latex]

[latex]=-\frac{2114 ×10^{3} J mol^{-1}}{8.314 J K^{-1} mol^{-1} ×298.15 k}-\frac{242.6×10^{3} J mol^{-1} }{8.314 J K^{-1} mol^{-1}}[/latex]

[latex]×\left( \frac{1}{T_{f}}-\frac{1}{298.15 K}\right)[/latex]

[latex]\ln K_{P}\left( 800. K \right)[/latex]

[latex]=-\frac{211.4 ×10^{3} J mol^{-1}}{8.314 J K^{-1}×298.15 K} -\frac{242.6×10^{3} J mol^{-1}}{8.314 J K^{-1} mol^{-1}}[/latex]

[latex]×\left( \frac{1}{800. K} -\frac{1}{298.15 K}\right)=-23.888[/latex]

[latex]K_{P}\left( 800. K \right)=4.22×10^{-11}[/latex]

The values for [latex]K_{P}[/latex] at 1500. and 2000. K are [latex]1.03×10^{-3}[/latex] and 0.134, respectively.

b. The value of [latex]\alpha[/latex] at 2000. K is given by

[latex]\alpha=\sqrt{\frac{k_{P}\left( T \right)}{K_{P}\left( T \right)+4\frac{P}{P°}}}=\sqrt{\frac{0.134}{0.134+4 ×0.01}}=0.878[/latex]

The values of [latex]\alpha[/latex] at 1500. and 800. K are 0.159 and [latex]3.23×10^{-5}[/latex], respectively.

Question 6.11: Using the result of Example Problem 6.10 and the data tables...

Related Answered Questions

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) ...

Verified Answer:

a. Using data from Table 4.1 (see Appendix A, Data Tables), calculate KP at 298.15 K for the reaction (CO(g)+H2O(l)→CO2(g)+H2(g). b. Based on the value that you obtained for part (a), do you expect the mixture to consist mainly of CO2(g) and H2(g) or mainly of CO(g) +H2O(l) at equilibrium? ...

Verified Answer:

Verified Answer:

Verified Answer:

Calculate the conventional molar Gibbs energy of (a) Ar(g) and (b) H2O(l) at 1 bar and 298.15 K. ...

Verified Answer:

Verified Answer:

Verified Answer:

Using the preceding discussion and the tabulated values of ΔG°f and ΔH°f in Appendix A, calculate the CO2(g) pressure in equilibrium with a mixture of CaCO3(s) and CaO(s) at 1000.,1100.,and 1200. K. ...

Verified Answer:

NO2(g) and N2O4(g) are in a reaction vessel with partial pressures of 0.350 and 0.650 bar, respectively, at 298 K. Is this system at equilibrium? If not, will the system move toward reactants or products to reach equilibrium? ...

Verified Answer:

Calculate ΔG°R for the reaction in Example Problem 6.6 at 325 K. Is the reaction spontaneous at this temperature? ...

Verified Answer: