91Ó°ÊÓ

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.

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.

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

Carbon's half-reactions are as follows:

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

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

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

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${∆}_r{G}^{°}$:

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

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

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

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

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

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

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

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