Question

The strongest known chemical bond is that in carbon monoxide, $\mathbf{\text{CO}}$, with bond enthalpy of$\mathbf{1}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{\text{kJ}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}$. Furthermore, the entropy increase in a gaseous dissociation of the kind$\mathbf{\text{AB}}\mathbf{\leftrightharpoons }\mathbf{\text{A}}\mathbf{+}\mathbf{\text{B}}$is about$\mathbf{110}\mathbf{}\mathbf{\text{J}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathbf{\text{K}}}^{\mathbf{-}\mathbf{1}}$. These factors establish a temperature above which there is essentially no chemistry of molecules. Show why this is so and find the temperature.

Short answer

When $∆H°$and $∆S°$are both positive, as in this case$∆G°$ is zero.

The final temperature $T=9454\text{K}$.

Step-by-step solution

Given data

$\text{CO}$, with bond enthalpy of $1.05\times {10}^3\text{kJ}{\text{mol}}^{-1}$.

$\text{AB}\leftrightharpoons \text{A}+\text{B}$With entropy change$110\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$ .

 Step 2: Concept used

The energy is associated with a chemical reaction that can be used to do work. The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system.

Calculationof final temperature 

The change in Gibbs free energy$\Delta G°$for a system depends upon the change in enthalpy$\Delta H$and the change in entropy$\Delta S$according to the following equation:

$\Delta G=\Delta H-T\Delta S$ ............ (1)

The reaction is spontaneous at all temperatures when$∆H$$∆H$is negative and$∆S$ is positive.

The reaction is not spontaneous when$∆H$ is positive and $∆S$is negative.

In the given data $∆H°$and $∆S°$are both positive, as in this case$∆G°$ is zero.

$\Delta G=\Delta H-T\Delta S$

$0=\Delta H-T\Delta S$

$T\Delta S=\Delta H$

$T=\frac{\Delta H}{\Delta S}$

............. (2)

$\Delta H=1.05\times {10}^3{\text{kJmol}}^{-1}$

$\Delta S=110\times {10}^3{\text{Jmol}}^{-1}{\text{K}}^{-1}$

Substitute these values in equation (2):

$$\begin{gathered} T=\frac{1.05\times {10}^3{\text{kJmol}}^{-1}}{110{\text{Jmol}}^{-1}{\text{K}}^{-1}} \\ =9545\text{K} \end{gathered}$$