Q16.11 CYL - Chemistry (OpenStax)

Question

Calculate the free energy change for the same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?

Step-by-Step Solution

Answer

The calculated boiling point of CS₂ is \(323K\).

Step1: Define the enthalpy of the reaction

The change in Gibbs free energy is as follows:

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

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

 

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

Step 2: Determine the estimate the boiling pont of CS 2

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

\(C{S_{2(l)}} \to C{S_{2(g)}}\)

 

Solving for enthalpy change of vaporization:

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

 

Solving for entropy change of vaporization:

\(\Delta {S_{vap}} = \Delta S_{f\left( {C{S_{2(g)}}} \right.}^o - \Delta S_{f\left( {C{S_{2(l)}}} \right.}^o = 237.8 - 151 = 86.8\frac{J}{{{\rm{ mole }} \cdot K}}\)

 

Solving for boiling point using the entropy definition for phase changes: 

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

\(\begin{array}{l}{T_b} = \frac{{\Delta {H_{vap}}}}{{\Delta {S_{vap}}}}\\ = \frac{{28000\frac{J}{{{\rm{ mole }}}}}}{{86.8\frac{J}{{{\rm{ mole }} \cdot K}}}}\\ = 323K\end{array}\).