Question

The compound IrH₃(CO)(P(C₆H₅)₃)₂ exists in two forms: the meridional (“mer”) and facial (“fac”). At 25°C in a nonaqueous solvent, the reaction mer→ fac has a rate constant of 2.33 s^-1, and the reaction fac → mer has a rate constant of 2.10 s^-1 . What is the equilibrium constant of the mer-to-fac reaction at 25°C?

Short answer

Equilbrium constant K value is 1.11

Step-by-step solution

Rate constant for equilbrium reactions;

For a reaction in chemical equilibrium rate of forward reaction is equal to the rate of backward reaction,so the rate constant of forward reaction is eqaul to rate constant of backward reaction according to the principle of detailed balance and hence equilbrium constant is the defined as the ratio between forward and reverse equilibrium constants.

$Mer-I{r}_3\left(Co\right){\left(p{\left(C_6{H}_5\right)}_3\right)}_2\rightleftharpoons \mathrm{Fac}-I{r}_3\left(Co\right){\left(p{\left(C_6{H}_5\right)}_3\right)}_2\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(1\right)$

Where the forward rate constant k is equal to the rate constant k_-1 of the backward reaction, if the equilibrium constant of the reaction is K then,

It can be defined as ratio between the forward rate constant and the rate constant of backward reaction.

$K={\frac{k}{k}}_{-1}.......................\left(2\right)$

k is the rate consatnt of forward reaction

k_-1 is the rate constant of backward reacttion

  Rate constant value:

From the above given question rate conatant value of backward reaction can be sorted out in the following way

$$\begin{gathered} K={\frac{k}{k}}_{-1}.......................\left(2\right) \\ Substitutingk=2.32{,}^{-1}{k}_{-1}\hspace{0.17em}=2.10s^{-1} \\ K=\frac{2.32s^{-1}}{2.10s^{-1}} \\ K\hspace{0.17em}=1.11 \end{gathered}$$

Therefore the equilbrium constant value is 1.11