Q20.84 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Among the many complex ions of cobalt are the following: 

$\mathbf{\text{Co}}{{\mathbf{\left(}{\mathbf{\text{NH}}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{6}}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathbf{\text{en}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{\text{Co}}{{\mathbf{\left(}\mathbf{\text{en}}\mathbf{\right)}}_{\mathbf{3}}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{\text{NH}}}_{\mathbf{3}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}$ where "en" stands for ethylenediamine, ${\mathbf{\text{H}}}_{\mathbf{2}}{\mathbf{\text{NCH}}}_{\mathbf{2}}{\mathbf{\text{CH}}}_{\mathbf{2}}{\mathbf{\text{NH}}}_{\mathbf{2}}$. Six $\mathbf{\text{Co}}\mathbf{-}\mathbf{\text{N}}$ bonds are broken and six $\mathbf{\text{Co}}\mathbf{-}\mathbf{\text{N}}$ bonds are formed in this reaction, so $\mathit{\Delta }{\mathit{H}}_{\mathbf{\text{rxn}}}^{\mathbf{°}}\mathbf{\approx }\mathbf{0}$; yet $\mathit{K}\mathbf{>}\mathbf{1}$ . What are the signs of $\mathit{\Delta }{\mathit{S}}^{\mathbf{°}}$ and $\mathit{\Delta }{\mathit{G}}^{\mathbf{°}}$? What drives the reaction?

Step-by-Step Solution

Answer

We can estimate that based on the supplied reaction. $\Delta S_{rxn}^{°}>0$  and  $\Delta G^{°}<0$.

Since entropy is the driving force behind the reaction $T\cdot \Delta S_{rxn}>>\Delta H_{rxn}$

Step 1: Definition of Delta

It is used in chemistry to indicate a change in enthalpy as well as the addition of heat to the reaction. A triangle symbol is used to represent it.

Step 2: Calculate &#916; n r x n

Because all of the components on both sides of the reaction - reactants and products are dissolved in water, they are all present in the aqueous solution. There is no change in physical state within the response. We can compare the number of particles before and after the reaction to study the effect on the process's entropy change.

Therefore,

$\begin{array}{l}\Delta n_{rxn}=\sum n_{\text{products }}-\sum n_{\text{reactants }}\\ \Delta n_{rxn}=(n\left(\text{Co}{\left(en\right)}_3^3\right)+n\left({\text{NH}}_3\right))-\left(n\left(\text{Co}{\left({\text{NH}}_3\right)}_6^{3+}\right)+n\left(en\right)\right)\\ \Delta n_{rxn}=(1 \text{mol}+6 \text{mol})-1 \text{mol}+3 \text{mol})\\ \Delta n_{rxn}=3 \text{mol}\end{array}$

The forward reaction increases the number of particles in the system by three.

Step 3: What derives the reaction

Resulting in increased disorder and higher entropy:

$\Delta S_{rnn}^{°}>0$

To estimate the sign of $\Delta G^{°}$ , we can use the equation $\Delta G^{°}=\Delta H_{rxn}^{°}-T\cdot \Delta S_{rxn}^{°}$

Since $\Delta H_{rxn}^{°}\approx 0$,

$\begin{array}{l}\Delta G^{°}=0-T\cdot \Delta S_{rxn}^{°}\\ \Delta G^{°}=-T\cdot \Delta S_{rxn}^{°}\end{array}$

As entropy of the reaction is positive,

$-T\cdot \Delta S_{Txn}^{°}<0$

Therefore, Gibb's Free energy would also be negative:

$\Delta G^{°}<0$

The reaction is driven by entropy since $T\cdot \Delta S_{rxn}>>\Delta H_{rxn}$