Question

Question: The ${\mathbf{\text{N}}}_{\mathbf{2}}\mathbf{\text{O}}$molecule has the structure$\mathbf{\text{N}}\mathbf{-}\mathbf{\text{N}}\mathbf{-}\mathbf{\text{O}}$. In an ordered crystal of${\mathbf{\text{N}}}_{\mathbf{2}}\mathbf{\text{O}}$, the molecules are lined up in a regular fashion, with the orientation of each determined by its position in the crystal. In a random crystal (formed on rapid freezing), each molecule has two equally likely orientations.

(a) Calculate the number of microstates available to a random crystal of${\mathit{N}}_{\mathbf{\text{A}}}$(Avogadro's number) of molecules.

(b) Calculate the entropy change when$\mathbf{1}\mathbf{.00}\mathbf{}\mathbf{\text{mol}}$of a random crystal is converted to an ordered crystal.

Short answer

(a) The number of microstates $=2^{6.022\times {10}^{23}}$.

(b) The change in entropy $\Delta S=-5.76{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

Step-by-step solution

Given data

We are observing an ${\text{N}}_2\text{O}$molecule. The structure:$\text{N}-\text{N}-\text{O}$ In a random crystal each molecule has two equally likely orientations.

Concept of vapor pressure

a) Microstates are the number of different possible arrangements of molecular position and kinetic energy at a particular thermodynamic state.

(b) Entropy Change is the phenomenon that 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. A thermodynamic system that has more randomness means it has high entropy.

Calculate the number of microstates

(a)

The number of orientations$\left(v\right)=2$.

Avogadro's number: $6.022\times {10}^{23}{\text{mol}}^{-1}$.

The no. of microstatesdata-custom-editor="chemistry" $=v^{N_A}$

The no. of microstates$v^{N_A}=2^{6.022\times {10}^{23}}$

Calculate the entropy

(b)

The change between the entropy of the ordered and the random crystal is:

$\Delta S=S_o-S_r$ ............... (1)

where $S_o=0$.

Formula to calculate entropy change is $S=k_B\ln w$.

Where$w$ is the number of microstates. So, it can be written as: $S=k_B\ln v^{N_A}$.

Calculate the entropy of the random crystal as follows:

$\begin{array}{c}S=k_B\ln v^{N_A}\\ =k_B\ln 2^{N_A}\\ =k_B\times N_A\times \ln 2\\ =R\times 0.693\\ =5.76{\text{Jmol}}^{-1}{\text{K}}^{-1}\end{array}$

Calculate the entropy change

Calculating the entropy change

$S_o=0$.

$S_r=5.76{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

Substitute these values in equation (1)

$\Delta S=0-5.76$

$\Delta S=-5.76{\text{Jmol}}^{-1}{\text{K}}^{-1}$