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Chapter 14: Problem 28

Consider a hypothetical reaction between \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) that is first order in A, zero order in B, and second order in C. (a) Write the rate law for the reaction. (b) How does the rate change when \([A]\) is doubled and the other reactant concentrations are held constant? (c) How does the rate change when [B] is tripled and the other reactant concentrations are held constant? (d) How does the rate change when \([\mathrm{C}]\) is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are tripled? \((f)\) By what factor does the rate change when the concentrations of all three reactants are cut in hal??

Short Answer

Expert verified
The rate law for the given reaction is \(Rate=k[A]^1[B]^0[C]^2\). (b) Doubling the concentration of A doubles the rate. (c) Tripling the concentration of B has no effect on the rate. (d) Tripling the concentration of C increases the rate by a factor of 9. (e) Tripling the concentrations of all reactants increases the rate by a factor of 27. (f) Cutting the concentrations of all reactants in half decreases the rate by a factor of \(\frac{1}{8}\).

Step by step solution

01

(a) Rate Law

The rate law is a mathematical formula that describes how the rate of a reaction depends on the concentrations of the reactants. In this case, the reaction is first order in A, zero order in B, and second order in C. So the rate law for this reaction can be written as: \[Rate = k[A]^1[B]^0[C]^2\]
02

(b) Doubling Concentration of A

To determine how the rate changes when the concentration of A is doubled, let's use the rate law and assume that the concentration of A is \(2[A]\), while the concentrations of B and C are held constant: \[New\ Rate = k[2A]^1[B]^0[C]^2\] Since A is first order, the effect on the rate is proportional to the change in the concentration of A. Therefore, the rate will also be doubled when the concentration of A is doubled.
03

(c) Tripling Concentration of B

Now we will determine the rate when the concentration of B is tripled and the concentrations of A and C are held constant: \[New\ Rate = k[A]^1[3B]^0[C]^2\] As B is zero order, the change in its concentration does not affect the rate. Thus, the rate remains unchanged when the concentration of B is tripled.
04

(d) Tripling Concentration of C

Next, we will investigate how the rate changes when the concentration of C is tripled, while the concentrations of A and B are held constant: \[New\ Rate = k[A]^1[B]^0[3C]^2\] Since C is second order, the effect on the rate is proportional to the square of the change in the concentration of C. Therefore, the rate will increase by a factor of \(3^2 = 9\) when the concentration of C is tripled.
05

(e) Tripling All Reactant Concentrations

Let's find out how the rate changes when the concentrations of all three reactants are tripled. We can rewrite the rate law with this change implemented: \[New\ Rate=k[3A]^1[3B]^0[3C]^2\] Calculate the ratio of the new rate to the original rate law: \[\frac{New\ Rate}{Rate}=\frac{k[3A]^1[3B]^0[3C]^2}{k[A]^1[B]^0[C]^2}=3^1\times1^0\times3^2=3\times9\] Thus, the rate changes by a factor of 27 when the concentrations of all three reactants are tripled.
06

(f) Halving All Reactant Concentrations

Finally, we will determine how the rate changes when the concentrations of all three reactants are cut in half: \[New\ Rate = k[\frac{1}{2}A]^1[\frac{1}{2}B]^0[\frac{1}{2}C]^2 \] Calculate the ratio of the new rate to the original rate law: \[\frac{New\ Rate}{Rate}=\frac{k[\frac{1}{2}A]^1[\frac{1}{2}B]^0[\frac{1}{2}C]^2}{k[A]^1[B]^0[C]^2}=\frac{1}{2}^1\times1^0\times\frac{1}{2}^2=\frac{1}{2}\times\frac{1}{4}\] Therefore, the rate changes by a factor of \(\frac{1}{8}\) when the concentrations of all three reactants are cut in half.

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Key Concepts

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

Rate Law
The rate law of a chemical reaction is a mathematical expression that illustrates how the reaction rate is influenced by the concentration of reactants. For the given exercise involving the reaction of substances A, B, and C, the rate law is derived based on the reaction's orders: it is first order in A, zero order in B, and second order in C. This leads to a rate law given by \[ \text{Rate} = k[A]^1[B]^0[C]^2 \].
  • The symbol \(k\) represents the rate constant, a unique value for a specific reaction at a particular temperature.
  • The exponents (1 for A, 0 for B, 2 for C) tell you the order of the reaction with respect to each reactant.
  • An order of 1 indicates a linear relationship with the rate, while an order of 0 means the concentration has no effect, and an order of 2 means the rate is affected by the square of the concentration change.
Understanding the rate law helps predict how changes in reactant concentrations will affect the reaction rate.
Order of Reaction
The order of reaction refers to the power to which the concentration of a reactant is raised in the rate law, giving insight into how the concentration affects the reaction rate. In our case:- **First Order with Respect to A**: This means if the concentration of A changes, the rate changes proportionally. For example, doubling \([A]\) will double the reaction rate since \[ \text{New Rate} = k[2A]^1 = 2 imes \text{Original Rate} \].- **Zero Order with Respect to B**: Here, changes in \([B]\) do not affect the reaction rate. Tripling \([B]\), or any change, results in no change in rate as \[ \text{New Rate} = k[A]^1[3B]^0[C]^2 = \text{Original Rate} \].
- **Second Order with Respect to C**: The rate is exponentially affected by the concentration of C. Tripling \([C]\) makes the rate rise by a factor of nine because \[ \text{New Rate} = k[A]^1[B]^0[3C]^2 = 9 imes \text{Original Rate} \].
These orders combine to determine overall how a reaction progresses with changing reactants.
Effect of Concentration Change
The effect of changing reactant concentrations is integral to understanding and controlling reaction rates. By manipulating concentrations, you can either speed up or slow down a reaction.
  • **Doubling Concentration of A** will increase the rate, because the reaction is first order with respect to A.
  • **Changing Concentration of B** has no effect on rate since the reaction is zero order with respect to B.
  • **Tripling Concentration of C** will significantly raise the rate as the reaction is second order with respect to C. A change here is quite impactful.
  • **Tripling all Reactant Concentrations**: When you simultaneously triple all reactants, the rate increases by a factor that accounts for each order: here, 27 times the original rate (3 for A, 1 for B, and 9 for C).
  • **Halving all Reactant Concentrations** results in a large decrease in rate 鈥� by a factor of \(\frac{1}{8}\) 鈥� representing the compounded decrease from cutting each concentration in half.
By studying these effects, one can efficiently tune the conditions to achieve desired reaction speeds.

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Most popular questions from this chapter

Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

(a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important?

Molecular iodine, \(\mathrm{I}_{2}(\mathrm{~g})\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{MI}_{2}\) at this temperature, how much will remain after \(5.12 \mathrm{~s}\) assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

Americium-241 is used in smoke detectors. It has a first-order rate constant for radioactive decay of \(k=1.6 \times 10^{-3} \mathrm{yr}^{-1}\). By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of \(k=0.011\) day \(^{-1}\). (a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a \(1.00\)-mg sample of each isotope remains after 3 half-lives? (d) How much of a 1.00-mg sample of each isotope remains after 4 days?

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\). and a frequency factor of \(A=6.0 \times 10^{8} M^{-1} \mathrm{~s}^{-1}\). The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100{ }^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the shape for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.
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