# 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.

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Question: 6.11

## Using the result of Example Problem 6.10 and the data tables, consider the dissociation equilibrium Cl2(g)→←2Cl(g).a. Calculate KP at 800. K, 1500. K, and 2000. K for P=0.010 bar. b. Calculate the degree of dissociation α at 300. K, 1500. K, and 2000. K. ...

a.   \Delta G°_{R}=2\Delta G°_{f}\left( Cl,...
Question: 6.9

## 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? ...

a.  \ln K_{P}=-\frac{1}{RT}\Delta G°_{R}[/l...
Question: 6.16

## In the previous Example Problem, it was assumed that CP=CV. a. Use equation 3.38, CP,m=Cv,m +TVm β²/κ, and the experimentally determined value CP,m=27.98 J mol-¹ to obtain a value for CV,m for Hg(l) at 300. K.b. Did the assumption that CP,m≈CV,m in Example Problem 6.15 introduce an appreciable ...

a.   C_{P,m}-C_{V,m}=\frac{TV_{m}\beta^{2}}...
Question: 6.15

## A 1000. g sample of liquid Hg undergoes a reversible transformation from an initial state Pi = 1.00 bar , Ti=300. K to a final state Pf =300. bar, Tf=600. K. The density of Hg is 13,534 kg m-³,β =1.81× 10^-4 K-¹, CP=27.98 J mol-¹ K-¹, and κ =3.91 × 10^-6 bar-¹, in the range of the variables in this ...

Because U is a state function, \Delta U[/la...
Question: 6.17

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

a. Because Ar $\left( g\right)$ is an...
Question: 6.14

## At 298 K, the thermal expansion coefficient and the isothermal compressibility of liquid water are β =2.04 × 10-^4 K-¹ and κ =45.9 × 10-^6 bar-¹.a. Calculate (∂U/∂V)T  for water at 320. K and Ρ= 1.00 bar. b. If an external pressure equal to (∂U/∂V)T were applied to 1.00 m³ of liquid water at 320.K ...

a. \left( \frac{\partial U}{\partial V} \ri...
Question: 6.13

## At 298.15 K ΔG°f (C,graphite) =0, and ΔG°f (C,diamond)  =2.90 KJ mol-¹ Therefore, graphite is the more stable solid phase at this temperature at P=P°=1 bar . Given that the densities of graphite and diamond are 2.25 and 3.52 kg L, respectively, at what pressure will graphite and diamond be in ...

At equilibrium \Delta G =G\left( C, graphit...
Question: 6.12

## 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. ...

For the reaction CaCo_{3}\left( s \right) \...
Question: 6.8

## 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? ...

The reaction of interest is 2 NO_{2}\left( ...
To calculate $\Delta G°_{R}$ at the e...