Question

The rate for the reaction is ${\mathbf{\text{OH}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{\text{Aq}}\mathbf{\right)}\mathbf{\text{+HCN}}\mathbf{\left(}\mathbf{A}\mathbf{q}\mathbf{\right)}{\mathbf{\text{→H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}{\mathbf{\text{+CN}}}^{\mathbf{\text{-}}}\mathbf{\left(}\mathbf{A}\mathbf{q}\mathbf{\right)}$ is first order in both ${\mathbf{\text{OH}}}_{\mathbf{2}}$ and HCN concentrations and the rate constant k at 25°C is 3.7 × 10⁹ L mol^-1. Suppose 0.500 L of a 0.0020 M NaOH solution is rapidly mixed with the same volume of a 0.0020 M HCN solution. Calculate the time (in seconds) required for the OH^- concentration to decrease to a value of 1.0 × 10^-4 M

Short answer

The time required to decrease of OH^-concentration is $2.432\times {10}^{-6}s$

Step-by-step solution

Rate constant of a reaction

Rate constant of a reaction is the expression that depicts the relationship between the rate of chemical reaction and concentration of reacting substances. The rate constant of second order reaction is given by

$k=\frac{1}{t}\left[\frac{1}{C}-\frac{1}{C_t}\right]..........\left(1\right)$

Where k is the rate constant of the reaction,C₀ and C_t are initial and final concendration of the reactents.

Calculation of molarity:

Molarity calculation when the base and acid is mixed is given by the following formulae

$$\begin{gathered} {\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2..........\left(2\right) \\ \text{MolarityofNaOHsolution}{\text{M}}_1=0.0020M \\ \text{VolumeofNaOHsolution}{\text{V}}_1=0.500\text{L} \\ MolarityofHCNsolution{\text{M}}_2=? \\ {\text{VolumeofHCNsolutionV}}_2=1\text{Liter} \\ {\text{M}}_1{\text{V}}_1={\text{M}}_2{V}_2 \\ 0.0020M\frac{mol}{\text{L}}\times 0.500L={\text{M}}_2\times 1\text{L} \\ {\text{M}}_2\text{=0.001}\frac{\text{mol}}{\text{L}} \end{gathered}$$

Calculation of  Rate constant of the reaction:

$$\begin{gathered} \text{k=}\frac{1}{t}\left[\frac{1}{C}-\frac{1}{C_t}\right]..........\left(1\right) \\ t=\frac{1}{k}\left[\frac{1}{C}-\frac{1}{C_t}\right] \\ t=\frac{1}{3.7\times {10}^9{\displaystyle \frac{\text{L}}{\text{mol.s}}}}\left[\frac{1}{1.0\times {10}^{-4}{\displaystyle \frac{\text{mol}}{L}}}-\frac{1}{0.001{\displaystyle \frac{\text{mol}}{L}}}\right] \\ t=2.432\times {10}^{-6}\mathrm{s} \end{gathered}$$

Therefore the time required to decrease of OH^-concentration is $2.432\times {10}^{-6}\mathrm{s}$