Question

In the activity, select CH3Cl. Click on the Run button, and observe the graph of standard molar entropy versus temperature. Observe the equilibrium position, in which the solid and liquid phases exist together. Notice that the entropy increases from 79 J⋅mol−1 K−1 to 116 J⋅mol−1 K−1 as CH3Cl melts.

Calculate the enthalpy of fusion (ΔHfus) at this equilibrium.

Express the change in enthalpy (ΔHfus) to two significant figure

Answer 1

Concepts and reason

The concept used to solve this question is calculating the standard Gibbs free energy.

Mathematically, the relation between standard Gibbs free energy and enthalpy is defined as follows:

${\rm{\Delta G = \Delta {\rm H} - T\Delta S}}$

Here, ${\rm{\Delta G}}$ is the free energy change, ${\rm{\Delta H}}$ is the enthalpy change, T is absolute temperature, and ${\rm{\Delta S}}$ is entropy change.

Fundamentals

To calculate enthalpy of fusion, it is needed to calculate Gibbs free energy change and entropy change at equilibrium position. Calculate entropy change from the given values of entropy increases form ${\rm{79 J}}{\rm{.mo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{.}}{{\rm{K}}^{{\rm{ - 1}}}}$ to ${\rm{116 J}}{\rm{.mo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{.}}{{\rm{K}}^{{\rm{ - 1}}}}$. At equilibrium Gibbs free energy change is zero.

From these values, calculate enthalpy of fusion using the following formula:

${\rm{\Delta G = \Delta {\rm H} - T\Delta S}}$ …… (1)

Here, ${\rm{\Delta G}}$ is the free energy change, ${\rm{\Delta H}}$ is the enthalpy change, T is absolute temperature, and ${\rm{\Delta S}}$ is entropy change.

$\begin{array}{l}\\{\rm{\Delta S = }}\left( {{\rm{116 - 79}}} \right){\rm{J}}{\rm{.mo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{.}}{{\rm{K}}^{{\rm{ - 1}}}}\\\\{\rm{ = 37 J}}{\rm{.mo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{.}}{{\rm{K}}^{{\rm{ - 1}}}}\\\end{array}$

And at equilibrium change in free energy is ‘Zero’.

Calculate enthalpy of fusion as follows:

${\rm{\Delta G = \Delta {\rm H} - T\Delta S}}$

Substitute the values of ${\rm{\Delta G}}$, ${\rm{\Delta S}}$, and T in the above equation.

$\begin{array}{l}\\{\rm{\Delta G = \Delta {\rm H} - T\Delta S}}\\\\{\rm{\Delta {\rm H}}} = {\rm{\Delta G}} + {\rm{T\Delta S}}\\\\{\rm{\Delta {\rm H} = 0 + }}\left( {{\rm{175}}{\rm{.8K \times }}\left( {{\rm{37}}} \right){\rm{J}}{\rm{.mo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{.}}{{\rm{K}}^{{\rm{ - 1}}}}} \right)\\\\{\rm{\Delta {\rm H} = 6,504 J/mol}}\\\\{\rm{ = 6}}{\rm{.5}} \times {\rm{1}}{{\rm{0}}^3}{\rm{ J/mol}}\\\end{array}$

Therefore, the enthalpy of fusion in two significant figures is .

Ans:

Therefore, the enthalpy of fusion in two significant figures is .