Q16.10 CYL
Question
Use the information in Appendix G to estimate the boiling point of CS2.
Step-by-Step Solution
VerifiedThe estimated boiling point of CS2 is 323 K.
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
\({\bf{\Delta S = }}\frac{{{\bf{\Delta H}}}}{{\bf{T}}}\)
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
\(\Delta {S_{vap}} = \frac{{\Delta {H_{vap}}}}{{{T_b}}}\)
\(\begin{array}{l}{{\rm{T}}_{\rm{b}}}{\rm{ = }}\frac{{{\rm{\Delta }}{{\rm{H}}_{{\rm{vap}}}}}}{{{\rm{\Delta }}{{\rm{S}}_{{\rm{vap}}}}}}\\{\rm{ = }}\frac{{{\rm{28000}}\frac{{\rm{J}}}{{{\rm{ mole }}}}}}{{{\rm{86}}{\rm{.8}}\frac{{\rm{J}}}{{{\rm{ mole \times K}}}}}}\\{\rm{ = 323K}}\end{array}\)