Problem 90 - Chemistry: The Central Science

Question

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(28{ }^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.0 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2} \mathrm{~s} \mathrm{} 0.025 \mathrm{M}\), what is the rate of formation of \(\mathrm{Cl}^{-}\)?

Step-by-Step Solution

Answer

The rate of formation of Cl- in the given reaction at the specified conditions is \(3.5 \times 10^{-6}\ \mathrm{M\ s^{-1}}\).

Step 1: Write the rate law for the reaction

Since it's given that the reaction is first order with respect to both reactants (H2S and Cl2), we can write the rate law for this reaction as follows: \[Rate = k[\mathrm{H}_{2} \mathrm{~S}][\mathrm{Cl}_{2}]\]

Step 2: Substitute the values

Next, substitute the given values for k, concentration of H2S, and concentration of Cl2. The rate constant k is given as 3.5 x 10^{-2} M^{-1} s^{-1}, and the concentrations of H2S and Cl2 are given as 2.0 x 10^{-4} M and 0.025 M, respectively. \[Rate = (3.5 \times 10^{-2} \mathrm{M^{-1}s^{-1}})(2.0 \times 10^{-4} \mathrm{M})(0.025 \mathrm{M})\]

Step 3: Calculate the rate of the reaction

Now, multiply the values as required and calculate the rate of the reaction. \[Rate = (3.5 \times 10^{-2})(2.0 \times 10^{-4})(0.025)\] \[Rate = 1.75 \times 10^{-6} \mathrm{M\ s^{-1}}\]

Step 4: Calculate the rate of formation of Cl-

Since we have the rate of the reaction, we can find the rate of formation of Cl-. In the given reaction, for every 1 mole of Cl2 reacting, 2 moles of Cl- are formed. Therefore, the rate of formation of Cl- will be twice the rate of the reaction: \[Rate_{formation\ of\ Cl^{-}} = 2 \times Rate\] \[Rate_{formation\ of\ Cl^{-}} = 2 \times 1.75 \times 10^{-6} \mathrm{M\ s^{-1}}\] \[Rate_{formation\ of\ Cl^{-}} = 3.5 \times 10^{-6}\ \mathrm{M\ s^{-1}}\]

Step 5: Report the result

The rate of formation of Cl- in the given reaction at the specified conditions is 3.5 x 10^{-6} M s^{-1}.

Key Concepts

Rate Law

In chemical kinetics, the rate law is an equation that expresses the rate of a reaction as a function of the concentration of its reactants. Every reaction has its unique rate law, which must be determined experimentally. The form of the rate law can vary depending on the reaction order with respect to each reactant, which might include zero, first, second, and higher-order reactions.

For the reaction involving hydrogen sulfide (\[\mathrm{H}_{2} \mathrm{~S}(a q)\]) and chlorine (\[\mathrm{Cl}_{2}(g)\]), the rate law is determined to be first order with respect to each reactant. This implies that the reaction rate is directly proportional to the concentration of the reactants involved.

Hence, the rate law for this reaction can be expressed as:
\[Rate = k[\mathrm{H}_{2} \mathrm{~S}][\mathrm{Cl}_{2}]\]

First Order Reaction

A first order reaction is one where the rate depends linearly on the concentration of one reactant. This means if you double the concentration, the rate of reaction also doubles. In our exercise, the given reaction is first order for each reactant, hydrogen sulfide and chlorine.

This accelerates the reaction rate linearly with an increase in the concentration of \[\mathrm{H}_{2} \mathrm{~S}(a q)\], or \[\mathrm{Cl}_{2}(g)\]. By understanding this relationship, one can predict the reaction behavior upon changes in concentrations.

• First order with respect to \[\mathrm{H}_{2} \mathrm{~S}\]: Changes in \[\mathrm{H}_{2} \mathrm{~S}\] affect the rate proportionally.
• First order with respect to \[\mathrm{Cl}_{2}\]: Changes in \[\mathrm{Cl}_{2}\] affect the rate proportionally.

Rate Constant

The rate constant, denoted by the symbol \(k\), is a proportionality factor in the rate law. In the context of our problem, the rate constant for the disappearance of hydrogen sulfide (\[\mathrm{H}_{2} \mathrm{~S}\]) is given as \(3.5 \times 10^{-2} \, \mathrm{M}^{-1} \, \mathrm{s}^{-1}\).

The units of the rate constant depend on the overall order of the reaction. Because it is a second order reaction (first order + first order = second order), the units are \(\mathrm{M}^{-1} \mathrm{s}^{-1}\).

• A larger rate constant indicates a faster reaction at constant concentrations.
• The rate constant is only affected by changes in temperature or the presence of a catalyst.

Reaction Rate Calculation

Calculating the reaction rate involves using the rate law and plugging in the current concentrations of the reactants. In our example, the concentrations given are:

• \(\mathrm{H}_{2} \mathrm{~S} = 2.0 \times 10^{-4} \mathrm{M}\)
• \(\mathrm{Cl}_{2} = 0.025 \mathrm{M}\)

By substituting these values into the rate law:
\[Rate = (3.5 \times 10^{-2} \, \mathrm{M}^{-1} \, \mathrm{s}^{-1})(2.0 \times 10^{-4} \, \mathrm{M})(0.025 \, \mathrm{M})\]
Perform the multiplication:\[Rate = 1.75 \times 10^{-6} \, \mathrm{M} \cdot \mathrm{s}^{-1}\]
The resulting rate is the rate of disappearance for \[\mathrm{H}_{2} \mathrm{~S}\], which is also used to find the rate of formation of \[\mathrm{Cl}^{-}\] by considering stoichiometry.