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Chapter 12: Problem 29

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

Short Answer

Expert verified
a. rate of a chemical reaction: \(M/s\) b. rate constant for a zero-order rate law: \(M/s\) c. rate constant for a first-order rate law: \(s^{-1}\) d. rate constant for a second-order rate law: \(M^{-1}s^{-1}\) e. rate constant for a third-order rate law: \(M^{-2}s^{-1}\)

Step by step solution

01

Identifying the required units for the rate of a chemical reaction

The rate of a chemical reaction is given by the change in concentration of a reactant or product per unit time. Since the concentration is expressed in moles/L (M) and time in seconds (s), the units for the rate of a chemical reaction will simply be the concentration units (moles/L) divided by the time units (seconds), resulting in M/s.
02

Identifying the required units for a zero-order rate law

A zero-order reaction has a rate law in the form of: Rate = k[A]^0, where Rate is the rate of the reaction, k is the rate constant, and [A] is the concentration of a reactant. Since [A]^0 is equal to 1, the rate law simplifies to Rate = k. Therefore, the units for the rate constant (k) in a zero-order rate law will be the same as the units for the rate of a chemical reaction, which is M/s.
03

Identifying the required units for a first-order rate law

A first-order reaction has a rate law in the form of: Rate = k[A], where Rate is the rate of the reaction, k is the rate constant, and [A] is the concentration of a reactant. To determine the units for k, we can rearrange the rate law: k = Rate / [A]. Since Rate has the units M/s and [A] has the units M, the units for k in a first-order rate law will be (M/s)/(M), which simplifies to 1/s or s^{-1}.
04

Identifying the required units for a second-order rate law

A second-order reaction has a rate law in the form of: Rate = k[A]^2, where Rate is the rate of the reaction, k is the rate constant, and [A] is the concentration of a reactant. To determine the units for k, we can rearrange the rate law: k = Rate / ([A]^2). Since Rate has the units M/s and [A]^2 has the units M^2, the units for k in a second-order rate law will be (M/s)/(M^2), which simplifies to L/mol*s or M^{-1}s^{-1}.
05

Identifying the required units for a third-order rate law

A third-order reaction has a rate law in the form of: Rate = k[A]^3, where Rate is the rate of the reaction, k is the rate constant, and [A] is the concentration of a reactant. To determine the units for k, we can rearrange the rate law: k = Rate / ([A]^3). Since Rate has the units M/s and [A]^3 has the units M^3, the units for k in a third-order rate law will be (M/s)/(M^3), which simplifies to L^2/mol^2*s or M^{-2}s^{-1}. In conclusion, the units for each parameter are as follows: a. rate of a chemical reaction: M/s b. rate constant for a zero-order rate law: M/s c. rate constant for a first-order rate law: s^{-1} d. rate constant for a second-order rate law: M^{-1}s^{-1} e. rate constant for a third-order rate law: M^{-2}s^{-1}

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Most popular questions from this chapter

Define stability from both a kinetic and thermodynamic perspective. Give examples to show the differences in these concepts.

The rate law for the reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ at some temperature is $$ \text {Rate} =-\frac{\Delta[\text { NOBr }]}{\Delta t}=k[\mathrm{NOBr}]^{2} $$ a. If the half-life for this reaction is 2.00 s when \([\mathrm{NOBr}]_{0}=\) \(0.900 M,\) calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to 0.100\(M ?\)

The decomposition of NH3 to N2 and H2 was studied on two surfaces: Without a catalyst, the activation energy is 335 \(\mathrm{kJ} / \mathrm{mol}\) . a. Which surface is the better heterogeneous catalyst for the decomposition of \(\mathrm{NH}_{3} ?\) Why? b. How many times faster is the reaction at 298 \(\mathrm{K}\) on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor \(A\) is the same for each reaction. c. The decomposition reaction on the two surfaces obeys a rate law of the form $$ |text {Rate} =k \frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{H}_{2}\right]} $$ How can you explain the inverse dependence of the rate on the \(\mathrm{H}_{2}\) concentration?

For the reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C},\) explain at least two ways in which the rate law could be zero order in chemical A.

A popular chemical demonstration is the "magic genie" procedure, in which hydrogen peroxide decomposes to water and oxygen gas with the aid of a catalyst. The activation energy of this (uncatalyzed) reaction is 70.0 \(\mathrm{kJ} / \mathrm{mol}\) . When the catalyst is added, the activation energy (at \(20 .^{\circ} \mathrm{C} )\) is 42.0 \(\mathrm{kJ} / \mathrm{mol}\) . Theoretically, to what temperature ( \((\mathrm{C})\) would one have to heat the hydrogen peroxide solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at \(20 .^{\circ} \mathrm{C} ?\) Assume the frequency factor \(A\) is constant, and assume the initial concentrations are the same.
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