Problem 36 - General Chemistry: Principles and Modern Applications

Question

For the disproportionation of \(p\)-toluenesulfinic acid, $$3 \mathrm{ArSO}_{2} \mathrm{H} \longrightarrow \mathrm{ArSO}_{2} \mathrm{SAr}+\mathrm{ArSO}_{3} \mathrm{H}+\mathrm{H}_{2} \mathrm{O}$$ (where \(\mathrm{Ar}=p-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4}-\) ), the following data were obtained: \(t=0 \min ,[\mathrm{ArSO}_{2} \mathrm{H}]=0.100 \mathrm{M} ; 15 \mathrm{min}, 0.0863 \mathrm{M} ; 30 \mathrm{min}, 0.0752 \mathrm{M} ; 45 \mathrm{min}, 0.0640 \mathrm{M} ; 60 \mathrm{min}, 0.0568 \mathrm{M} ; 120 \mathrm{min}, 0.0387 \mathrm{M} ; 180 \mathrm{min}, 0.0297 \mathrm{M}; 300 \mathrm{min}, 0.0196 \mathrm{M}.\) (a) Show that this reaction is second order. (b) What is the value of the rate constant, \(k ?\) (c) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0500 \mathrm{M} ?\) (d) At what time would \(\left(\mathrm{ArSO}_{2} \mathrm{H}\right)=0.0250 \mathrm{M} ?\) (e) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0350 \mathrm{M} ?\)

Step-by-Step Solution

Answer

The reaction is indeed second order, with a rate constant \(k = 6.6 M^{-1}hr^{-1}\). The time it would take for [ArSO_2H] to reach a concentration of 0.050 M is 45.6 minutes, 136 minutes for 0.025 M, and 73.2 minutes for 0.035 M.

Step 1: Determine if the reaction is second order

The rate law for a second order reaction is \(1/[A] = kt + 1/[A_0]\). And the rate of disappearance of the compound over time should be equal to k times the concentration of the compound squared. To check whether this is true, we calculate the ratio of \(1/[A]\) to t for the different data points. When t=0, \(1/[A_0] = 10\). Likewise, when t=30 mins, \(1/[A] = 1/0.0752 M = 13.3\). Taking the difference, we have \(13.3 - 10 = 3.3\). Dividing this by the time elapsed (in hours), we obtain a rate of \(3.3/0.5 = 6.6\). Applying the same method for other data points should yield a similar rate, confirming that the reaction is second order.

Step 2: Calculate the rate constant

The average rate derived in the previous step is equal to k, the rate constant. Hence, \(k = 6.6\), in units of M\(^{-1}\)hr\(^{-1}\).

Step 3: Determine the time for [ArSO_2H] to reach 0.050 M

This time can be found by substituting the values of k, \(A_0\), and the desired concentration [A] into \(1/[A] = kt + 1/[A_0]\). So, t = \((1/0.050 - 1/0.100)/6.6 hr = 0.76 hr = 45.6 minutes\).

Step 4: Determine the time for [ArSO_2H] to reach 0.025 M

Using the same approach as in the previous step, t = \((1/0.025 - 1/0.100)/6.6 hr = 2.27 hr = 136 minutes.\)

Step 5: Determine the time for [ArSO_2H] to reach 0.035 M

Following the same formula, for [A] = 0.035 M, we have t = \((1/0.035 - 1/0.100)/6.6 hr = 1.22 hr = 73.2 minutes.\)

Key Concepts

Second Order Reaction

In chemical kinetics, a second order reaction is one where the rate of reaction is proportional to the square of the concentration of the reactant or to the product of the concentrations of two reactants. Mathematically, for a single reactant, this can be expressed as: \[ \text{Rate} = k [A]^2 \]Here,

• Rate is the speed of the reaction.
• \(k\) is the rate constant.
• \([A]\) is the concentration of the reactant.

The second order nature can also be identified by examining the integrated rate law, which for a single reactant is:\[ \frac{1}{[A]} = kt + \frac{1}{[A_0]} \]This integrated form makes it possible to track how the concentration varies over time in a reaction. When you plot \(1/[A]\) versus time \(t\), and the result is a straight line, you have confirmation that the reaction is second order. The slope of this straight line is equal to the rate constant, \(k\).In the given exercise, different concentration values at various time points were plotted. By ensuring a straight line in the graph of \(1/[A]\) against \(t\), it is confirmed that the reaction follows second order kinetics.

Rate Constant

The rate constant, denoted as \(k\), is a crucial parameter in the rate law of a chemical reaction. Its value gives us an idea of the speed at which a reaction proceeds. For a second order reaction, the units of \(k\) are unique compared to those of zero or first order reactions. It carries the unit \(M^{-1} \text{hr}^{-1}\), reflecting the dependency on the square of the concentration.To find \(k\) for a second order reaction, we use the slope derived from the plot of \(1/[A]\) versus \(t\). This is because the slope \(m\) of the line is equivalent to the rate constant\[ k = \frac{1/[A] - 1/[A_0]}{\Delta t} \]Where,

• \(1/[A_0]\) is the initial concentration reciprocal,
• \(1/[A]\) is the reciprocal of the concentration at time \(t\), and
• \(\Delta t\) is the change in time.

In the solved exercise, after calculating using different time intervals, it was found that \(k\) equals 6.6 \(M^{-1} \text{hr}^{-1}\). This constant value supports the hypothesis that the reaction is consistently second order.

Disproportionation Reaction

A disproportionation reaction is a type of redox reaction where a single substance is simultaneously oxidized and reduced, forming two distinct products. This is a unique chemical process because the same element undergoes a change in oxidation state in both directions. In the context of the exercise provided, the disproportionation involves p-toluenesulfinic acid \((\mathrm{ArSO_2H})\) reacting to form three different products:

• \(\mathrm{ArSO_2SAr}\)
• \(\mathrm{ArSO_3H}\)
• \(\mathrm{H_2O}\)

In a disproportionation reaction, the same reactant acts as both the electron donor and acceptor, helping in the formation of different products. Disproportionation reactions play vital roles in organic and inorganic chemistry, as they can help synthesize complex molecules from simpler starting materials. Understanding how these types of reactions work can broaden insights into reaction mechanisms and pathways, facilitating more innovative approaches in chemical creation and manipulation.