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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 — by a factor of \(\frac{1}{8}\) — 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

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) According to the collision model, why does temperature affect the value of the rate constant? (c) Does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature?

You perform a series of experiments for the reaction \(A \longrightarrow B+C\) and find that the rate law has the form rate \(=k[\mathrm{~A}]^{x}\). Determine the value of \(x\) in each of the following cases: (a) There is no rate change when \([\mathrm{A}]_{0}\) is tripled. (b) The rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled. (c) When \([\mathrm{A}]_{0}\) is doubled, the rate increases by a factor of 8 .

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}{ }^{-}(a q)+\mathrm{H}^{+}(a q)\). In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Indicate whether each statement is true or false. (a) If you compare two reactions with similar collision factors, the one with the larger activation energy will be faster. (b) A reaction that has a small rate constant must have a small frequency factor. (c) Increasing the reaction temperature increases the fraction of successful collisions between reactants.

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

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