Chapter 11, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

The complete rate law for the hydrolysis of benzene sulfonyl chloride in presence of fluoride ion has to be derived.

Explanation of Solution

The rate is affected by the concentration of Benzenesulfonyl chloride $\left(BSC\right)$, but it is kept constant in this experiment.  So its role in representing the residual rate is not obvious.  Therefore considering this a rate law can be written as follows,

  $rate=2.4\times {10}^{-7}mol{L}^{-1}{s}^{-1}+k\left[BSC\right]{\left[F^{-}\right]}^i$

Where $iandk$ are presently unknown.

In order to study the effect of $\left[F^{-}\right]$ in the rate of the reaction, residual rate has to be subtracted from the observed rate.

    $\begin{array}{cccc}Experiment& \begin{array}{l}\left[F^{-}\right]\times {10}^2\\ \left(mol/L\right)\end{array}& \begin{array}{l}Initialrate\times {10}^7\\ \left(mol{L}^{-1}{s}^{-1}\right)\end{array}& \begin{array}{l}Adjustedinitialrate\times {10}^7\\ \left(mol{L}^{-1}{s}^{-1}\right)\end{array}\\ 1& 0& \text{2}\text{.4}& 2.4-2.4=0\\ 2& 0.5& \text{5}\text{.4}& 5.4-2.4=3.0\\ 3& 1.0& \text{7}\text{.9}& 7.9-2.4=5.5\\ 4& 2.0& \text{13}\text{.9}& 13.9-2.4=11.5\\ 5& 3.0& \text{20}\text{.2}& 20.2-2.4=17.8\\ 6& 4.0& \text{25}\text{.2}& 25.2-2.4=22.8\\ 7& 5.0& \text{32}\text{.0}& 232.0-2.4=29.6\end{array}$

Therefore the adjusted rate law can be written in the following form.

    Adjusted rate $=k\left[BSC\right]{\left[F^{-1}\right]}^i$

The experiments $2to3,3to4,and4to6$ has to be compared for getting the order of $F^{-}.$  each of this shows a factor two increase in the concentration as shown below.

    $\begin{array}{c}2to3:\frac{5.5}{3.0}=1.8\\ \\ 3to4:\frac{11.5}{5.5}=2.1\\ \\ 4to6:\frac{22.8}{11.5}=2.0\end{array}$

It is clear that the concentration change and the adjusted rate change are almost equal, so the fluoride ion concentration and adjusted rate are proportional.  Though it is only an approximate relationship, a graph can be drawn by taking log of the rate on y axis and $\log \left[F^{-}\right]$ on x axis.  The graph will be linear and the slope will give the order of the reaction.

    $\begin{array}{cc}\log \left[F^{-}\right]& \begin{array}{l}\log \left(adjustedinitialrate\right)\\ \left(mol{L}^{-1}{s}^{-1}\right)\end{array}\\ -2.3& -6.52\\ -2.00& -6.26\\ -1.70& -5.939\\ -1.52& -5.750\\ -1.40& -5.642\\ -1.30& -5.529\end{array}$

Figure 1

Therefore it is clear that the reaction is first order with respect to fluoride ion in the adjusted rate law.

So the rate can be written s follows,

    $rate=2.4\times {10}^{-7}mol{L}^{-1}{s}^{-1}+k\left[BSC\right]\left[F^{-}\right]$

(b)

Interpretation Introduction

Interpretation:

The rate constant has to be calculated for the hydrolysis of benzene sulfonyl chloride in presence of fluoride ion.

Explanation of Solution

The rate constant can be determined using the adjusted rate law as follows,

    $k=\frac{adjustedrate}{\left[BSC\right]\left[F^{-}\right]}$

The rate constant can be calculated for the all the experiments with non-zero fluoride ion can be calculated as follows,

    $k_2=\frac{adjustedrate}{\left[BSC\right]\left[F^{-}\right]}=\frac{3.0\times {10}^{-7}mol{L}^{-1}{s}^{-1}}{\left(2\times {10}^{-4}M\right)\left(0.5\times {10}^{-2}mol{L}^{-1}\right)}=0.3M^{-1}{s}^{-1}$

    $k_3=\frac{5.5\times {10}^{-7}mol{L}^{-1}{s}^{-1}}{\left(2\times {10}^{-4}M\right)\left(1\times {10}^{-2}mol{L}^{-1}\right)}=0.3M^{-1}{s}^{-1}$

    $k_4=\frac{11.5\times {10}^{-7}mol{L}^{-1}{s}^{-1}}{\left(2\times {10}^{-4}M\right)\left(0.5\times {10}^{-2}mol{L}^{-1}\right)}=0.3M^{-1}{s}^{-1}$

    $k_5=\frac{17.8\times {10}^{-7}mol{L}^{-1}{s}^{-1}}{\left(2\times {10}^{-4}M\right)\left(3.0\times {10}^{-2}mol{L}^{-1}\right)}=0.3M^{-1}{s}^{-1}$

    $k_6=\frac{22.8\times {10}^{-7}mol{L}^{-1}{s}^{-1}}{\left(2\times {10}^{-4}M\right)\left(4.0\times {10}^{-2}mol{L}^{-1}\right)}=0.3M^{-1}{s}^{-1}$

    $k_7=\frac{29.6\times {10}^{-7}mol{L}^{-1}{s}^{-1}}{\left(2\times {10}^{-4}M\right)\left(5.0\times {10}^{-2}mol{L}^{-1}\right)}=0.3M^{-1}{s}^{-1}$

    $\begin{array}{cccc}Experiment& \begin{array}{l}\left[F^{-}\right]\times {10}^2\\ \left(mol/L\right)\end{array}& \begin{array}{l}Adjustedinitialrate\times {10}^7\\ \left(mol{L}^{-1}{s}^{-1}\right)\end{array}& \begin{array}{l}\text{adjusted}\text{rate}\text{constant}\\ \left(M^{-1}{s}^{-1}\right)\end{array}\\ 2& 0.5& 3.0& 0.3\\ 3& 1.0& 5.5& 0.3\\ 4& 2.0& 11.5& 0.3\\ 5& 3.0& 17.8& 0.3\\ 6& 4.0& 22.8& 0.3\\ 7& 5.0& 29.6& 0.3\end{array}$

Therefore the average value of the rate constant is $0.3Lmo{l}^{-1}{s}^{-1}.$