Problem 143 - Chemistry: An Atoms-Focused Approach

Question

Adding NH \(_{3}\) to the stack gases at an electric power generating plant can reduce \(\mathrm{NO}_{x}\) emissions. This selective noncatalytic reduction (SNR) process depends on the reaction between \(\mathrm{NH}_{2}\) (an odd- electron compound) and NO: $$\mathrm{NH}_{2}(g)+\mathrm{NO}(g) \rightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ The following kinetic data were collected at \(1200 \mathrm{K}\) $$\begin{array}{cllc}\text { Experiment } & \left[\mathrm{NH}_{2}\right]_{0}(\mathrm{M}) & [\mathrm{NO}]_{0}(M) & \text { Rate }(M / \mathrm{s}) \\\1 & 1.00 \times 10^{-5} & 1.00 \times 10^{-5} & 0.12 \\\\\hline 2 & 2.00 \times 10^{-5} & 1.00 \times 10^{-5} & 0.24 \\\\\hline 3 & 2.00 \times 10^{-5} & 1.50 \times 10^{-5} & 0.36 \\\\\hline 4 & 2.50 \times 10^{-5} & 1.50 \times 10^{-5} & 0.45 \\\\\hline\end{array}$$ a. What is the rate law for the reaction? b. What is the value of the rate constant at \(1200 \mathrm{K} ?\)

Step-by-Step Solution

Answer

Question: Determine the rate law and the rate constant at 1200 K for the reaction between NH\(_{2}\) and NO using the given kinetic data. Answer: The rate law for the reaction is rate = k[NH\(_{2}\)][NO]. The rate constant at 1200 K is \(1.2\times10^6\) M\(^{-1}\)s\(^{-1}\).

Step 1: Write a generic rate law

First, let's write a generic rate law for the reaction considering the concentration of the reactants: $$rate = k[\mathrm{NH}_{2}]^n [\mathrm{NO}]^m$$ Here, k is the rate constant, n is the reaction order with respect to NH\(_{2}\), and m is the reaction order with respect to NO.

Step 2: Determine the reaction order with respect to NH\(_{2}\)

We will use the information from the table to determine the reaction order with respect to NH\(_{2}\). Let's compare experiments 1 and 2. Experiment 1: $$rate_1 = k[\mathrm{NH}_{2}]_{1}^n[\mathrm{NO}]_{1}^m = k(1.00\times10^{-5})^n(1.00\times10^{-5})^m = 0.12$$ Experiment 2: $$rate_2 = k[\mathrm{NH}_{2}]_{2}^n[\mathrm{NO}]_{2}^m = k(2.00\times10^{-5})^n(1.00\times10^{-5})^m = 0.24$$ Dividing the rate equation for experiment 2 by the rate equation for experiment 1, we get: $$\frac{rate_2}{rate_1} = \frac{(2.00\times10^{-5})^n}{(1.00\times10^{-5})^n}$$ By plugging in the values of \(rate_1\) and \(rate_2\), we get: $$\frac{0.24}{0.12} = 2 = (2)^n$$ Which means that \(n=1\). Thus, the reaction is first order with respect to NH\(_{2}\).

Step 3: Determine the reaction order with respect to NO

Now let's use the same approach to determine the reaction order with respect to NO. Let's compare experiments 2 and 3. Experiment 2 was already calculated above. Experiment 3: $$rate_3 = k[\mathrm{NH}_{2}]_{3}^n[\mathrm{NO}]_{3}^m = k(2.00\times10^{-5})^1(1.50\times10^{-5})^m = 0.36$$ Dividing the rate equation for experiment 3 by the rate equation for experiment 2, we get: $$\frac{rate_3}{rate_2} = \frac{(1.50\times10^{-5})^m}{(1.00\times10^{-5})^m}$$ By plugging in the values of \(rate_2\) and \(rate_3\), we get: $$\frac{0.36}{0.24} = 1.5=(1.5)^m$$ Which means that \(m=1\). Thus, the reaction is first order with respect to NO.

Step 4: Write the rate law with the determined reaction orders

Now that we have determined the reaction orders, we can write the rate law for the reaction: $$rate = k[\mathrm{NH}_{2}]^1[\mathrm{NO}]^1$$ Or simply: $$rate = k[\mathrm{NH}_{2}][\mathrm{NO}]$$ This is the answer to part (a) of the problem.

Step 5: Calculate the rate constant at 1200 K

To find the rate constant k, we can use the rate law and the data from any of the experiments. As an example, we can use Experiment 1 as follows: $$rate = k[\mathrm{NH}_{2}][\mathrm{NO}]$$ $$0.12 = k(1.00\times10^{-5})(1.00\times10^{-5})$$ $$k = \frac{0.12}{(1.00\times10^{-5})(1.00\times10^{-5})}$$ Calculate the value of k: $$k = 1.2 \times 10^6 (\mathrm{M}^{-1} \mathrm{s}^{-1})$$ The value of the rate constant at 1200 K is \(1.2\times10^6\) M\(^{-1}\)s\(^{-1}\). This is the answer to part (b) of the problem.

Key Concepts

Rate Law

Understanding rate law is essential when studying chemical kinetics. It gives us a relationship between the concentration of the reactants and the rate of reaction. For the reaction involving NH\_2 and NO, the rate law is determined based on experiments. The general form is expressed as \(rate = k[\text{NH}\_2]^n [\text{NO}]^m\). Here, \(k\) is a constant called the rate constant, while \(n\) and \(m\) are the reaction orders with respect to NH\_2 and NO, respectively.
By looking at the change in rate with varying concentrations of each reactant separately, one can understand how changes in concentration affect the reaction rate. This way, we can write the specific rate law for the reaction: \(rate = k[\text{NH}\_2][\text{NO}]\). As determined in the original exercise, the reaction is first order for both NH\_2 and NO, which makes analysis straightforward.

Reaction Order

The reaction order shows how the concentration of reactants affects the rate of reaction. It is depicted by the exponents \(n\) and \(m\) in the rate law equation. In chemical kinetics, determining the correct reaction order is crucial to understanding how a reaction progresses over time.
For NH\_2 and NO, experimentation revealed both to exhibit a first order. This means the rate of reaction is directly proportional to the concentration of each reactant. If you double the concentration of NH\_2 or NO while keeping the other constant, the rate of reaction will also double. Building on this fundamental, one can evaluate and predict the rates under different scenarios using simple algebraic manipulations. Understanding reaction order helps in scaling reactions, optimizing conditions for industrial processes, and predicting reaction behavior.

Rate Constant

A rate constant \(k\) is a proportionality factor in the rate law equation, central to quantifying the speed of a reaction. Unlike reaction order, the rate constant's value is influenced by factors such as temperature. It provides a numerical value to rate laws, allowing direct calculations of the reaction rate at a given set of conditions.
The value of \(k\) is determined experimentally and serves as a benchmark for comparing rates of reactions under similar conditions. In the presented exercise, using the data from Experiment 1, \(k\) was calculated as \(1.2 \times 10^6\,\text{M}^{-1}\text{s}^{-1}\). The units of \(k\) reflect the overall reaction order and help in correcting calculations regarding reaction rates. Understanding and determining the rate constant allows chemists to make accurate predictions about reaction behaviors in different setups and temperatures, bringing theory into practical applications.