Question

The compound$\mathbf{\text{Pt}}{\left({\text{NH}}_3\right)}_{\mathbf{2}}{\mathbf{\text{I}}}_{\mathbf{2}}$comes in two forms, the cis and the trans, which differ in their molecular structure. The following data are available:

Combine these data with data from Appendix D to compute the standard entropies of both of these compounds at$\mathbf{25}\mathbf{°}\mathbf{\text{C}}$

Short answer

Standard entropy of cis compound$\Delta S_f^{°}$, cis${\left(\text{Pt}{\left(\text{NH}\right)}_3\right)}_2{\text{I}}_2=216.27\text{J}{\text{K}}^{-1}$

Standard entropy of trans compound$\Delta S_f^{°}$ cis ${\left(\text{Pt}{\left(\text{NH}\right)}_3\right)}_2{\text{I}}_2=219.46\text{J}{\text{K}}^{-1}$

Step-by-step solution

Given data

The standard entropies of the reactants:

$\Delta S_f^{°}(\text{Pt},s)=41.63ȷ{\text{K}}^{-1}$$

$\Delta S_f^{°}\left({\text{N}}_2,g\right)=191.50\text{J}{\text{K}}^{-1}$

$\Delta S_f^{°}\left({\text{H}}_2,g\right)=130.51\text{J}{\text{K}}^{-1}$

$\Delta S_f^{°}\left({\text{I}}_2,g\right)=116.14\text{J}{\text{K}}^{-1}$

Concept used

The entropy content of one mole of pure substance at a standard condition of pressure and any relevant temperature is known as the standard molar entropy. To calculate the reaction's standard entropy, multiply the total of the absolute entropies of the reactants and products by the relevant stoichiometric coefficients for each. The result is the difference between the two.

The reaction

The reaction of formation of $\text{Pt}{\left({\text{NH}}_3\right)}_2{I}_2$rom its elements is:

$$\begin{gathered} \text{Pt}\left(\text{s}\right)+{\text{N}}_2\left(g\right)+3{\text{H}}_2\left(g\right)+{\text{I}}_2(\text{s})\to \text{Pt}{\left({\text{NH}}_3\right)}_2{\text{I}}_2\left(\text{cis} \\ \text{Pt}\right(\text{s})+{\text{N}}_2(g)+3{\text{H}}_2(g)+{\text{I}}_2(\text{s})\to \text{Pt}{\left({\text{NH}}_3\right)}_2{\text{I}}_2(\text{trans}) \end{gathered}$$

Calculate the entropy change of cis and trans

The standard entropy of the cis and the trans structure has to be calculated first.

To obtain the$\Delta S_f^{°}{\left(\text{Pt}{\left(\text{NH}\right)}_3\right)}_2{\text{I}}_2$from the formula for the Gibb's free energy change:

$\Delta G_f^{°}=\Delta H_f^{°}-T\Delta S_f^{°}$

This would mean that:

$\Delta S_f^{°}=\frac{\Delta H_{f-}^{°}\Delta G_f^{°}}{T}$

Calculating the entropy change for cis:

$\Delta S°\left(\text{cis}\right)=\frac{-286.56+130.25}{298}$

$\Delta S°\left(\text{cis}\right)=-524.53\text{J}{\text{mol}}^{-1}$

Calculating the standard entropy change for trans:

$\Delta S°\left(\text{trans}\right)=\frac{-316.94+161.505}{298}$

$\Delta S°\left(\text{trans}\right)=-521.61\text{J}{\text{mol}}^{-1}$

Calculate the standard entropy of cis and trans compounds 

The standard entropy for these reactions would be:

$\Delta S°=\Delta S_f^{°}{\left(\text{Pt}{\left(\text{NH}\right)}_3\right)}_2{\text{I}}_2-\Delta S_f^{°}\text{(Pt},s)+\Delta S_f^{°}\left({\text{N}}_2,g\right)+3\Delta S_f^{°}\left({\text{H}}_2,g\right)+\Delta S_f^{°}\left({\text{I}}_2,g\right)$......... (1)

$\Delta S_f^{°}(\text{Pt},s)=41.63ȷ{\text{K}}^{-1}$$

$\Delta S_f^{°}\left({\text{N}}_2,g\right)=191.50\text{J}{\text{K}}^{-1}$

$\Delta S_f^{°}\left({\text{H}}_2,g\right)=130.51\text{J}{\text{K}}^{-1}$

$\Delta S_f^{°}\left({\text{I}}_2,g\right)=116.14\text{J}{\text{K}}^{-1}$

• Calculating the standard entropy of the cis structure:

$\Delta S°\left(\text{cis}\right)=-524.53\text{J}{\text{mol}}^{-1}$

Substitute the values in equation (1)

$$\begin{gathered} \Delta S°\left(\text{cis}\right)=\Delta S_f^{°},\text{cis}{\left(\text{Pt}{\left(\text{NH}\right)}_3\right)}_2{\text{I}}_2-\Delta S_f^{°}(\text{Pt},s)+\Delta S_f^{°}\left({\text{N}}_2,g\right)+3\Delta S_f^{°}\left({\text{H}}_2,g\right)+\Delta S_f^{°}\left({\text{I}}_2,g\right) \\ \Delta S_f^{°},\text{cis}{\left(\text{Pt}{\left(\text{NH}\right)}_3\right)}_2{\text{I}}_2=\Delta S°\left(\text{cis}\right)+\Delta S_f^{°}(\text{Pt},s)+\Delta S_f^{°}\left({\text{N}}_2,g\right)+3\Delta S_f^{°}\left({\text{H}}_2,g\right)+\Delta S_f^{°}\left({\text{I}}_2,g\right) \\ =-524.53+(41.63+191.50+3\times 130.51+116.14) \\ =216.27\text{J}{\text{K}}^{-1} \end{gathered}$$

• Calculating the standard entropy of the trans structure:

$\Delta S°\left(\text{trans}\right)=-521.61\text{J}{\text{mol}}^{-1}$

Substitute the values in equation (1)

$$\begin{gathered} \Delta S°\left(\text{trans}\right)=\Delta S_f^{°}\text{, trans}{\left(\text{Pt}{\left(\text{NH}\right)}_3\right)}_2{\text{I}}_2-\Delta S_f^{°}(\text{Pt},s)+\Delta S_f^{°}\left({\text{N}}_2,g\right)+3\Delta S_f^{°}\left({\text{H}}_2,g\right)+\Delta S_f^{°}\left({\text{I}}_2,g\right) \\ \Delta S_f^{°}\text{,trans}{\left(\text{Pt}{\left(\text{NH}\right)}_3\right)}_2{\text{I}}_2=-521.61+(41.63+191.50+3\times 130.51+116.14) \\ \Delta S_f^{},\text{trans}{\left(\text{Pt}{\left(\text{NH}\right)}_3\right)}_2{\text{I}}_2=219.46\text{J}{\text{K}}^{-1} \end{gathered}$$