Question

The decomposition of ammonia is first order with respect to \(\mathrm{NH}_{3}\). (Compare with Study Question 58.) $$2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g})$$ (a) What is the rate equation for this reaction? (b) Calculate the rate constant, \(k\), given the following data: $$\begin{array}{lc}\hline\left[\mathrm{NH}_{3}\right](\mathrm{mol} / \mathrm{L}) & \text { Time }(\mathrm{s}) \\\\\hline 0.67 & 0 \\\0.26 & 19 \\\\\hline\end{array}$$ (c) Determine the half-life of \(\mathrm{NH}_{3}\)

Short answer

(a) Rate equation: \( \, \text{Rate} = k [\mathrm{NH}_3] \, \); (b) \( \, k \approx 0.066 \, \mathrm{s}^{-1} \, \); (c) Half-life \( \, \approx 10.5 \, \mathrm{s} \, \).

Step-by-step solution

Identify Rate Equation

Given that the decomposition of ammonia is first order with respect to \(\mathrm{NH}_{3}\), the rate equation can be expressed as \( \text{Rate} = k [\mathrm{NH}_{3}] \), where \(k\) is the rate constant and \([\mathrm{NH}_{3}]\) is the concentration of ammonia.

Use First Order Integrated Rate Law

For a first order reaction, the integrated rate law is given by \( \ln ([\mathrm{NH}_{3}]_t/[\mathrm{NH}_{3}]_0) = -kt \), where \([\mathrm{NH}_{3}]_t\) is the concentration at time \(t\), and \([\mathrm{NH}_{3}]_0\) is the initial concentration.

Calculate Rate Constant \(k\)

Substitute the given values into the integrated rate equation: \([\mathrm{NH}_{3}]_0 = 0.67\ \mathrm{mol/L}\), \([\mathrm{NH}_{3}]_t = 0.26\ \mathrm{mol/L}\), and \(t = 19\ \mathrm{s}\). \[ \ln(0.26/0.67) = -k(19) \] Solve for \(k\): \[ k = -\frac{\ln(0.26/0.67)}{19} \approx 0.066\ \mathrm{s}^{-1} \]

Determine Half-Life

The half-life for a first order reaction is calculated using the formula \( t_{1/2} = \frac{0.693}{k} \). Substitute the found rate constant: \[ t_{1/2} = \frac{0.693}{0.066} \approx 10.5\ \mathrm{s} \]

Key concepts

First Order Reaction

In chemical kinetics, a first order reaction depends linearly on the concentration of one reactant. This means that if the concentration of the reactant doubles, the rate of the reaction also doubles. For the decomposition of ammonia, which produces nitrogen and hydrogen gas, the reaction is first order with respect to omenclature{ ext{NH}_{3}}{Ammonia.}. Thus, the rate equation is simple and is expressed as \( \text{Rate} = k [\mathrm{NH}_{3}] \), where \( k \) is the rate constant and \([\mathrm{NH}_{3}] \) is the concentration.

Rate Constant

The rate constant, \( k \), is a crucial factor in determining how fast a reaction proceeds. For a first order reaction like the decomposition of ammonia, \( k \) has units of \( \mathrm{s}^{-1} \). To find \( k \), you can use experimental data for concentration and time. By substituting the initial and final concentrations and the corresponding time into the integrated rate law, you can solve for \( k \). This gives you insight into the speed of the reaction under specific conditions.

Integrated Rate Law

The integrated rate law for a first order reaction links the concentration of reactants to the time that has passed. It is mathematically expressed as:

• \( \ln\left( \frac{[\mathrm{NH}_{3}]_t}{[\mathrm{NH}_{3}]_0} \right) = -kt \)

Here, \([\mathrm{NH}_{3}]_t\) is the concentration at time \( t \), \([\mathrm{NH}_{3}]_0\) is the initial concentration, and \( k \) is the rate constant. This formula helps predict concentrations at different times, which is essential for understanding how quickly a reaction will proceed and adjust conditions accordingly.

Half-Life Calculation

The half-life of a reaction is the time it takes for the concentration of a reactant to reduce to half of its initial value. For first order reactions, the half-life is constant and is calculated using the formula:

• \( t_{1/2} = \frac{0.693}{k} \)

This means that regardless of the starting concentration, the time it takes for half of the ammonia to decompose is consistent. The half-life provides a convenient way to quantify the reaction's progress and compare it with others under similar conditions.