Question

The disproportionation of to graphite and ${\mathbf{\text{CO}}}_{\mathbf{\text{2}}}$is thermodynamically favoured but slow. (a) What does this mean in terms of the magnitudes of the equilibrium constant (K), rate c$\mathbf{\text{CO}}$onstant (k), and activation energy $\left({\text{E}}_{\text{a}}\right)$? (b) Write a balanced equation for the disproportionation of $\mathbf{\text{CO}}$. (c) Calculate at ${\mathbf{\text{K}}}_{\mathbf{\text{c}}}$. (d) Calculate at${\mathbf{\text{K}}}_{\mathit{P}}$ .

Short answer

a) High activation energy, low-rate constant and equilibrium constant.

b) The equation is$\text{2CO}\to {\text{C( graphite ) + CO}}_{\text{2}}$ .

c) The value is ${\text{K}}_{\text{eq}}{\text{= 2.5×10}}^{\text{30}}$.

d) The value is${\text{K}}_{\text{p}}{\text{= 6.2×10}}^{\text{33}}$ .

Step-by-step solution

Define equilibrium

When the concentrations of reactants and products are constant and their ratio does not change, a chemical reaction is in equilibrium. When a chemical reaction does not convert all reactants to products, it achieves equilibrium. Many reactions achieve a state of balance or dynamic equilibrium, in which both reactants and products are present.

Explanation

a) We can state that the value of the equilibrium constant is over one if the reaction is thermodynamically favoured yet sluggish because the products are favoured over the reactants. The equilibrium constant has no relation with the rate of chemical transformations; instead, it is linked to the activities of products and reactants and can't tell us anything about the reaction's kinetics.

The rate constant, on the other hand, can provide additional information about the reaction's dynamics. We can expect low values of the rate constant, k, if the reaction is kinetically advantageous but slow.

Finally, the activation energy,, is inversely proportional to the rate constant: the higher the activation energy ${\text{E}}_{\text{a}}$, the lower the rate constant, and hence the slower the reaction.

Therefore, low-rate constant and equilibrium constant values, high activation energy value.

Explanation

b) The disproportionation of $CO$balanced equation is:

$\text{2CO}\to {\text{C( graphite ) + CO}}_{\text{2}}$

Carbon's half-reactions are as follows:

$$\begin{gathered} {\text{C}}^{\text{II}}\to {\text{C}}^{\text{IV}}{\text{+ 2e}}^{\text{-}} \\ {\text{C}}^{\text{II}}{\text{+ 2e}}^{\text{-}}\to {\text{C}}^{\text{o}} \end{gathered}$$

Therefore, the equation is$\text{2CO}\to {\text{C( graphite ) + CO}}_{\text{2}}$ .

Explanation

c) Before calculating${\text{K}}_{\text{c}}$, keep in mind that there are two gaseous and one solid components. ${\text{K}}_{\text{c}}$will not be affected by the graphite.

$$\begin{gathered} \text{T = 298K} \\ {\text{K}}_{\text{c}}\text{=}\frac{\text{[C( graphite )]}\left[{\text{CO}}_{\text{2}}\right]}{{\text{[CO]}}^{\text{2}}} \\ \text{=}\frac{\left[{\text{CO}}_{\text{2}}\right]}{{\text{[CO]}}^{\text{2}}} \end{gathered}$$

We must adopt a thermodynamical approach to this problem because we don't have any of the concentrations. If we know the value of $∆G$or the change in Gibb's energy, we may calculate ${\text{K}}_{\text{eq}}$:

${\text{lnK}}_{\text{eq}}\text{= -}\frac{{∆}_r{G}^{°}}{{\displaystyle \text{RT}}}$

Appendix C data can be used to determine${∆}_r{G}^{°}$:

$$\begin{gathered} {∆}_r{G}^{°}={∆}_r{G}^{°}\text{( products )}-{∆}_r{G}^{°}\text{( reactants )} \\ {∆}_r{G}^{°}=\left[1mol\times 0kj/mol+1mol\times \left(-394.4kj/mol\right)\right]-\left[2mol\times \left(-110.50kj/mol\right)\right] \\ {∆}_r{G}^{°}=\text{- 394.4kJ +221.0kJ} \\ {∆}_r{G}^{°}=\text{- 173.4kJ} \end{gathered}$$

We can now calculate${\text{K}}_{\text{eq}}$:

$$\begin{gathered} {\text{lnK}}_{\text{eq}}\text{= -}\frac{{∆}_r{G}^{°}}{RT} \\ l{\text{nK}}_{\text{eq}}\text{= -}\frac{{\displaystyle \text{- 173.4kJ}}}{{\displaystyle {\text{8.314×10}}^{\text{- 3}}{\text{kJK}}^{\text{- 1}}{\text{mol}}^{\text{- 1}}\text{×298K}}} \\ l{\text{nK}}_{\text{eq}}\text{= -69.99} \\ {\text{K}}_{\text{eq}}{\text{= -e}}^{\text{69.99}} \\ {\text{K}}_{\text{eq}}{\text{= 2.5×10}}^{30} \end{gathered}$$

Therefore, the value is${\text{K}}_{\text{eq}}{\text{= 2.5×10}}^{30}$ .

Explanation

d) Solid components, like ${\text{K}}_{\text{c}}$, are not included in${\text{K}}_p$.

${\text{K}}_{\text{p}}\text{=}\frac{\text{p}\left({\text{CO}}_{\text{2}}\right)}{{\text{p}}^{\text{2}}\text{(CO)}}$

Using the ideal gas law, we can connect ${\text{K}}_{\text{c}}$and ${\text{K}}_p$:

$$\begin{gathered} \text{pV = nRT} \\ \frac{\text{n}}{\text{V}}\text{=}\frac{\text{p}}{\text{RT}} \\ \text{c =}\frac{\text{p}}{\text{RT}} \\ {\text{K}}_{\text{c}}\text{=}\frac{{\text{K}}_{\text{p}}}{\text{RT}} \\ {\text{K}}_{\text{p}}{\text{=K}}_{\text{c}}{\text{×RTK}}_{\text{p}} \\ =2.5\times {10}^{30}\times {\text{8.314K}}^{\text{- 1}}{\text{mol}}^{-1}\text{×298K} \\ {\text{K}}_{\text{p}}{\text{= 6.2×10}}^{33} \end{gathered}$$

Therefore, the value is${\text{K}}_{\text{p}}{\text{= 6.2×10}}^{33}$