Problem 98 - 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 \(\mathrm{Cl}_{2}\) is \(0.025 \mathrm{M},\) what is the rate of formation of \(\mathrm{Cl}^{-} ?\)

Step-by-Step Solution

Answer

The rate of formation of Cl⁻ is \(3.5 \times 10^{-6}\) M/s.

Step 1: Write the rate equation

Given that the reaction is first order in both reactants H₂S and Cl₂, we can write the rate equation as: $$ rate = k[\mathrm{H}_{2}\mathrm{S}][\mathrm{Cl}_{2}] $$ where \(k\) is the rate constant, [\(\mathrm{H}_{2}\mathrm{S}\)] and [\(\mathrm{Cl}_{2}\)] represent the concentrations of H₂S and Cl₂, respectively.

Step 2: Calculate the rate of consumption of H₂S

We are given the rate constant \(k = 3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{\; s}^{-1}\), the concentration of H₂S, [\(\mathrm{H}_{2}\mathrm{S}\)] \(= 2.0 \times 10^{-4} \mathrm{M}\), and the concentration of Cl₂, [\(\mathrm{Cl}_{2}\)] \(= 0.025 \mathrm{M}\). We can now calculate the rate of disappearance (or consumption) of H₂S using the rate equation: $$ rate = (3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{\; s}^{-1})\left(2.0 \times 10^{-4} \mathrm{M}\right)\left(0.025 \mathrm{M}\right) $$ Calculate the rate: $$ rate = 1.75 \times 10^{-6} \mathrm{M} \cdot \mathrm{s}^{-1} $$ So, the rate of consumption of H₂S is \(1.75 \times 10^{-6}\) M/s.

Step 3: Find the rate of formation of Cl⁻

Looking at the balanced reaction equation, we can see that the stoichiometric coefficient of H₂S is 1 and the stoichiometric coefficient of Cl⁻ is 2. This means that for each molecule of H₂S consumed, 2 molecules of Cl⁻ will be formed. Therefore, we can use stoichiometry to find the rate of formation of Cl⁻. $$ \text{Rate of formation of Cl}^{-} = 2 \times \text{Rate of consumption of H}_{2}\text{S} $$ Substitute the rate of consumption of H₂S from step 2: $$ \text{Rate of formation of Cl}^{-} = 2 \times (1.75 \times 10^{-6} \mathrm{M} \cdot \mathrm{s}^{-1}) $$ Calculate the rate of formation of Cl⁻: $$ \text{Rate of formation of Cl}^{-} = 3.5 \times 10^{-6} \mathrm{M} \cdot \mathrm{s}^{-1} $$ Therefore, the rate of formation of Cl⁻ is \(3.5 \times 10^{-6}\) M/s.

Key Concepts

First Order Reactions

In chemistry, first order reactions are a specific type of reaction where the rate of the reaction depends linearly on the concentration of a single reactant. This means that if the concentration of this reactant is doubled, the rate of reaction doubles too.

For a reaction

• involving a single reactant A, the rate is directly proportional to the concentration of A.

The general form for a first order reaction rate is given by:\[rate = k[ ext{A}]\]where:- \( k \) is the rate constant for the reaction, depending on factors such as temperature and the nature of the reactants.- \([ ext{A}]\) is the concentration of the reactant.

In the reaction given in our exercise, involving hydrogen sulfide and chlorine, each reactant follows first order kinetics. That means each of their concentrations independently affects the reaction rate. This simplifies calculations, as the rate can be expressed as:\[rate = k[ ext{H}_2 ext{S}][ ext{Cl}_2]\]Understanding the characteristics of first order reactions helps in predicting how changes in concentration affect reaction rates.

Rate Equations

The rate equation is crucial in the study of reaction kinetics. In essence, it provides a mathematical relationship between the speed of a reaction and the concentrations of the reactants. This helps chemists to determine how various factors influence the speed of chemical reactions.

Every chemical reaction has a unique rate equation based on the mechanism of the reaction, noted as:\[rate = k[ ext{A}]^m[ ext{B}]^n\]Here:- \( m \) and \( n \) are the order of the reaction with respect to reactants A and B, respectively.- The overall order of the reaction is \( m+n \).

In our specific exercise, since the reaction is first order with respect to both hydrogen sulfide (\( ext{H}_2 ext{S}\)) and chlorine (\( ext{Cl}_2\)), both \( m \) and \( n \) are 1, and the overall rate equation can be stated as:\[rate = k[ ext{H}_2 ext{S}][ ext{Cl}_2]\]where the rate constant \( k \) is influenced by temperature and specific reaction conditions. Understanding and using the rate equation allows us to calculate the rate of the formation or disappearance of different products and reactants.

Stoichiometry

Understanding stoichiometry is essential in chemical reactions as it refers to the measurement of reactants and products using balanced chemical equations. It tells us the relationship between the quantities of reactants and products.

In a chemical reaction, coefficients in the balanced equation provide the stoichiometric ratios, which can be used to convert between amounts of reactants and products. In our exercise, the balanced reaction is:\[ ext{H}_2 ext{S}(aq) + ext{Cl}_2(aq) \rightarrow ext{S}(s) + 2 ext{H}^+(aq) + 2 ext{Cl}^-(aq)\]Here, the stoichiometric coefficients can be used to determine that:

• For every 1 molecule of \( ext{H}_2 ext{S}\) that reacts, 2 molecules of \( ext{Cl}^-\) ions are formed.

Applying this stoichiometry to rates allows us to scale the rate calculations: If the rate of disappearance of \( ext{H}_2 ext{S}\) is \(1.75 \times 10^{-6} \) M/s, the formation of \( ext{Cl}^-\) proceeds at two times this rate, given the 1:2 ratio—thus, \(3.5 \times 10^{-6} \) M/s.

Stoichiometry is thus a powerful tool to predict how changes in the quantity of reactants affect the quantity of products.