Question

Question: Two different crystalline forms of sulphur are the rhombic form and the monoclinic form. At atmospheric pressure, rhombic sulphur undergoes a transition to monoclinic when it is heated above $\mathbf{368}\mathbf{.5}\mathbf{}\mathbf{\text{K}}$.

role="math" localid="1663510441629" $\mathit{S}\mathbf{(}\mathit{s}\mathbf{,}\mathit{r}\mathit{h}\mathit{o}\mathit{m}\mathit{b}\mathit{i}\mathit{c}\mathbf{)}\mathbf{\to }\mathit{S}\mathbf{(}\mathit{s}\mathbf{,}\mathbf{\text{monoclinic}}\mathbf{)}$

(a) What is the sign of the entropy change $\mathbf{\left(}\mathit{\Delta }\mathit{S}\mathbf{\right)}$for this transition?

(b) $\mathit{\Delta }\mathit{H}$for this transition is $\mathbf{400}\mathbf{}\mathbf{\text{J}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}$. Calculate $\mathit{\Delta }\mathit{S}$for the transition.

Short answer

(a) ΔS will have positive sign ( + ).

(b) $\Delta S=1.09{\text{JK}}^{-1}{\text{mol}}^{-1}$.

Step-by-step solution

Given data

$S(s,rhombic)\to S(s,\text{monoclinic})$.

Temperature is $368.5\text{K}$.

ΔH (Change in enthalpy) for this transition is $400\text{J}{\text{mol}}^{-1}$.

Concept of entropy

Entropy Change is the phenomenon which is the measure of change of disorder or randomness in a thermodynamic system. It is related to the conversion of heat or enthalpy in work. More randomness in a thermodynamic system indicates high entropy.

If the entropy of a system rises, ΔS is positive. If the entropy of a system drops, ΔS is negative.

The sign of entropy change

a)

Sulphur undergoes a transition from rhombic to monoclinic at atmospheric pressure. Rhombic sulphur was heated, the average kinetic energy per molecule was increased and the range of momenta available to the molecule was increased.

Because of this $\Omega$for gas increases.

$\Delta S=k\ln \Omega$

The entropy of the gas has to be increased.

ΔS will have a positive sign ( + ).

Calculate the entropy change

b)

The entropy change has been calculated as follows:

$\Delta S=\frac{\Delta H}{T}$ .............. (1)

Where, ( ΔH )- the change in enthalpy and (T)- the average temperature of the object

$\begin{array}{l}\Delta H=400{\text{Jmol}}^{-1}\\ \text{T}=368.5\text{K}\end{array}$

Substitute these values in equation (1)

$\begin{array}{c}\Delta S=\frac{400{\text{Jmol}}^{-1}}{3685\text{K}}\\ =1.09{\text{JK}}^{-1}{\text{mol}}^{-1}\end{array}$