91影视

The initial rate of forward and backward reactions can differ depending on the reaction conditions for a reversible reaction. But after some time, the system enters the state of a dynamic equilibrium in which the forward and backward reactions occur at the same rate.

Therefore, there will be no change in the concentration of reactants and products. The reaction quotient determined at this state is the equilibrium constant for the reaction.

The equilibrium constant for a reversible reaction is the ratio of the concentration of products to the reactants at the equilibrium state. It will be constant for a reaction at a particular temperature irrespective of the initial concentration of the reactants.

Temperature is the only factor that changes the equilibrium constant of a reaction. The equilibrium constant can also be calculated in terms of partial pressures of the gaseous reactants and products for reactions involving gases.

Consider ${}^{{\mathbf{K}}_{\mathbf{P}}}$and ${}^{{\mathbf{K}}_{\mathbf{P}}^{\mathbf{\text{'}}}}$are the equilibrium constant values of a reaction at ${}^{{\mathbf{T}}_{\mathbf{1}}}$and ${}^{{\mathbf{T}}_{\mathbf{2}}}$temperatures, respectively. Then, the relation between these values can be given as:

$\mathbf{log}\mathbf{}\left(\frac{K_P^{\text{'}}}{K_P}\right)\mathbf{}\mathbf{=}\frac{{\mathbf{螖贬}}_{\mathbf{rxn}}^{\mathbf{0}}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{}\mathbf{R}}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Where ${}^{{\mathbf{螖贬}}_{\mathbf{rxn}}^{\mathbf{0}}}$is the standard enthalpy of reaction, and R is the universal gas constant with the value 8.314 J/mol/K.

Substituting the known values for the reaction,

$$\begin{gathered} \mathbf{log}\mathbf{}\mathbf{\left(}\frac{{\mathbf{K}}_{\mathbf{P}}^{\mathbf{\text{'}}}}{\mathbf{1}\mathbf{.}\mathbf{80}}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{-}\frac{\mathbf{3}\mathbf{.}\mathbf{32}\mathbf{}\mathit{k}\mathit{j}\mathbf{.}\mathit{m}\mathit{o}\mathit{l}}{\mathbf{2}\mathbf{.}\mathbf{303}\mathbf{脳}\left(8.314脳{10}^{-3}\mathrm{kJ}.{\mathrm{mol}}^{-1}.K^{-1}\right)}\mathbf{(}\frac{\mathbf{1}}{\mathbf{298}\mathbf{K}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{500}\mathbf{K}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02266} \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{{\mathbf{K}}_{\mathbf{P}}^{\mathbf{\text{'}}}}{\mathbf{1}\mathbf{.}\mathbf{80}}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{02266}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{949} \end{gathered}$$

The equilibrium constant at 500 K can be calculated as:

$$\begin{gathered} \mathbf{K}{\mathbf{\text{'}}}_{\mathbf{P}\mathbf{}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{949}\mathbf{脳}\mathbf{1}\mathbf{.}\mathbf{80} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{708} \end{gathered}$$

The equilibrium constant at 500 K is 1.708.