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Hydrogen sulfide is oxidized by chlorine in aqueous solution. $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{HCl}(a q) $$ The experimental rate law is $$ \text { Rate }=k\left[\mathrm{H}_{2} \mathrm{~S}\right]\left[\mathrm{Cl}_{2}\right] $$ What is the reaction order with respect to \(\mathrm{H}_{2} \mathrm{~S} ?\) with respect to \(\mathrm{Cl}_{2}\) ? What is the overall order?

Short Answer

Expert verified
The order with respect to \(\mathrm{H}_{2}\mathrm{~S}\) is 1, for \(\mathrm{Cl}_{2}\) is 1, and the overall order is 2.

Step by step solution

01

Understand the Reaction

The equation provided represents the reaction of hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{~S}\)) with chlorine (\(\mathrm{Cl}_{2}\)) to yield sulfur (\(\mathrm{S}\)) and hydrochloric acid (\(\mathrm{HCl}\)).
02

Analyze the Rate Law

The given rate law is \( \text{Rate} = k[\mathrm{H}_{2}\mathrm{~S}][\mathrm{Cl}_{2}] \). This equation tells us how the rate of the reaction depends on the concentration of the reactants \(\mathrm{H}_{2}\mathrm{~S}\) and \(\mathrm{Cl}_{2}\).
03

Identify Reaction Orders for Each Reactant

From the rate law, the reaction order with respect to a reactant is the exponent of its concentration in the rate equation. Both \([\mathrm{H}_{2}\mathrm{~S}]\) and \([\mathrm{Cl}_{2}]\) have an exponent of 1 in the rate law, hence each has a reaction order of 1.
04

Calculate Overall Reaction Order

The overall reaction order is the sum of the individual orders of each reactant. Therefore, the overall order is \(1 + 1 = 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Order
In chemical kinetics, the reaction order is a crucial concept that tells us how changes in concentrations of reactants influence the rate of a reaction. For any chemical reaction, the reaction order with respect to a specific reactant shows how the rate depends on the concentration of that reactant.
Every term in the rate law equation has an exponent associated with it, indicating the reaction order of the reactant it corresponds to.
For example, in the reaction between hydrogen sulfide and chlorine, represented by the rate law \( \text{Rate} = k[\mathrm{H}_{2}\mathrm{~S}][\mathrm{Cl}_{2}] \), the concentration of hydrogen sulfide is raised to the power of 1.
This implies:
  • The reaction order with respect to \( \mathrm{H}_{2}\mathrm{~S} \) is 1.
  • Similarly, the reaction order with respect to \( \mathrm{Cl}_{2} \) is also 1.
Understanding reaction orders helps in predicting how the rate will change with varying reactant concentrations.
Rate of Reaction
The rate of reaction measures how quickly the reactants are converted to products in a chemical reaction. It is typically expressed in terms of concentration change over time.
The rate is influenced by several factors including the concentration of reactants, temperature, and presence of catalysts.
For our specific reaction between hydrogen sulfide and chlorine, the rate of reaction is governed by the experimental rate law:
  • \( \text{Rate} = k[\mathrm{H}_{2}\mathrm{~S}][\mathrm{Cl}_{2}] \)
This expression tells us:
  • The rate is directly proportional to the concentration of \( \mathrm{H}_{2}\mathrm{~S} \) and \( \mathrm{Cl}_{2} \).
  • If the concentration of either reactant is increased, the rate will increase linearly, as both reactants have an order of 1.
The rate constant \( k \) remains unchanged unless conditions such as temperature are altered.
Experimental Rate Law
An experimental rate law is an empirical equation derived from experimental data that describes the rate of a chemical reaction.
It provides a mathematical relationship between the rate of reaction and the concentrations of reactants.
In the case we're analyzing, the experimental rate law for the reaction of hydrogen sulfide with chlorine is:
  • \( \text{Rate} = k[\mathrm{H}_{2}\mathrm{~S}][\mathrm{Cl}_{2}] \)
This empirical equation indicates:
  • Each reactant affects the rate of the reaction independently, according to its concentration raised to the power of its reaction order.
  • The rate law does not necessarily reflect the stoichiometry of the reaction but results from experimental observations.
Understanding experimental rate laws is vital for predicting reaction behavior under different conditions and allows for further optimization of chemical reactions in practical applications.

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Most popular questions from this chapter

You are running the reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow \mathrm{C}+3 \mathrm{D}\). Your lab partner has conducted the first two experiments to determine the rate law for the reaction. He has recorded the initial rates for these experiments in another data table. Come up with some reactant concentrations for Experiment 3 that will allow you to determine the rate law by measuring the initial rate. $$ \begin{array}{lll} \text { Experiment } & \text { Concentration } & \text { Concentration } \\ \text { Number } & \text { of A }(M) & \text { of B }(M) \\ 1 & 1.0 & 1.0 \\ 2 & 2.0 & 1.0 \\ 3 & & \end{array} $$

The following values of the rate constant were obtained for the decomposition of nitrogen dioxide at various temperatures. Plot the logarithm of \(k\) versus \(1 / T\) and from the graph obtain the energy of activation. $$ \begin{array}{ll} \text { Temperature }\left({ }^{\circ} \mathbf{C}\right) & \boldsymbol{k}(\mathbf{L} / \mathbf{m o l} \cdot \mathbf{s}) \\ 320 & 0.527 \\ 330 & 0.776 \\ 340 & 1.121 \\ 350 & 1.607 \end{array} $$

Sketch a potential-energy diagram for the reaction of nitric oxide with ozone. $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ The activation energy for the forward reaction is \(10 \mathrm{~kJ} ;\) the \(\Delta H^{\circ}\) is \(-200 \mathrm{~kJ}\). What is the activation energy for the reverse reaction? Label your diagram appropriately.

A second-order decomposition reaction run at \(550{ }^{\circ} \mathrm{C}\) has a rate constant of \(3.1 \times 10^{-2} \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})\). If the initial concentration of the reactant is \(0.10 M\), what is the concentration of this reactant after \(1.0 \times 10^{2} \mathrm{~s} ?\) What is the half-life of this reaction under these conditions?

Cyclobutane, \(\mathrm{C}_{4} \mathrm{H}_{8}\), consisting of molecules in which four carbon atoms form a ring, decomposes, when heated, to give ethylene. $$ \mathrm{C}_{4} \mathrm{H}_{8}(g) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(g) $$ The reaction is first order. In an experiment, the initial concentration of cyclobutane was \(0.00150 M\). After heating at \(450^{\circ} \mathrm{C}\) for \(455 \mathrm{~s}\), this was reduced to \(0.00119 \mathrm{M}\). What was the concentration of cyclobutane after a total of \(750 \mathrm{~s}\) ?

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