Problem 27
Question
What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law
Step-by-Step Solution
Verified Answer
a. The units of the rate of a chemical reaction are M/s.
b. The units of the rate constant for a zero-order rate law are M/s.
c. The units of the rate constant for a first-order rate law are \(s^{-1}\).
d. The units of the rate constant for a second-order rate law are \(L * \text{mol}^{-1} * \text{s}^{-1}\).
e. The units of the rate constant for a third-order rate law are \(L^2 * \text{mol}^{-2} * \text{s}^{-1}\).
1Step 1: Identify the rate of a chemical reaction
The rate of a chemical reaction is defined as the change in concentration of one of the reactants or products over a change in time. Mathematically, we can write the rate as: \(\text{rate} = -\frac{1}{\Delta t} \Delta [A]\) or \(\frac{1}{\Delta t} \Delta [B]\), where \([A]\) and \([B]\) represent the concentrations of the reactants.
2Step 2: Determine the units
Given the concentration is provided in moles per liter (M) and time in seconds (s), the rate will have the units of M/s.
#b-e. Units of Rate Constants for different order rate laws#
3Step 1: Using the general rate law equation
The general rate law equation is given by: rate = k * [A]^m * [B]^n, where k is the rate constant, [A] and [B] are concentrations of reactants, and m and n are the orders of the reaction concerning the reactants.
4Step 2: Calculate the rate constant units for different orders
b. For a zero-order rate law (\(m+n=0\)), the rate law equation becomes: rate = k. So the rate constant units will be the same as the rate, which is M/s.
c. For a first-order rate law (\(m+n=1\)), the rate law equation becomes: rate = k * [A]. Thus, the rate constant units are \(\frac{\text{M}}{\text{s} * \text{M}}\), which simplifies to \(s^{-1}\).
d. For a second-order rate law (\(m+n=2\)), the rate law equation becomes: rate = k * [A]^2 or rate = k * [A] * [B]. In this case, the rate constant units are \(\frac{\text{M}}{\text{s} * \text{M}^2}\), which simplifies to \(L * \text{mol}^{-1} * \text{s}^{-1}\).
e. For a third-order rate law (\(m+n=3\)), the rate law equation becomes rate = k * [A]^3 or rate = k * [A]^2 * [B]. In this case, the rate constant units are \(\frac{\text{M}}{\text{s} * \text{M}^3}\), which simplifies to \(L^2 * \text{mol}^{-2} * \text{s}^{-1}\).
Key Concepts
Chemical Reaction RatesZero-Order Rate LawFirst-Order Rate LawSecond-Order Rate LawThird-Order Rate Law
Chemical Reaction Rates
Understanding the rate of a chemical reaction is central to the study of chemical kinetics. This concept refers to how fast a reactant is used up or how fast a product forms in a chemical reaction. It's usually expressed as a change in concentration over time. For example, if the concentration of a reactant decreases over time, then the rate at which the reaction occurs is calculated based on this change. A general expression for this rate is:
\[\text{rate} = -\frac{1}{\Delta t}\Delta [A]\]
or
\[\text{rate} = \frac{1}{\Delta t}\Delta [B]\]
where \([A]\) and \([B]\) are the concentrations of the reactants, and \(\Delta\) signifies the change over time \(\Delta t\). In simpler terms, the negative sign is usually used when we are looking at the disappearance of a reactant, and we ignore it when we describe the formation of a product. We measure the rate in moles per liter per second (M/s), providing us insight into the dynamics of the chemical process.
\[\text{rate} = -\frac{1}{\Delta t}\Delta [A]\]
or
\[\text{rate} = \frac{1}{\Delta t}\Delta [B]\]
where \([A]\) and \([B]\) are the concentrations of the reactants, and \(\Delta\) signifies the change over time \(\Delta t\). In simpler terms, the negative sign is usually used when we are looking at the disappearance of a reactant, and we ignore it when we describe the formation of a product. We measure the rate in moles per liter per second (M/s), providing us insight into the dynamics of the chemical process.
Zero-Order Rate Law
The zero-order rate law describes a reaction where the rate is independent of the concentration of the reactant(s). This scenario might seem counterintuitive because we might expect that having more reactant molecules would lead to a faster reaction. However, in some cases, such as when a catalyst surface is fully saturated with reactant molecules, additional increases in concentration don't affect the rate. When we write the rate equation as
\[\text{rate} = k\]
for a zero-order reaction, \(k\) is the rate constant, which has units of M/s. This simplifies the understanding of the mechanism because the rate will always be constant as long as we have some reactant present. It's important to remember that these reactions will stop once the reactant is used up, as opposed to merely slowing down as the reactants are depleted.
\[\text{rate} = k\]
for a zero-order reaction, \(k\) is the rate constant, which has units of M/s. This simplifies the understanding of the mechanism because the rate will always be constant as long as we have some reactant present. It's important to remember that these reactions will stop once the reactant is used up, as opposed to merely slowing down as the reactants are depleted.
First-Order Rate Law
In first-order reactions, the rate is directly proportional to the concentration of a single reactant. This is commonly observed in many natural and industrial processes, including radioactive decay and certain organic reactions. The rate law for a first-order reaction looks like this:
\[\text{rate} = k[A]\]
where \([A]\) is the concentration of the reactant and \(k\) is the rate constant. The units of the rate constant for first-order reactions are inverse seconds \(s^{-1}\), which signal that the rate will halve for every halving in the concentration of the reactant. This kind of behaviour is often advantageous to understand and predict how a single reactant can control the speed of a reaction.
\[\text{rate} = k[A]\]
where \([A]\) is the concentration of the reactant and \(k\) is the rate constant. The units of the rate constant for first-order reactions are inverse seconds \(s^{-1}\), which signal that the rate will halve for every halving in the concentration of the reactant. This kind of behaviour is often advantageous to understand and predict how a single reactant can control the speed of a reaction.
Second-Order Rate Law
Second-order reactions are characterized by the reaction rate being proportional to the square of a reactant's concentration or the product of two reactants' concentrations. These reactions are described by the rate law
\[\text{rate} = k[A]^2\]
or alternatively
\[\text{rate} = k[A][B]\]
in cases where two different reactants are involved. The rate constant \(k\) has units of \(L \cdot \text{mol}^{-1} \cdot \text{s}^{-1}\), indicating that the reaction rate increases quadratically with an increase in concentration of a single reactant or linearly with the concentration of two reactants. Analyzing such reactions often requires careful consideration of how concentrations interact with each other to affect the reaction speed.
\[\text{rate} = k[A]^2\]
or alternatively
\[\text{rate} = k[A][B]\]
in cases where two different reactants are involved. The rate constant \(k\) has units of \(L \cdot \text{mol}^{-1} \cdot \text{s}^{-1}\), indicating that the reaction rate increases quadratically with an increase in concentration of a single reactant or linearly with the concentration of two reactants. Analyzing such reactions often requires careful consideration of how concentrations interact with each other to affect the reaction speed.
Third-Order Rate Law
A third-order reaction involves either the cubic dependence on the concentration of a single reactant or a combination where the rate is proportional to the product of three reactant concentrations. These cases are exemplified by:
\[\text{rate} = k[A]^3\]
or
\[\text{rate} = k[A]^2[B]\]
or any other combination where the sum of the powers equals three. The rate constant \(k\) in third-order reactions has units of \(L^2 \cdot \text{mol}^{-2} \cdot \text{s}^{-1}\). Due to the high order, even small changes in reactant concentration can lead to large changes in the reaction rate, making these reactions sensitive to concentration variations. Understanding third-order reactions is often necessary for complex chemical processes where multiple reactants are involved.
\[\text{rate} = k[A]^3\]
or
\[\text{rate} = k[A]^2[B]\]
or any other combination where the sum of the powers equals three. The rate constant \(k\) in third-order reactions has units of \(L^2 \cdot \text{mol}^{-2} \cdot \text{s}^{-1}\). Due to the high order, even small changes in reactant concentration can lead to large changes in the reaction rate, making these reactions sensitive to concentration variations. Understanding third-order reactions is often necessary for complex chemical processes where multiple reactants are involved.
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