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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. The units for the rate of a chemical reaction are mol/Ls. b. The units for the rate constant of a zero-order rate law are mol/Ls. c. The units for the rate constant of a first-order rate law are s⁻¹. d. The units for the rate constant of a second-order rate law are L/mol·s. e. The units for the rate constant of a third-order rate law are L²/mol²·s.

Step by step solution

01

a. Units for the rate of a chemical reaction

The rate of a chemical reaction (represented as 'r') is usually expressed as the change in concentration (∆[A]) of the reacting species A over the change in time (∆t). Thus: \(r = \frac{-\Delta{[A]}}{\Delta{t}}\) The units for the change in concentration are moles per liter (mol/L) and time in seconds (s). Therefore, the unit for the rate of a chemical reaction will be in mol/Ls.
02

b. Units for the rate constant of a zero-order rate law

In a zero-order rate law, the rate of the reaction (r) is directly proportional to the concentration of the reactant [A]^0: \(r = k [A]^0\) Since [A]^0 = 1, the rate constant k has the same units as the rate of reaction. Thus, the units for a zero-order rate constant k are mol/Ls.
03

c. Units for the rate constant of a first-order rate law

In a first-order rate law, the rate of the reaction (r) is directly proportional to the concentration of the reactant [A]: \(r = k [A]\) To find the units of the rate constant k, we rearrange the equation: \(k = \frac{r}{[A]}\) Since the units of the rate are mol/Ls and the units of the concentration are mol/L, the units for the rate constant k in a first-order rate law are s⁻¹.
04

d. Units for the rate constant of a second-order rate law

In a second-order rate law, the rate of the reaction (r) is directly proportional to the square of the concentration of the reactant [A]^2: \(r = k [A]^2\) To find the units of the rate constant k, we rearrange the equation: \(k = \frac{r}{[A]^2}\) Since the units of the rate are mol/Ls and the units of the concentration squared are (mol/L)^2, the units for the rate constant k in a second-order rate law are L/mol·s.
05

e. Units for the rate constant of a third-order rate law

In a third-order rate law, the rate of the reaction (r) is directly proportional to the cube of the concentration of the reactant [A]^3: \(r = k [A]^3\) To find the units of the rate constant k, we rearrange the equation: \(k = \frac{r}{[A]^3}\) Since the units of the rate are mol/Ls and the units of the concentration cubed are (mol/L)^3, the units for the rate constant k in a third-order rate law are L²/mol²·s.

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Most popular questions from this chapter

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \mathrm{~kJ} / \mathrm{mol}\). In the presence of a catalyst at \(37^{\circ} \mathrm{C}\), the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

The activation energy for the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is \(125 \mathrm{~kJ} / \mathrm{mol}\), and \(\Delta E\) for the reaction is \(-216 \mathrm{~kJ} / \mathrm{mol}\). What is the activation energy for the reverse reaction \(\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow\right.\) \(\left.\mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?\)

Draw a rough sketch of the energy profile for each of the following cases: a. \(\Delta E=+10 \mathrm{~kJ} / \mathrm{mol}, E_{\mathrm{a}}=25 \mathrm{~kJ} / \mathrm{mol}\) b. \(\Delta E=-10 \mathrm{~kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{~kJ} / \mathrm{mol}\) c. \(\Delta E=-50 \mathrm{~kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{~kJ} / \mathrm{mol}\)

In the Haber process for the production of ammonia, $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ what is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?

Consider the reaction $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ If, in a certain experiment, over a specific time period, \(0.0048 \mathrm{~mol}\) \(\mathrm{PH}_{3}\) is consumed in a 2.0-L container each second of reaction, what are the rates of production of \(\mathrm{P}_{4}\) and \(\mathrm{H}_{2}\) in this experiment?

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