Question

The rate for the reaction

${\mathbf{\text{OH}}}^{\mathbf{-}}\mathbf{}\left(\text{Aq}\right)\mathbf{+}\mathbf{}\mathbf{\text{N}}{{\mathbf{\text{H}}}_{\mathbf{4}}}^{\mathbf{+}}\mathbf{}\left(\text{Aq}\right)\mathbf{}\mathbf{\to }\mathbf{}{\mathbf{\text{H}}}_{\mathbf{2}}\mathit{O}\left(\text{l}\right)\mathbf{}\mathbf{+}\mathbf{}{\mathbf{\text{NH}}}_{\mathbf{3}}\left(\text{Aq}\right)$

is first order in both OH₂ and concentrations, and the rate constant k at 20°C is 3.4 × 10¹⁰ L mol^-1. Suppose 1.00 L of a 0.0010 M NaOH solution is rapidly mixed with the same volume of 0.0010 M NH₄Cl solution. Calculate the time (in seconds) required for the OH₂concentration to decrease to a value of 1.0 × 10^-5M.

Short answer

${\text{Thetimerequiredforthedecreaseofhydroxideionconcendrationis2.9×10}}^{-6}\text{s}$.

Step-by-step solution

Step-1: Rate expression.

Integrated rate expression for the second order reaction in which both the reactants are of first order is given by

$\frac{1}{\nu }\left(\frac{1}{C}-\frac{1}{C_0}\right)=kt........\left(1\right)$

Where C₀ is the initial concentration and C_t is the final concentration ,k is the rate constant ,t is the time in seconds.

Step-2:  Concentration of NaOH.

1.00L volume of 0.0010M is mixed with 1.00L of 0.0010M of NH₄Cl,the volume of the solution becomes double (2.00L),so the initial concentration of NaOH is

$\begin{array}{l}\text{InitialconcendrationofNaOH}\left(C_0\right)=\frac{0.0010\text{M}}{2}\\ \left(C_0\right)=0.0005\text{M}\end{array}$

Step-3: Time required for the reaction:

According to given data

$\begin{array}{l}\text{Initialconcendration}\left(C_0\right)=0.0005M\\ \left(C_0\right)=0.0005mol{\text{L}}^{-1}\\ \text{Finalconcendration}\left(C_t\right)=1.0\times {10}^{-5}\text{M}\\ \left(C_t\right)=1.0\times {10}^{-5}{\text{molL}}^{-1}\\ K=3.4\times {10}^{10}L{\text{mol}}^{-1}{s}^{-1}\\ {\text{νOH}}^{-},\text{N}{{\text{H}}_4}^{+}=1\\ \text{Substitutingallthesevaluesinequation-1}\\ \frac{1}{\nu }\left(\frac{1}{C}-\frac{1}{C_0}\right)=kt........\left(1\right)\\ \frac{1}{1}\left(\frac{1}{1.0\times {10}^{-5}mol{L}^{-1}}-\frac{1}{0.0005mol{L}^{-1}}\right)=3.4\times {10}^{10}{\text{Lmol}}^{-1}{s}^{-1}\\ \frac{1}{1}\left(\frac{1}{1.0\times {10}^{-5}}-\frac{1}{0.0005}\right)\frac{1}{mol{L}^{-1}}=3.4\times {10}^{10}{\text{Lmol}}^{-1}{\text{s}}^{-1}\times t\\ t=\frac{1}{1\times 3.4\times {10}^{10}{\text{Lmol}}^{-1}{s}^{-1}}\left(\frac{1}{0.000001}-\frac{1}{0.0005}\right)\frac{1}{{\text{molL}}^{-1}}\\ t=0.29\left(100000-2000\right)\times {10}^{-10}\text{s}\\ t=2.9\times {10}^{-6}\text{s}\\ {\text{Thetimerequiredforthedecreaseofhydroxideionconcendrationis2.9×10}}^{-6}\text{s}\end{array}$