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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

The decomposition of \(\mathrm{NO}_{2}(g)\) occurs by the following bimolecular elementary reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The rate constant at 273 \(\mathrm{K}\) is \(2.3 \times 10^{-12} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) , and the activation energy is 111 \(\mathrm{kJ} / \mathrm{mol}\) . How long will it take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of 2.5 \(\mathrm{atm}\) to 1.5 \(\mathrm{atm}\) at \(500 . \mathrm{K}\) ? Assume ideal gas behavior.

The rate law of a reaction can only be determined from experiment. Two experimental procedures for determining rate laws were outlined in Chapter 12. What are the two procedures and how are they used to determine the rate laws?

For enzyme-catalyzed reactions that follow the mechanism $$ \begin{aligned} \mathrm{E}+\mathrm{S} & \rightleftharpoons \mathrm{E} \cdot \mathrm{S} \\ \mathrm{E} \cdot \mathrm{S} & \rightleftharpoons \mathrm{E}+\mathrm{P} \end{aligned} $$ a graph of the rate as a function of [S], the concentration of the substrate, has the following appearance: Note that at higher substrate concentrations the rate no longer changes with [S]. Suggest a reason for this.

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?

A certain first-order reaction is 45.0\(\%\) complete in 65 s. What are the values of the rate constant and the half-life for this process?
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