Question

The evaporation of a 120 nm film of \(n\) -pentane from a single crystal of aluminum oxide is zero order with a rate constant of \(1.92 \times 10^{13}\) molecules/cm \(^{2} \cdot \mathrm{s}\) at 120 \(\mathrm{K}.\) \begin{equation} \begin{array}{l}{\text { a. If the initial surface coverage is } 8.9 \times 10^{16} \text { molecules/cm', how long }} \\ {\text { will it take for one- half of the film to evaporate? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. What fraction of the film is left after } 10 \text { s? Assume the same initial }} \\ {\text { coverage as in part a. }}\end{array} \end{equation}

Short answer

The time needed for half of the film to evaporate is calculated by dividing half of the initial surface coverage by twice the rate constant. After calculating, the time \( t \) is found to be approximately 231.25 seconds. After 10 seconds, the fraction of the film left is approximately 0.99979.

Step-by-step solution

Understand Zero Order Kinetics

Zero order kinetics means the rate of evaporation is constant and does not depend on the amount of substance present. The rate equation for zero order kinetics is given by \[ -\frac{d[\text{n-pentane}]}{dt} = k \] where \( k \) is the rate constant and \( [\text{n-pentane}] \) denotes the concentration of n-pentane in molecules per unit area.

Solve for Evaporation Time for Half the Film

To find the time required for half of the film to evaporate, integrate the rate equation from the initial coverage to half of that value. The integrated form of the zero order rate law is: \[ [\text{n-pentane}] = -kt + [\text{n-pentane}]_0 \] Set \( [\text{n-pentane}] \) to \( \frac{1}{2}[\text{n-pentane}]_0 \) and solve for \( t \):\[ \frac{1}{2}[\text{n-pentane}]_0 = -kt + [\text{n-pentane}]_0 \]\[ \frac{1}{2}[\text{n-pentane}]_0 = [\text{n-pentane}]_0 - kt \]\[ kt = \frac{1}{2}[\text{n-pentane}]_0 \]\[ t = \frac{1}{2k}[\text{n-pentane}]_0 \]

Calculate Time Using the Given Values

Substitute the given values of the initial surface coverage and the rate constant into the equation from Step 2:\[ t = \frac{1}{2(1.92 \times 10^{13})}(8.9 \times 10^{16}) \]\[ t = \frac{1}{2(1.92 \times 10^{13})}\times 8.9 \times 10^{16} \text{ seconds} \]Calculate the time \( t \) to find out how long it will take for half the film to evaporate.

Calculate the Fraction of Film Left After 10 Seconds

Using the zero order rate law, calculate the remaining surface coverage after 10 seconds:\[ [\text{n-pentane}] = -kt + [\text{n-pentane}]_0 \]Substitute \( k = 1.92 \times 10^{13} \) molecules/cm\(^2\)\( \cdot \)s, \( t = 10 \) s, and \( [\text{n-pentane}]_0 = 8.9 \times 10^{16} \) molecules/cm\(^2\) into the equation:\[ [\text{n-pentane}] = -(1.92 \times 10^{13} \times 10) + 8.9 \times 10^{16} \]Calculate to find the remaining surface coverage, then divide by the initial surface coverage to get the fraction of the film left.

Key concepts

Chemical Kinetics

Chemical kinetics is the branch of physical chemistry that studies the rates of chemical processes. At its core, kinetics is concerned with how different variables such as concentration, temperature, and catalysts affect the speed of chemical reactions. Understanding the kinetics of a reaction provides insight into the reaction mechanism and the sequence of steps that lead to the formation of products.

For example, in the context of the evaporation of n-pentane from an aluminum oxide surface, chemical kinetics enables us to predict the rate at which the evaporation occurs. By knowing the reaction rate, scientists and engineers can design processes to optimize the use of materials or control the rate of a reaction to better suit industrial needs.

Rate Constant

The rate constant, symbolized as 'k,' is a proportionality factor that links the reaction rate to the concentrations of reactants in the rate law of a chemical reaction. For a reaction of zero order, like the evaporation of n-pentane, the rate constant k is constant and the reaction rate does not depend on the concentration of the reactant.

The value of k provides crucial information: it denotes the speed of a particular zero order process at a given temperature, with units that are specific to the reaction's order. In our exercise, the rate constant for the evaporation of n-pentane is given as a value of \(1.92 \times 10^{13}\) molecules/cm\(^2\)·s at 120K. This high value illustrates how rapidly the evaporation proceeds at this low temperature.

Evaporation Kinetics

Evaporation kinetics, especially concerning thin films, refers to the rate at which a liquid phase on a surface transitions into a gas phase. This is an important area in material science as it can influence the manufacturing processes for coatings, thin films, and various nanotechnologies. The kinetics of evaporation impacts not only the rate of film mass loss but also the film's uniformity and thickness over time.

Focusing on zero order kinetics in evaporation, such as that of n-pentane on an aluminum oxide surface, implies that the evaporation rate remains constant rather than decreasing as the film becomes thinner. This situation often occurs when the system is not limited by reactant concentration but could be influenced by surface factors or diffusion rates at interfaces, which are paramount in industrial applications like drying paint or microchip fabrication.