Question

Make a graph of $[\mathrm{A}]$ versus time for zero-, first-, and second-order reactions. From these graphs, compare successive half-lives.

Short answer

In summary, for zero-order reactions, we have a straight line with a negative slope. Successive half-lives are equal, with a constant decrease in reactant concentration per half-life. For first-order reactions, we have an exponential decay curve. Successive half-lives are also equal, but the concentration reduces by half relative to the previous concentration. Lastly, for second-order reactions, we have a curve that increases in steepness over time. Successive half-lives are not equal and increase in length over time, indicating a slower reactant consumption as the concentration decreases.

Step-by-step solution

Understanding the Zero-, First-, and Second-order reactions

For this problem, it is important to know the meaning and distinction between zero-, first-, and second-order reactions. Zero-order reactions are characterized by a constant rate of change regardless of the reactant concentration. First-order reactions have a rate of change that depends linearly on the concentration of the reactant A, while in second-order reactions, the rate of change depends on the square of the concentration of the reactant A.

Formulate the equations for each reaction order

For each reaction order, we use the integrated rate laws formulated below: Zero-order: $$[A]=kt+[A{]}_0$$ First-order: $$[A]=[A{]}_0{e}^{-kt}$$ Second-order: $$\frac{1}{[A]}=kt+\frac{1}{[A{]}_0}$$

Setup the graphs

We will plot three graphs, one for each reaction order, in order to visualize how [A] changes over time for each type of reaction. Label the x-axis as 'Time' and the y-axis as 'Concentration of A'.

Plot the Zero-order reaction

With the given equation for the zero-order reaction, $[A]=kt+[A{]}_0$, we plot a straight line with a negative slope. The slope represents the rate constant, k.

Plot the First-order reaction

With the given equation for the first-order reaction, $[A]=[A{]}_0{e}^{-kt}$, we plot an exponential decay curve. The concentration of A decreases over time, but the decrease is dependent on the concentration of A.

Plot the Second-order reaction

With the given equation for the second-order reaction, $\frac{1}{[A]}=kt+\frac{1}{[A{]}_0}$, we would plot a curved line that increases in steepness over time, representing a greater rate of reaction as the concentration of A increases.

Comparing successive half-lives for the three reaction orders

With the graphs plotted, we can now compare successive half-lives. - For a zero-order reaction, the successive half-lives are equal; it means that in equal intervals of time, the concentration of A is reduced by the same amount. - For a first-order reaction, the successive half-lives are also equal; however, the concentration is reduced by half relative to the previous concentration (i.e., half of what remained, not a fixed amount). - For a second-order reaction, the successive half-lives are not equal. They are found to be longer as the reaction proceeds, indicating that less reactant is being consumed in a certain amount of time as the concentration decreases. In conclusion, both zero- and first-order reactions have equal successive half-lives, but the absolute change in reactant concentration for each half-life is different. Zero-order reactions lose a fixed amount of reactant per half-life, while first-order reactions lose half of the remaining reactant per half-life. Lastly, the second-order reaction half-lives are not equal and increase in length over time.

Key concepts

Zero-Order Reactions

Zero-order reactions are chemical processes where the rate of reaction is independent of the reactant concentration. This means that even as the concentration of reactants changes, the rate at which products form remains constant. Mathematically, this relationship is expressed as $$\frac{d[A]}{dt}=-k$$ where $$A$$ represents the concentration of the reactant and $$k$$ is the rate constant.

The integrated rate law for zero-order reactions is given by: $$[A]=-kt+[A{]}_0$$ where $$[A{]}_0$$ is the initial concentration at time zero. This linear equation suggests that when we plot the concentration of A versus time, we get a straight line with a negative slope equivalent to the rate constant $$k$$.

For zero-order reactions, the half-life, which is the time taken for half of the reactant to be consumed, is calculated using: $$t_{1/2}=\frac{[A{]}_0}{2k}$$. It is important to note that for zero-order reactions, the half-lives are not the same throughout the reaction because the rate doesn't depend on the concentration of A.

First-Order Reactions

In first-order reactions, the rate at which reactants transform into products is directly proportional to the concentration of a single reactant. This type of reaction fits the rate law $$\frac{d[A]}{dt}=-k[A]$$, which tells us that as the concentration of the reactant decreases, the rate of reaction decreases proportionally.

The integrated rate law for a first-order reaction is $$\ln ([A])=-kt+\ln ([A{]}_0)$$ or equivalently, $$[A]=[A{]}_0{e}^{-kt}$$. When this is graphed, the concentration of A decreases exponentially over time.

The half-life of a first-order reaction is a constant, unique property. It is given by $$t_{1/2}=\frac{\ln (2)}{k}$$. This value does not depend on the concentration of reactants and remains the same throughout the process. Hence, every half-life period, the concentration of the reactant reduces by exactly half, no matter how much reactant remains.

Second-Order Reactions

Second-order reactions exhibit rates that are proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants. The rate equation for a reaction that is second-order in A can be written as $$\frac{d[A]}{dt}=-k[A{]}^2$$.

The integrated rate law for second-order reactions can be expressed as $$\frac{1}{[A]}=kt+\frac{1}{[A{]}_0}$$. If we plot $$\frac{1}{[A]}$$ against time, we obtain a linear graph, where the slope is equal to the rate constant $$k$$.

The second-order reaction's half-life is not constant and is dependent on the concentration of reactants. It is calculated by the formula: $$t_{1/2}=\frac{1}{k[A{]}_0}$$. As the reaction proceeds and the concentration of the reactant decreases, the half-life increases, leading to a longer duration for half of the reactant to be used up.

Integrated Rate Laws

Integrated rate laws are equations that relate the concentrations of reactants or products to time for a particular order of reaction. These equations are derived by integrating the differential rate laws, which provide the rate of reaction based on reactant concentration.

Integrated rate laws enable chemists to determine important characteristics of a reaction, like the rate constant $$k$$ and the half-life. They also allow for the plotting of concentration versus time data to gain insights into the order of the reaction:

• Zero-order reacts follow $$[A]=-kt+[A{]}_0$$.
• First-order reactions use $$\ln ([A])=-kt+\ln ([A{]}_0)$$ or $$[A]=[A{]}_0{e}^{-kt}$$.
• Second-order reactions correspond to $$\frac{1}{[A]}=kt+\frac{1}{[A{]}_0}$$.

Understanding these equations is pivotal for students not just in plotting graphs but also in making predictions about reaction behavior over time.

Reaction Half-Life

The concept of half-life is central to understanding how chemical reactions progress over time. A reaction's half-life is the period required for the amount or concentration of a reactant to decrease by half its initial value. The half-life can reveal the speed of a reaction and provides insight into its kinetics.

For each order of reaction, the half-life has a distinct equation:

• Zero-order: $$t_{1/2}=\frac{[A{]}_0}{2k}$$.
• First-order: $$t_{1/2}=\frac{\ln (2)}{k}$$.
• Second-order: $$t_{1/2}=\frac{1}{k[A{]}_0}$$.

The differences between these half-lives reflect the unique ways that reactant concentrations diminish over time, which is critical information for both theoretical studies and practical applications including drug design, food preservation, and more.