Problem 71 - Chemistry: An Atoms-Focused Approach

Question

Laughing Gas Nitrous oxide ( \(\mathrm{N}_{2} \mathrm{O}\) ) is used as an anesthetic (laughing gas) and in aerosol cans to produce whipped cream. It is a potent greenhouse gas and decomposes slowly to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{N}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ If the plot of \(\ln \left[\mathrm{N}_{2} \mathrm{O}\right]\) as a function of time is linear, what is the rate law for the reaction? b. How many half-lives will it take for the concentration of \(\mathrm{N}_{2} \mathrm{O}\) to reach \(6.25 \%\) of its original concentration?

Step-by-Step Solution

Answer

Answer: The reaction is first-order, and it takes approximately 4 half-lives for the nitrous oxide concentration to reduce to 6.25% of its initial concentration.

Step 1: Determine the order of the reaction

Since the plot of \(\ln [\mathrm{N}_2 \mathrm{O}]\) against time is linear, according to the integrated rate law for a first-order reaction: $$\ln [\mathrm{N}_2 \mathrm{O}] = -kt + \ln [\mathrm{N}_2 \mathrm{O}]_0$$ where \(k\) is the rate constant, \(t\) is the time, and \([\mathrm{N}_2 \mathrm{O}]_0\) is the initial concentration of \(\mathrm{N}_2 \mathrm{O}\). Thus, the reaction is first-order with respect to \(\mathrm{N}_2 \mathrm{O}\). The rate law for the reaction can be written as: $$\text{Rate} = k [\mathrm{N}_2 \mathrm{O}]$$

Step 2: Find the number of half-lives required for the concentration to reach \(6.25\%\) of its initial concentration

Let the initial concentration of \(\mathrm{N}_2 \mathrm{O}\) be \([\mathrm{N}_2 \mathrm{O}]_0\). We want to find the number of half-lives required for the concentration to reach \(6.25 \%\) of \([\mathrm{N}_2 \mathrm{O}]_0\). The relation between the initial, final concentration and number of half-lives is: $$[\mathrm{N}_2 \mathrm{O}] = [\mathrm{N}_2 \mathrm{O}]_0 \left(\frac{1}{2}\right)^n$$ where \(n\) is the number of half-lives. We are given that the final concentration is \(6.25\%\) of the initial concentration, so: $$0.0625 [\mathrm{N}_2 \mathrm{O}]_0 = [\mathrm{N}_2 \mathrm{O}]_0 \left(\frac{1}{2}\right)^n$$

Step 3: Solve for the number of half-lives (\(n\))

Next, we solve the equation for \(n\): $$0.0625 = \left(\frac{1}{2}\right)^n$$ $$\ln (0.0625) = \ln \left(\left(\frac{1}{2}\right)^n\right) = n \ln \left(\frac{1}{2}\right)$$ $$n = \frac{\ln (0.0625)}{\ln (1/2)}$$ Calculating the value of \(n\): $$n \approx 4$$ Thus, it will take approximately 4 half-lives for the concentration of \(\mathrm{N}_2 \mathrm{O}\) to reach \(6.25 \%\) of its original concentration.

Key Concepts

Rate Law

A rate law is a mathematical equation that describes the speed of a chemical reaction. It shows how the rate depends on the concentration of the reactants. In the case of nitrous oxide (\( \mathrm{N}_2 \mathrm{O} \)), we discovered that it follows a first-order reaction. This means that its rate depends linearly on the concentration of nitrous oxide.
For first-order reactions, the rate law can be represented as:

• Rate = \( k[\mathrm{N}_2 \mathrm{O}] \)

Here, \( k \) is the constant rate of reaction, and \( [\mathrm{N}_2 \mathrm{O}] \) is the concentration of nitrous oxide. The beauty of this equation is its simplicity. The reaction speed changes directly with the concentration. Whenever you double the \([\mathrm{N}_2 \mathrm{O}]\), the rate of reaction will also double, assuming other conditions remain constant.
Identifying reaction orders like this helps us understand the kinetics of the reaction. Knowing how fast a reaction proceeds can change how we approach processes involving chemicals, especially when safety or efficiency is a concern.

Half-Life

The concept of half-life is crucial in understanding how reactions proceed over time, especially as it applies to first-order kinetics. The half-life (\( t_{1/2} \)) of a reaction is the time it takes for the concentration of a substance to reduce to half its initial value. In a first-order reaction, the half-life is constant and doesn't depend on the initial concentration.
For first-order reactions, the half-life is given by:

• \( t_{1/2} = \frac{0.693}{k} \)

This steady measure contributes to predicting how long a reaction will last or estimating when a certain concentration will be achieved. For example, to decrease the nitrous oxide concentration to 6.25% of its initial concentration, it must undergo multiple half-lives.
Since 6.25% equates to about one-sixteenth of the original concentration, it takes \( 4 \) half-lives to reach this level. Each halving reduces the concentration to half of what it was at the beginning of that interval. Understanding this helps quantify chemical processes within set timelines.

Nitrous Oxide Decomposition

Nitrous oxide (\( \mathrm{N}_2 \mathrm{O} \)), commonly known as laughing gas, undergoes decomposition to form nitrogen (\( \mathrm{N}_2 \)) and oxygen (\( \mathrm{O}_2 \)). This decomposition is of environmental interest due to its contribution as a greenhouse gas. The breakdown reaction can be expressed as:

• \( 2 \mathrm{N}_2 \mathrm{O} \rightarrow 2 \mathrm{N}_2 + \mathrm{O}_2 \)

The reaction's nature is slow, and, being a first-order reaction, the degradation rate is linked to the nitrous oxide concentration within the environment or a system.
This knowledge is particularly useful not only in chemical industries where \( \mathrm{N}_2 \mathrm{O} \) is utilized but also in environmental science. Monitoring its decomposition informs strategies to regulate its atmospheric levels. Its linear degradation pattern allows creators of aerosols and other products to predict the amount of gas remaining over time under controlled conditions, helping in environmental management and industrial applications.