Question

In a study of the reaction of pyridine (C₅H₅N) with methyl iodide (CH₃I) in a benzene solution, the following set of initial reaction rates was measured at 25°C for different initial concentrations of the two reactants:

(a) Write the rate expression for this reaction.

(b) Calculate the rate constant k and give its units.

(c) Predict the initial reaction rate for a solution in which [C₅H₅N] is 5.0 × 10^-5 M and [CH₃I] is 2.0 × 10^-5 M.

Short answer

The initial rate of the reaction is$7.5\times {10}^{-8}mo{l}^{-1}L{s}^{-1}{\left(mol{L}^{-1}\right)}^2$

Step-by-step solution

STEP-1: Rate of reaction:

The rate for the following reaction as given in the question is given by

$\begin{array}{l}{C}_5{H}_{5}N+C{H}_{3}I\to products\\ Rate=k{\left[C_5{H}_{5}N\right]}^m{\left[C{H}_{3}I\right]}^n.....\left(1\right)\end{array}$

Where k is the rate constant of reaction while m and n are the orders of the reaction.

STEP:2  Order of  the reaction with respect to m:

The rate expression for the second initial concentration is given as

$3.0\times {10}^{-6}=k{\left(2.00\times {10}^{-4}\right)}^m{\left(2.00\times {10}^{-4}\right)}^n.......\left(2\right)$

The rate expression for third initial concentration is given as

$6.0\times {10}^{-6}=k{\left(2.00\times {10}^{-4}\right)}^m{\left(4.00\times {10}^{-4}\right)}^n.......\left(3\right)$

Divide the equation 3 by 2

$\begin{array}{c}\frac{6.0\times {10}^{-6}}{3.0\times {10}^{-6}}=k\frac{{\left(2.00\times {10}^{-4}\right)}^m}{{\left(2.00\times {10}^{-4}\right)}^m}\frac{{\left(4.00\times {10}^{-4}\right)}^n}{{\left(2.00\times {10}^{-4}\right)}^n}.......\left(4\right)\\ 2=2^n\\ n=1\end{array}$

STEP-3:  Order of a reaction with respect to n:

For the first initial concentration the rate expression is given as

$7.5\times {10}^{-7}=k{\left(1.00\times {10}^{-4}\right)}^m{\left(1.00\times {10}^{-4}\right)}^n.......\left(5\right)$

For third initial concentration the rate expression is given as

$3.0\times {10}^{-6}=k{\left(2.00\times {10}^{-4}\right)}^m{\left(2.00\times {10}^{-4}\right)}^n.......\left(6\right)$

Divide the equation 5 by 4

$$\begin{gathered} \frac{7.5\times {10}^{-7}}{3.0\times {10}^{-6}}=\frac{k{\left(1.00\times {10}^{-4}\right)}^m}{k{\left(2.00\times {10}^{-4}\right)}^m}\frac{{\left(1.00\times {10}^{-4}\right)}^n}{{\left(2.00\times {10}^{-4}\right)}^n}.......\left(7\right) \\ 0.25={\left(\frac{1}{2}\right)}^m{\left(\frac{1}{2}\right)}^1 \\ \mathrm{Since}\mathrm{n}=1 \\ 0.25={\left(\frac{1}{2}\right)}^{\mathrm{m}+1} \end{gathered}$$

Apply logarithm on both sides

$\begin{array}{c}log0.25=\left(m+1\right)log\frac{1}{2}\\ -0.6020=\left(m+1\right)\left(-0.3010\right)\\ m=1\\ \end{array}$

Hence the order of the reaction with respect to pyridine (m) is$\frac{1}{2}$.

STEP-4:  Rate expression

As the reaction is first order in each reactant ,overall rate expression is given as

$rate=k\left[C_5{H}_{5}N\right]\left[C{H}_{3}I\right]........\left(8\right)$

STEP-5:  Rate constant value

To find the value of rate constant K substitute the concentration of reactants in the expression 4

$\begin{array}{c}7.5\times {10}^{-7}=k{\left(1.00\times {10}^{-4}\right)}^{2}mo{l}^2{L}^{-2}\\ k=\frac{7.5\times {10}^{-7}mol{L}^{-1}{s}^{-1}}{{\left(1.00\times {10}^{-4}\right)}^{2}mo{l}^2{L}^{-2}}\\ k=\frac{7.5\times {10}^{-7}}{{10}^{-8}}mo{l}^{\left(1-2\right)}{L}^{\left(-1+2\right)}{s}^{-1}\\ k=75mo{l}^{-1}L{s}^{-1}\end{array}$

STEP-6: Rate of  a reaction

Substituting the concentration of pyridine as$\left(5.00\times {10}^{-5}\right)$ and methyl iodide concentration as $2.0\times {10}^{-5}$and K value as 75 mol^-1Ls^-1

$\begin{array}{l}Rate=7.5\times 10mo{l}^{-1}L{s}^{-1}\times 5.0\times {10}^{-5}M\times 2.0\times {10}^{-5}M\\ Rate=7.5\times {10}^{-8}mo{l}^{-1}L{s}^{-1}{\left(mol{L}^{-1}\right)}^2\end{array}$

Hence the initial rate is$7.5\times {10}^{-8}mo{l}^{-1}L{s}^{-1}{\left(mol{L}^{-1}\right)}^2$.