Question

You perform a series of experiments for the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) 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 . [Section 14.3\(]\)

Short answer

For the given cases, we find the values of $x$ to be: (a) $x=0$ (b) $x=2$ (c) $x=3$

Step-by-step solution

(a) Find x when rate is unchanged as [A]₀ is tripled

In this case, we are given that there's no change in rate when the initial concentration of [A] is tripled. Let the initial rate be: rate₁ = k[ A₁ ]^x When [A]₀ is tripled, the new rate remains the same: rate₂ = k[ 3A₁ ]^x Since the rate is unchanged, rate₁ = rate₂. Let's set up the equation and solve for x: k[ A₁ ]^x = k[ 3A₁ ]^x First, divide both sides of the equation by k to eliminate it: [ A₁ ]^x = [ 3A₁ ]^x Now, let's divide both sides of the equation by [A₁]^x: 1 = 3^x Taking the logarithm of both sides: x = log(1) / log(3) As log(1) = 0, we have x = 0.

(b) Find x as rate increases by a factor of 9 when [A]₀ is tripled

In this case, the rate increases by a factor of 9 when the initial concentration of [A] is tripled. Let the initial rate be: rate₁ = k[ A₁ ]^x The new rate is 9 times the initial rate when [A]₀ is tripled: rate₂ = 9k[ A₁ ]^x = k[ 3A₁ ]^x Let's set up the equation with the given conditions and solve for x: 9 * k[ A₁ ]^x = k[ 3A₁ ]^x First, divide both sides of the equation by k to eliminate it: 9[ A₁ ]^x = [ 3A₁ ]^x Now, let's divide both sides of the equation by [A₁]^x: 9 = 3^x Taking the logarithm of both sides: x = log(9) / log(3) As log(9) = 2, we have x = 2.

(c) Find x as rate increases by a factor of 8 when [A]₀ is doubled

In this case, the rate increases by a factor of 8 when the initial concentration of [A] is doubled. Let the initial rate be: rate₁ = k[ A₁ ]^x The new rate is 8 times the initial rate when [A]₀ is doubled: rate₂ = 8k[ A₁ ]^x = k[ 2A₁ ]^x Let's set up the equation with the given conditions and solve for x: 8 * k[ A₁ ]^x = k[ 2A₁ ]^x First, divide both sides of the equation by k to eliminate it: 8[ A₁ ]^x = [ 2A₁ ]^x Now, let's divide both sides of the equation by [A₁]^x: 8 = 2^x Taking the logarithm of both sides: x = log(8) / log(2) As log(8) = 3, we have x = 3. So, the values of x for cases a, b and c are 0, 2, and 3, respectively.

Key concepts

Chemical Kinetics

Understanding the pace at which chemical reactions occur is the essence of chemical kinetics. This branch of chemistry is concerned with the rates of chemical processes, which are influenced by various factors such as concentration of reactants, temperature, and presence of catalysts.

In determining the rate law like in the exercise provided, chemists experiment with varying concentrations of reactants to observe how the rate of formation of products changes. The rate law itself is a mathematical expression encapsulating this relationship, essentially predicting the speed of a reaction for given concentrations of reactants.

Reaction Rate

The reaction rate tells us how rapidly a reaction proceeds. In other words, it's a measure of the change in concentration of reactants or products per unit time. For the reaction \( A \longrightarrow B + C \) given in the exercise, the rate can be represented as the increase in concentration of products B and C over time, or the decrease in concentration of reactant A.

Exercise Improvement Advice

To facilitate understanding, when visualizing reaction rates, one might think of it as a 'speedometer' for chemical reactions, providing real-time data on how fast reactants are being converted into products. Understanding the provided step-by-step solution requires a knack for observing how changes in concentration impact this 'speed,' which is precisely what was observed in the experiment described in the exercise.

Rate Constant

At the heart of the rate law equation lies the rate constant, denoted as \( k \). This is a proportionality constant that links the rate of the reaction to the concentrations of the reactants raised to a power, often representative of the reaction order with respect to each reactant.

The rate constant is independent of the concentration of reactants, but it varies with temperature. In the exercise solutions, we see how the rate constant remains the same when deriving the reaction order \( x \) by comparing the initial and altered rates. Another key point is that the value of \( k \) aids in determining how sensitive a reaction is to changes in reactant concentration.

Concentration Dependence

Often, the rate of a chemical reaction depends heavily on the concentration of reactants. This is known as concentration dependence.

The rate law exemplifies this dependency through the exponent \( x \) placed on the reactant's concentration, as seen in the rate equation rate \( = k[A]^x \). The varying values of \( x \) in different scenarios in the exercise solutions— zero when the rate doesn't change, two when the rate increases ninefold, and three when it increases eightfold—illustrate how the reaction's rate can vary with the concentration of reactants. This can tell us whether the reaction is zero, first, or second order with respect to a given reactant.