A reaction \(A+B \longrightarrow C\) obeys the following rate law: Rate \(=k[\mathrm{~B}]^{2}\). (a) If \([\mathrm{A}]\) is doubled, how will the rate change? Will the rate constant change? Explain. (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B}\) ? What is the overall reaction order? (c) What are the units of the rate constant?

Short Answer

Expert verified
When the concentration of A is doubled, the rate of the reaction will not change, as A is not present in the rate law equation (Rate = \(k[\mathrm{B}]^{2}\)). The rate constant (k) depends on the temperature and the nature of the reaction, not on the concentration of reactants, so it will not change either. The reaction order of A is 0, and the reaction order of B is 2. The overall reaction order is the sum of the reaction orders, which is 2. The units of the rate constant (k) are \(\mathrm{M}^{-1} \mathrm{s}^{-1}\).

Step by step solution

01

a) Effect of doubling A on the rate, and rate constant change

First, let's look at the given rate law: Rate = \(k[\mathrm{B}]^{2}\) If we double the concentration of A, the rate law equation remains the same: Rate \(= k[\mathrm{B}]^{2}\) As we can see, the concentration of A is not present in the rate law equation. That means the rate of the reaction is not affected by the concentration of A. So, if the concentration of A is doubled, the rate of the reaction will not change. Also, the rate constant (k) depends on the temperature and the nature of the reaction, not on the concentration of reactants. Therefore, the rate constant will not change when the concentration of A is doubled.
02

b) Reaction orders for A and B, and overall reaction order

The reaction order for each reactant is the power to which its concentration is raised in the rate law equation. In our case: Rate = \(k[\mathrm{A}]^{m}[\mathrm{B}]^{n}\) Comparing this with the given rate law (Rate = \(k[\mathrm{B}]^{2}\)), we can observe the following: 1. The concentration of A is not present in the rate law, so its reaction order (m) is 0: Reaction order of A: \(m = 0\) 2. The concentration of B is raised to the power of 2 in the rate law, so its reaction order (n) is 2: Reaction order of B: \(n = 2\) Now, to find the overall reaction order, we sum the reaction orders of A and B: Overall reaction order: \(m + n = 0 + 2 = 2\)
03

c) Units of the rate constant

To determine the units of the rate constant (k), we will use the given rate law and the fact that different units are associated with different reaction orders. Rate = \(k[\mathrm{B}]^{2}\) Dividing both sides of the equation by \([\mathrm{B}]^{2}\) we get: \(k = \frac{\mathrm{Rate}}{[\mathrm{B}]^{2}}\) Now, we'll express the quantities in their respective units: Rate: Molarity per unit time (M s⁻¹) Concentration of B: Molarity (M) So the equation becomes: \(k (\frac{\mathrm{~M} \cdot \mathrm{s}^{-1}}{\mathrm{M}^{2}}) = \frac{\mathrm{M} \cdot \mathrm{s}^{-1}}{\mathrm{M}^{2}}\) The units for the rate constant k are: Units of k: \(\mathrm{M}^{-1} \mathrm{s}^{-1}\)

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Key Concepts

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

Reaction Rate
The reaction rate refers to the speed at which reactants are converted into products in a chemical reaction. This rate can be influenced by several factors including reactant concentrations, temperature, and the presence of a catalyst. It is usually expressed as the change in concentration of a particular reactant or product over a specific period of time. For instance, if we refer to a simple reaction like \(A + B \longrightarrow C\), the reaction rate could be measured by how quickly \(A\) or \(B\) is decreasing or how quickly \(C\) is forming per unit time.

In practical terms, if a student were conducting an experiment where reactants are colored, they might observe the color fading as the reaction proceeds—a visual confirmation of the reaction rate. It's important to understand that this rate is not necessarily constant throughout a reaction, often changing as concentrations and conditions evolve.
Rate Constant
The rate constant, denoted by \(k\), is a proportionality constant in the rate law that provides a direct relationship between the reactant concentrations and the reaction rate. It's unique for every reaction and typically depends on temperature and catalyst presence — but not on reactant concentrations. A key takeaway for students is that the rate constant is unaffected by changes in concentrations of reactants.

It might help to think of the rate constant as the 'pace-setter' of a reaction under given conditions. If a reaction were a car, the rate constant would be akin to the car's engine quality—it determines how fast the car can potentially go, but the actual speed (reaction rate) also depends on how much you're stepping on the gas (reactant concentration).
Reaction Order
Reaction order is a term that helps us relate the concentrations of reactants to the rate of the reaction. It's given as an exponent in the rate law equation to which the concentration of a reactant is raised. The exponents can be whole numbers, fractions, or even zero. For instance, if we have a rate law where the concentration of \(B\) is squared, then \(B\) is second-order with respect to that reaction, and the overall reaction order is a sum of the orders with respect to each reactant.

To clarify with an example, in a rate law like Rate \(= k[\text{A}]^0[\text{B}]^2\), \(A\) is zero-order because its concentration has no effect on rate, as indicated by the exponent of 0, and \(B\) is second-order. The overall reaction order is the sum of these individual orders, giving us a reaction that is second-order overall.
Concentration Effect on Rate
In chemistry, how the concentration of a reactant affects the rate of reaction is key to understanding reaction kinetics. If all other conditions are held constant, increasing a reactant's concentration usually increases the reaction rate. This is because more reactant molecules are available to collide and react per unit volume. It's like increasing the number of runners in a race — the chance of finishers (reaction products) increases.

However, not all reactants in a rate law equation affect the rate. For example, if a rate law does not include a specific reactant, like \(A\) in our exercise, doubling its concentration won't impact the rate. This is a crucial concept that counters a common misconception among students: that all reactants equally influence the reaction rate.
Units of Rate Constant
The units of the rate constant are determined by the overall order of the reaction. For a reaction that is second-order overall, as in our example, the units for the rate constant will be \(M^{-1} s^{-1}\). In general, for a reaction of order \(n\), the rate constant will have the units of \((\text{concentration})^{1-n} (\text{time})^{-1}\).

Understanding the units of the rate constant is essential for solving problems relating to reaction rates and for checking if the rate constants are being used correctly in calculations. The units provide a way of ensuring that the rate law equation is dimensionally consistent, meaning that both sides of the rate equation have the same units.

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Most popular questions from this chapter

(a) A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{b})\) Another first-order reaction also has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathrm{c})\) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

Consider a hypothetical reaction between \(A, B,\) and \(C\) 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 \([\mathrm{A}]\) is doubled and the other reactant concentrations are held constant? (c) How does the rate change when \([\mathrm{B}]\) is tripled and the other reactant concentrations are held constant? (d) How does the rate change when [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 half?

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is catalyzed by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{array}{l} \mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{array} $$ (a) Show that the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (b) Why do we consider \(\mathrm{NO}_{2}\) a catalyst and not an intermediate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis?

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carboncontaining fragment such as \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2},\) and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a \(\mathrm{C}\) atom for \({ }^{\text {"R" }}\) ). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

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