Question

The reaction of OH^- with HCN in aqueous solution at 25°C has a forward rate constant k_f of 3.7 × 109 L mol-1. Using this information and the measured acid ionization constant of HCN (see Table 15.2), calculate the rate constant k_r in the first-order rate law$\text{rate}\hspace{0.17em}={\text{k}}_r\left[{\text{CN}}^{-}\right]$ for the transfer of hydrogen ions to CN^- from surrounding water molecules:${\text{H}}_2\text{O}\left(aq\right){\text{+CN}}^{\text{-}}\left(aq\right){\text{→OH}}^{\text{-}}\left(aq\right)+\text{HCN}\left(aq\right)\text{......}\left(1\right)$

Short answer

The rate constant of reverse reaction is 6x10⁴ s^-1

Step-by-step solution

Equilbrium constant:

Equilbrium is the condition where the rate of forward reaction is equal to the rate of backward reaction.For example in a reaction.

$$\begin{gathered} \text{A+B}\stackrel{}{\leftrightarrows C+D} \\ Rateofforwardreaction=k_f\left[A\right]\left[B\right] \\ Rateofbackwardreaction=k_b\left[A\right]\left[B\right] \\ \text{Wherekfandkbaretherateofforwardandbackwardreaction} \\ Atequilbrium \\ {k}_f\left[A\right]\left[B\right]=k_b\left[A\right]\left[B\right] \\ \frac{k_f}{k_b}\hspace{0.17em}=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]} \\ \frac{k_f}{k_b}=K \\ \text{WhereKistheequilbriumconstant} \end{gathered}$$

Rate constant of backward reaction

HCN has an ionization constant (K_a) of 6.17 x 10^-10

HCN is a weak acid. It ionizes in the following manner.

${\text{H}}_2\text{O}\left(aq\right){\text{+CN}}^{\text{-}}\left(aq\right){\text{→OH}}^{\text{-}}\left(aq\right)+\text{HCN}\left(aq\right)\text{......}\left(1\right)$

Ionaization of HCN is given as:

localid="1663682695686" $$\begin{gathered} {\text{k}}_a=\frac{\left[\text{CN}\right]\left[{\text{H}}_{\text{3}}\text{O}\right]}{\left[\text{HCN}\right]} \\ \text{Itcanbewrittenas} \\ \frac{\left[\text{HCN}\right]}{\left[\text{CN}\right]}=\frac{\left[{\text{H}}_{3}O\right]}{{\text{k}}_a} \\ {\text{OH}}^{-}\text{reactswith}C{N}^{-}to\text{form} \\ {\text{H}}_2\text{O}\left(aq\right){\text{+CN}}^{\text{-}}\left(aq\right){\text{→OH}}^{\text{-}}\left(aq\right)+\text{HCN}\left(aq\right)\text{......}\left(1\right) \end{gathered}$$

$$\begin{gathered} \text{Rateofforwardreaction}{k}_f=3.7\times {10}^9{\text{Lmol}}^{\text{-1}}{\text{s}}^{\text{-1}} \\ \text{Atequilbriumrateofforwardreactionisequaltorateofbackwardreaction} \\ {\text{k}}_f\left[\text{OH}\right]\left[{\text{CN}}^{-}\right]={\text{k}}_b\left[\text{CN}\right] \\ \text{Calculate}{k}_b\text{by}\text{substituting}\frac{\left[\text{HCN}\right]}{\left[{\text{CN}}^{-}\right]}\text{intheaboveexpression} \\ {k}_b=\frac{{\text{k}}_f\left[\text{OH}\right]\left[{\text{H}}_{\text{3}}\text{O}\right]}{{\text{k}}_a} \\ \left[{\text{OH}}^{-}\right]\left[{\text{H}}_{3}O\right]={10}^{-14} \end{gathered}$$

Substitute and calculate kb value:

$$\begin{gathered} {\mathrm{k}}_{\mathrm{b}}=\frac{\left(3.7\mathrm{x}{10}^9{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}\right)\left({10}^{-14}\right)}{6.17\mathrm{x}{10}^{-10}} \\ {\mathrm{k}}_{\mathrm{b}}=6.0\mathrm{x}{10}^4{\mathrm{s}}^{-1} \end{gathered}$$

The rate constant of reverse reaction is 6x10⁴ s^-1