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Chapter 16: Q16.10 CYL (page 902)

Use the information in Appendix G to estimate the boiling point of CS2.

Short Answer

Expert verified

The estimated boiling point of CS2is 323 K.

Step by step solution

01

Define the enthalpy of the reaction

The change in Gibbs free energy is as follows:

\({\bf{\Delta G = \Delta H - T\Delta S}}\)

where,

\({\bf{\Delta G }}\)= change in Gibbs free energy,

\({\bf{\Delta H}}\)= change in enthalpy,

T = absolute temperature in Kelvin, and

\({\bf{\Delta S}}\)= change in entropy.

The Gibbs free energy change is used to determine the spontaneity of a process. It is expressed in terms of the enthalpy and the entropy of a system.

Entropy is the degree of disorderness or randomness in a given system. The entropy change during a transition phase is expressed as

\({\bf{\Delta S = }}\frac{{{\bf{\Delta H}}}}{{\bf{T}}}\)

02

Determine the estimate the boiling point of CS2

Given:

\({\rm{\Delta }}{{\rm{H}}^ \circ }f({\rm{KJ/mol}})\)

\({{\rm{S}}^ \circ }f({\rm{J/Kmol}})\)

\({\rm{CS2}}\)(g)

115.3

237.8

\({\rm{CS2}}\)(l)

87.3

151

Carbon disulfide vaporization is given by

\({\rm{C}}{{\rm{S}}_{{\rm{2(l)}}}} \to {\rm{C}}{{\rm{S}}_{{\rm{2(g)}}}}\)

Solve for enthalpy change of vaporization

\(\Delta {H_{vap}} = \Delta H_{f\left( {C{S_{2(g)}}} \right.}^o - \Delta H_{f\left( {C{S_{2(l)}}} \right.}^o = 115.3 - 87.3 = 28\frac{{kJ}}{{ mole }}\)

Solve for entropy change of vaporization

\({\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}{\rm{ = \Delta S}}_{{\rm{f}}\left( {{\rm{C}}{{\rm{S}}_{{\rm{2(g)}}}}} \right.}^{\rm{o}}{\rm{ - \Delta S}}_{{\rm{f}}\left( {{\rm{C}}{{\rm{S}}_{{\rm{2(l)}}}}} \right.}^{\rm{o}}{\rm{ = 237}}{\rm{.8 - 151 = 86}}{\rm{.8}}\frac{{\rm{J}}}{{{\rm{ mole \times K}}}}\)

Solve boiling point using entropy definition for phase changes

\(\Delta {S_{vap}} = \frac{{\Delta {H_{vap}}}}{{{T_b}}}\)

\(\begin{array}{l}{{\rm{T}}_{\rm{b}}}{\rm{ = }}\frac{{{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}}}{{{\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}}}\\{\rm{ = }}\frac{{{\rm{28000}}\frac{{\rm{J}}}{{{\rm{ mole }}}}}}{{{\rm{86}}{\rm{.8}}\frac{{\rm{J}}}{{{\rm{ mole \times K}}}}}}\\{\rm{ = 323K}}\end{array}\)

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Most popular questions from this chapter

Calculate ΔG° using

(a) free energies of formation and

(b) enthalpies of formation and entropies(Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?

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(c) Hydrazine: \({{\bf{N}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(d) Hydrogen peroxide: \({{\bf{H}}_{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(l)}} \to {{\bf{H}}_{\bf{2}}}{\bf{O(g) + }}\frac{{\bf{1}}}{{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

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