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?

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.

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 (\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.