Question

In a study of ammonia production, an industrial chemist discovers that the compound decomposes to its elements \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in a first-order process. She collects the following data: $$ \begin{array}{llll} \text { Time (s) } & 0 & 1.000 & 2.000 \\ {\left[\mathrm{NH}_{3}\right](\mathrm{mol} / \mathrm{L})} & 4.000 & 3.986 & 3.974 \end{array} $$ (a) Use graphical methods to determine the rate constant. (b) What is the half-life for ammonia decomposition?

Short answer

Rate constant \( k = 0.002 \text{ s}^{-1} \), Half-life \( t_{1/2} = 346.5 \text{ s} \).

Step-by-step solution

- Understand the problem

The specified problem involves determining the rate constant and half-life for the decomposition of ammonia into nitrogen and hydrogen. The decomposition is said to follow a first-order process.

- Set up the first-order rate equation

For a first-order reaction, the rate equation is given by: \[\text{ln}([\text{NH}_3]_t) = -kt + \text{ln}([\text{NH}_3]_0)\] where \(\text{[NH}_3]_t\) is the concentration of ammonia at time \(t\), \(k\) is the rate constant, and \( \text{ln}\) is the natural logarithm.

- Calculate natural logarithms

Calculate \( \text{ln}([NH}_3])\) for each time value: At time \(t = 0 \) seconds: \(\text{ln}(4.000) = 1.386 \)At time \(t = 1.000 \) second: \(\text{ln}(3.986) = 1.384 \)At time \(t = 2.000 \) seconds: \(\text{ln}(3.974) = 1.382 \)

- Plot the data

Create a plot with \( \text{ln}([\text{NH}_3]) \) on the y-axis and time (in seconds) on the x-axis. The plot should result in a straight line if the reaction is first-order.

- Determine the rate constant

The slope of the straight line is equal to \( -k \). Using linear regression or a suitable method, find the slope of the line. The slope can be approximated by change in y over change in x: Slope \((-k)\) = \(\frac{1.382 - 1.386}{2 - 0} = \frac{-0.004}{2} = -0.002\)Since the slope is \( -k \), the rate constant \( k \) is 0.002 s\(^{-1}\).

- Calculate the half-life

For a first-order reaction, the half-life (\( t_{1/2} \)) is calculated using: \[ t_{1/2} = \frac{0.693}{k} \] Substitute the value of \( k = 0.002 \), \[ t_{1/2} = \frac{0.693}{0.002} = 346.5 \text{ seconds} \]

Key concepts

Ammonia Decomposition

The decomposition of ammonia (\text{NH}_3) into nitrogen (\text{N}_2) and hydrogen (\text{H}_2) is an example of a first-order chemical reaction. This means that the rate at which ammonia decomposes depends linearly on its concentration. In other words, as the concentration of ammonia decreases over time, the rate of decomposition decreases proportionately. Understanding this relationship is crucial, as it helps explain why the reaction rate slows down over time. To analyze this, chemists often use the first-order rate equation: \[ \text{ln}([\text{NH}_3]_t) = -kt + \text{ln}([\text{NH}_3]_0) \] Here, \[ [\text{NH}_3]_t \] is the concentration of ammonia at time \[ t \], \[ k \] is the rate constant, and \[ \text{ln} \] denotes the natural logarithm.

Rate Constant Calculation

The rate constant \[ k \] is an essential parameter that quantifies the speed of the reaction. To determine \[ k \] graphically, we first need to calculate the natural logarithm (ln) of the ammonia concentrations at different times:

• At time \[ t = 0 \] seconds: \[ \text{ln}(4.000) = 1.386 \]
• At time \[ t = 1.000 \] second: \[ \text{ln}(3.986) = 1.384 \]
• At time \[ t = 2.000 \] seconds: \[ \text{ln}(3.974) = 1.382 \]

Plotting these values with natural logarithms on the y-axis and time on the x-axis will yield a straight line if the reaction is first-order. The slope of this line is equal to \[ -k \]. Using linear regression or manual calculation, the slope can be approximated as \[ \frac{1.382 - 1.386}{2 - 0} = \frac{-0.004}{2} = -0.002 \]. Therefore, the rate constant \[ k \] is 0.002 s\textsuperscript{-1}. This value shows how fast the ammonia decomposes during the reaction.

Half-Life Determination

Half-life (\text{t}_{1/2}) is a vital concept in first-order reactions. It represents the time it takes for the concentration of a reactant to reduce to half of its initial value. For a first-order reaction, the half-life is given by the formula: \[ \text{t}_{1/2} = \frac{0.693}{k} \] We already know the rate constant \[ k = 0.002 \text{ s}^{-1} \]. Substituting this value into the formula, we get: \[ \text{t}_{1/2} = \frac{0.693}{0.002} = 346.5 \text{ seconds} \] Understanding the half-life is crucial because it provides insights into how quickly a reaction progresses. In this case, it takes approximately 346.5 seconds for half of the initial ammonia concentration to decompose into nitrogen and hydrogen. This consistent reduction over equal time intervals helps in predicting how long a reaction will proceed.