Express the units for rate constants when the concentrations are in moles per liter and time is in seconds for (a) zero-order reactions; (b) first-order reactions; (c) second-order reactions.

Short Answer

Expert verified
For zero-order reactions, the rate constant k has units M/s. For first-order reactions, k has units s^{-1}. For second-order reactions, k has units M^{-1}s^{-1}.

Step by step solution

01

Understanding Rate Law and Units

The rate of a chemical reaction can often be expressed by a rate law, which shows the relationship between the reaction rate and the concentrations of reactants. The rate constant, denoted as 'k', is a proportionality constant in this equation. The units of 'k' depend on the overall order of the reaction, which is determined by the sum of the exponents of the concentration terms in the rate law equation. Each order of reaction has its own units for the rate constant to ensure that the rate of the reaction has consistent units of concentration over time.
02

Zero-Order Reactions

For a zero-order reaction, the rate law has the form Rate = k[A]^0, where the concentration of reactant A does not affect the rate of the reaction. Since [A]^0 = 1, the rate is simply k and has units of concentration over time; thus, the units for k in this case would be moles per liter per second (M/s).
03

First-Order Reactions

For a first-order reaction, the rate law is Rate = k[A]^1, so the rate is directly proportional to the concentration of A. Here, the units of k must cancel out the concentration units, leaving only time^{-1}. Therefore, the units for k are s^{-1}.
04

Second-Order Reactions

In second-order reactions, the rate law is Rate = k[A]^2, indicating that the rate is proportional to the square of the concentration of reactant A. To ensure the rate has units of concentration/time, the units for k must be (concentration)^{-1} over (time), which in this case is M^{-1}s^{-1}.

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Key Concepts

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

Zero-Order Reactions
Zero-order reactions are chemical processes for which the rate is independent of the concentration of the reactant. This means that the rate of reaction remains constant, even when the concentration of the reactant changes. To visualize this, think of a saturated catalyst working at maximum capacity, no matter how much more reactant you add, it can't speed up the reaction.

In a mathematical form, the rate law for a zero-order reaction can be written as: \[ \text{Rate} = k[A]^0 \] with \( [A]^0 \) simply being 1. Therefore, the rate constant \( k \) in this situation has units of moles per liter per second (M/s), as it represents the rate at which the reactant is getting consumed over time.
First-Order Reactions
A first-order reaction depends linearly on the concentration of one reactant. Common examples include certain radioactive decay processes and the decomposition of substances when heated. In this case, the rate at which the reaction happens changes proportionally with changes in the concentration of the reactant.

For instance, if you double the amount of reactant, the rate of the reaction will also double. The rate law expression for a first-order reaction is: \[ \text{Rate} = k[A] \] The units for the rate constant \( k \) here need to be in inverse seconds (\( s^{-1} \)), indicating that it counteracts the molarity (M) unit of the reactant concentration, leaving a rate that is expressed in terms of time.
Second-Order Reactions
In second-order reactions, the reaction rate is proportional to the square of the concentration of one reactant, or to the product of the concentrations of two reactants. Hence, the impact on the rate is much more dramatic when compared to first-order reactions: doubling the concentration of a reactant quadruples the rate of reaction.

Representing this mathematically, we have: \[ \text{Rate} = k[A]^2 \text{ or } \text{Rate} = k[A][B] \] As such, the rate constant \( k \) in these reactions has the units of \( M^{-1}s^{-1} \), implying that it negates both concentration units of the reactants, thus leaving us with the rate's time dimension.
Rate Law
Rate law establishes a crucial relationship in chemical kinetics between the concentration of reactants and the rate of a chemical reaction. Different reactions follow different rate laws, which dictate how the rate responds to varying concentration levels of the reactants involved.

In essence, the rate law tells us 'how much' and 'how quickly' a reaction will occur under certain conditions, and it is often determined experimentally. A general form of the rate law can be written as: \[ \text{Rate} = k[A]^m[B]^n \] where \( m \) and \( n \) are the reaction orders with respect to reactants A and B, and \( k \) is the rate constant. It's crucial for students to understand that the values of \( m \) and \( n \) are not solely based on the stoichiometry of the reaction but are determined through careful experimentation.
Chemical Kinetics
Chemical kinetics is the field of chemistry that deals with understanding the speed or rate of a chemical reaction and what factors affect this rate. It helps us comprehend how quickly reactants turn into products and which conditions may alter this speed. This field incorporates concepts of reaction rates, reaction mechanisms, catalysis, and stability, providing a complete view of how and why reactions occur as they do.

Within chemical kinetics, scientists study the effect of temperature, concentration, surface area, and the presence of catalysts on the rates of chemical reactions. By understanding kinetics, chemists can design and control reactions to be more efficient, safer, and environmentally friendly, which is essential in industries such as pharmaceuticals, petrochemicals, and food technology.

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Most popular questions from this chapter

For the reversible, one-step reaction \(\mathrm{A}+\mathrm{B} \rightleftarrows \mathrm{C}+\mathrm{D}\), the forward rate constant is \(52.4 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\) and the rate constant for the reverse reaction is \(32.1 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\). The activation energy was found to be \(35.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the forward reaction and \(44.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the reverse reaction. (a) What is the equilibrium constant for the reaction? (b) Is the reaction exothermic or endothermic? (c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant?

Manganate ions, \(\mathrm{MnO}_{4}^{2-}\), react at \(2.0 \mathrm{~mol} \cdot \mathrm{L}^{-1} \cdot \mathrm{min}^{-1}\) in acidic solution to form permanganate ions and manganese(IV) oxide: \(3 \mathrm{MnO}_{4}{ }^{2-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{MnO}_{4}{ }^{-}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{~s})+\) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\). (a) What is the rate of formation of permanganate ions? (b) What is the rate of reaction of \(\mathrm{H}^{+}(\mathrm{aq})\) ? (c) What is the unique rate of the reaction?

Determine which of the following statements about catalysts are true. If the statement is false, explain why. (a) A heterogeneous catalyst works by binding one or more of the molecules undergoing reaction to the surface of the catalyst. (b) Enzymes are naturally occurring proteins that serve as catalysts in biological systems. (c) The equilibrium constant for a reaction is greater in the presence of a catalyst. (d) A catalyst changes the pathway of a reaction in such a way that the reaction becomes more exothermic.

The half-life for the second-order reaction of a substance A is \(50.5 \mathrm{~s}\) when \([\mathrm{A}]_{0}=0.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\). Calculate the time needed for the concentration of A to decrease to (a) one- sixteenth; (b) onefourth; (c) one-fifth of its original value.

The pre-equilibrium and the steady-state approximations are two different approaches to deriving a rate law from a proposed mechanism. For the following mechanism, determine the rate law (a) by the pre-equilibrium approximation and (b) by the steady-state approximation. (c) Under what conditions do the two methods give the same answer? (d) What will the rate laws become at high concentrations of \(\mathrm{Br}^{-}\)? $$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \text {(fast equilibrium) } \\ &\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O} \text { (slow) } \end{aligned} $$

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