Question

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 .

Short answer

In the given rate law, the values of x for each case are as follows: (a) \(x = 0\), (b) \(x = 2\), and (c) \(x = 3\).

Step-by-step solution

Case (a) - No rate change when \([\mathrm{A}]_{0}\) is tripled

Initially, the rate law is: rate \(= k[\mathrm{A}]^{x}\) When \([\mathrm{A}]_{0}\) is tripled, the new concentration \([\mathrm{A}]'_{0}\) is: \([\mathrm{A}]'_{0} = 3[\mathrm{A}]_{0}\) Now, let's calculate the new rate using the rate law: rate' \(= k[\mathrm{A}']^{x} = k(3[\mathrm{A}]_{0})^x\) According to the given condition, there is no rate change, which means rate' = rate. Therefore: \(k(\mathrm{3}[\mathrm{A}]_{0})^x = k[\mathrm{A}]_{0}^x\) Divide both sides by \(k[\mathrm{A}]_{0}^x\): \[\frac{\mathrm{(3}[\mathrm{A}]_{0})^^x}{[\mathrm{A}]_{0}^x} = 1\] Simplify the equation: \(3^x = 1\) Since \(3^0 = 1\), we have the value of x: \(x = 0\)

Case (b) - Rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled

As in case (a), let's calculate the new rate when \([\mathrm{A}]_{0}\) is tripled: rate' \(= k(3[\mathrm{A}]_{0})^x\) According to the given condition, the rate increases by a factor of 9, which means rate' = 9 * rate. Therefore: \(k(\mathrm{3}[\mathrm{A}]_{0})^x = 9k[\mathrm{A}]_{0}^x\) Divide both sides by \(k[\mathrm{A}]_{0}^x\): \[\frac{\mathrm{(3}[\mathrm{A}]_{0})^^x}{[\mathrm{A}]_{0}^x} = 9\] Simplify the equation: \(3^x = 9\) Since \(3^2 = 9\), we have the value of x: \(x = 2\)

Case (c) - Rate increases by a factor of 8 when \([\mathrm{A}]_{0}\) is doubled

As before, let's calculate the new rate when \([\mathrm{A}]_{0}\) is doubled: rate' \(= k(2[\mathrm{A}]_{0})^x\) According to the given condition, the rate increases by a factor of 8, which means rate' = 8 * rate. Therefore: \(k(\mathrm{2}[\mathrm{A}]_{0})^x = 8k[\mathrm{A}]_{0}^x\) Divide both sides by \(k[\mathrm{A}]_{0}^x\): \[\frac{\mathrm{(2}[\mathrm{A}]_{0})^^x}{[\mathrm{A}]_{0}^x} = 8\] Simplify the equation: \(2^x = 8\) Since \(2^3 = 8\), we have the value of x: \(x = 3\) Thus, in each case we have determined the values of x in the rate law: (a) \(x=0\) (b) \(x=2\) (c) \(x=3\)

Key concepts

Rate Law

Understanding the rate law is crucial for grasping the nuances of chemical kinetics, a branch of chemistry that studies the rates of chemical reactions. A rate law is a mathematical expression that describes the relationship between the reaction rate and the concentration of reactants. In simpler terms, it shows us how fast a reaction will occur given the amount of starting materials, known as reactants.

The general form of a rate law is: \(\text{rate} = k[\text{A}]^x[\text{B}]^y\cdots\), where \(k\) is the rate constant, \([\text{A}], [\text{B}], \ldots\) are the concentrations of reactants, and \(x, y, \ldots\) are the reaction orders with respect to each reactant. The overall order of the reaction is the sum of these individual orders (\(x+y+\ldots\)).

When analyzing a rate law, as in the exercise where the reaction rate was tested with varying concentrations of reactant \(A\), the goal was to determine the exponent \(x\), which signifies how the rate of the reaction changes with the concentration of \(A\). This involved observing changes in reaction rate when the concentration of \(A\) was altered, and meticulously evaluating the impact of this change.

Reaction Rate

The reaction rate tells us how quickly a chemical reaction occurs. It can be defined as the change in concentration of a reactant or product per unit time. For our exercise, the reaction between \(A\) and \(B\) to produce \(C\), we are interested in how the concentration of \(A\) decreases over time, as \(A\) is transformed into \(B\) and \(C\).

The rate can be affected by various factors like temperature, pressure, the presence of catalysts, and the concentration of reactants. In the exercise examples, we manipulated the concentration to observe changes in reaction rate, illuminating the direct influence concentration has.

Rate Measurement

Mathematically, reaction rate can be expressed as: \(\text{rate} = -\frac{\Delta[\text{A}]}{\Delta t}\) for a reactant, where \(\Delta[\text{A}]\) represents the change in concentration and \(\Delta t\) is the change in time. The negative sign indicates that the concentration of reactants diminishes over time. The experiments performed in the exercise aimed to determine how this rate changes under different initial concentrations of reactant \(A\).

Order of Reaction

The order of a reaction provides insight into the dependency of the reaction rate on the concentration of its reactants. It is a key concept in chemical kinetics, as it can be used to predict how changes in concentrations influence the speed of a reaction. In our exercise, the order with respect to reactant \(A\) was determined by observing how the reaction rate changed as the initial concentration of \(A\) was altered.

Orders of reaction are generally determined experimentally, and they can be integers, fractions, or even zero. A zero-order reaction means that the rate is independent of the concentration of the reactant, as was the case in part (a) of the exercise where tripling the concentration of \(A\) did not change the rate, yielding an order of zero (\(x=0\)).

A first-order reaction implies that the rate is directly proportional to one concentration term, while a second-order indicates dependency on either the square of one concentration term or on the product of two different concentrations. Our exercises (b) and (c) revealed that the rate law followed a second and a third-order reaction, respectively, with respect to \(A\), demonstrated by the proportional increase in rate to the increase in \(A\)'s concentration.

Understanding Reaction Orders

Grasping the concept of reaction order is essential for predicting and controlling chemical processes. This knowledge impacts a wide range of applications, from industrial synthesis to pharmaceuticals, where the efficiency and rate of reactions are crucial.