Problem 74 - Chemistry: An Atoms-Focused Approach

Question

Nitrous acid slowly decomposes to \(\mathrm{NO}, \mathrm{NO}_{2},\) and water in the following second-order reaction: $$2 \mathrm{HNO}_{2}(a q) \rightarrow \mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)$$ a. Use the following data to determine the rate constant for this reaction at \(298 \mathrm{K}:\) $$\begin{array}{cc} \text { Time (min) } & {\left[\mathrm{HNO}_{2}\right](\mu M)} \\\0 & 0.1560 \\\\\hline 1000 & 0.1466 \\\\\hline 1500 & 0.1424 \\\\\hline 2000 & 0.1383 \\\\\hline 2500 & 0.1345 \\\\\hline 3000 & 0.1309 \\\\\hline\end{array}$$ b. Determine the half-life for the decomposition of \(\mathrm{HNO}_{2}\). c. If the experiment that yielded the results in the preceding table had been continued for 3000 minutes more, what would the concentration of HNO \(_{2}\) have been?

Step-by-Step Solution

Answer

Answer: The rate constant for the reaction at 298 K is \(6.02 \times 10^{-5} \ M^{-1} min^{-1}\). The half-life for the decomposition of HNO2 is \(1.1 \times 10^4 \ min\). The predicted concentration of HNO2 after 3000 more minutes is \(0.1074 \ \mu M\).

Step 1: Part a: Determine the rate constant for the reaction at 298 K

For a second-order reaction, the rate law can be expressed as: $$\text {rate} = k[\mathrm{HNO}_2]^2$$ Where k is the rate constant at 298 K, which is what we have to find. Using the second-order integrated rate law, we can write: $$\frac{1}{[\mathrm{HNO}_2(t)]} - \frac{1}{[\mathrm{HNO}_2(0)]} = kt$$ We have the experimental values for \([\mathrm{HNO}_2]\) at different times, so we can use any of the data points to calculate k. Let's use the first data point of 1000 minutes: $$\frac{1}{0.1466} - \frac{1}{0.1560} = 1000k$$ Now, let's solve for k: $$k = \frac{\frac{1}{0.1466} - \frac{1}{0.1560}}{1000} = 6.02 \times 10^{-5} \ M^{-1} min^{-1}$$

Step 2: Part b: Determine the half-life for the decomposition of HNO2

For a second-order reaction, the half-life can be calculated using the following formula: $$t_{1/2} = \frac{1}{k[\mathrm{HNO}_2(0)]}$$ We already found the rate constant k, and we are given the initial concentration of HNO2 (\([\mathrm{HNO}_2(0)]\)). Now, we will substitute both values into the formula: $$t_{1/2} = \frac{1}{(6.02 \times 10^{-5})(0.1560)} = 1.1 \times 10^4 \ min$$

Step 3: Part c: Predict the concentration of HNO2 after 3000 more minutes

To predict the concentration of HNO2 after 3000 more minutes, we will use the second-order integrated rate law. Given the reaction has been running for 3000 minutes already, we need to calculate the new time which will be 6000 minutes: $$\frac{1}{[\mathrm{HNO}_2(6000)]} - \frac{1}{[\mathrm{HNO}_2(0)]} = k(6000)$$ Now, we will substitute the values into the equation, and solve for the concentration of HNO2 after 6000 minutes: $$\frac{1}{[\mathrm{HNO}_2(6000)]} - \frac{1}{0.1560} = (6.02 \times 10^{-5})(6000)$$ $$\frac{1}{[\mathrm{HNO}_2(6000)]} = \frac{1}{0.1560} + (6.02 \times 10^{-5})(6000)$$ $$[\mathrm{HNO}_2(6000)] = 0.1074 \ \mu M$$ So, the concentration of HNO2 after 3000 more minutes will be \(0.1074 \ \mu M\).

Key Concepts

Rate Constant Determination

When dealing with chemical reactions, the rate constant "k" is a crucial part of understanding how fast reactions occur. In our exercise, we need to determine this rate constant for a second-order reaction involving nitrous acid at 298 K. For second-order reactions, the rate is affected by the concentration of the reactant squared, which can be expressed as:
\[ \text{rate} = k[\mathrm{HNO}_2]^2 \]To find "k," we use the second-order integrated rate law:
\[ \frac{1}{[\mathrm{HNO}_2(t)]} - \frac{1}{[\mathrm{HNO}_2(0)]} = kt \]This formula allows us to relate the change in concentration over time to the rate constant. We use the concentrations at different times given in the problem to solve for "k." For instance, using the data point at 1000 minutes, we substitute the concentrations and time into the equation:

• Initial concentration \([\mathrm{HNO}_2(0)]=0.1560 \,\mu M\)
• Concentration at 1000 min \([\mathrm{HNO}_2(1000)]=0.1466 \,\mu M\)

Plug these into the equation to find "k":
\[ k = \frac{\frac{1}{0.1466} - \frac{1}{0.1560}}{1000} = 6.02 \times 10^{-5} \, M^{-1}\, min^{-1} \]Determining the rate constant helps describe how quickly the reaction progresses under the given conditions. This is foundational for predicting future or past concentrations within the system.

Half-Life Calculation

Half-life is an essential concept when discussing the kinetics of a reaction. For second-order reactions, the half-life is not constant like it is for first-order reactions but instead depends on the initial concentration of the reactants. You can calculate the half-life for a second-order reaction using the formula:
\[ t_{1/2} = \frac{1}{k[\mathrm{HNO}_2(0)]} \]With the rate constant "k" we determined earlier as \(6.02 \times 10^{-5} \, M^{-1}\, min^{-1}\), and the initial concentration \([\mathrm{HNO}_2(0)]=0.1560 \,\mu M\), we substitute these values into the formula:
\[ t_{1/2} = \frac{1}{(6.02 \times 10^{-5})(0.1560)} = 1.1 \times 10^4 \, min \]This calculation shows that the half-life is 11,000 minutes. This indicates the time it takes for the concentration of \([\mathrm{HNO}_2]\) to reduce to half its initial value of \(0.1560 \,\mu M\). The dependency on the initial concentration makes the second-order half-life unique, and understanding this helps us predict how long reactions might take to reach a certain extent.

Integrated Rate Law

Understanding the integrated rate law is fundamental to predicting the concentration of reactants over time in chemical reactions. For second-order reactions, this law is expressed as:
\[ \frac{1}{[\mathrm{HNO}_2(t)]} - \frac{1}{[\mathrm{HNO}_2(0)]} = kt \]Using this equation, you can calculate the concentration at any given time, provided you know the initial concentration and the rate constant "k." In the exercise, we already predicted the concentration of \([\mathrm{HNO}_2]\) after a total of 6000 minutes by rearranging the equation:
\[ \frac{1}{[\mathrm{HNO}_2(6000)]} = \frac{1}{0.1560} + (6.02 \times 10^{-5})(6000) \]Upon solving, we found that \([\mathrm{HNO}_2(6000)] = 0.1074 \ mu M\), confirming the usefulness of the integrated rate law in forecasting changes in concentration. By having such equations readily available, scientists and students can easily model and predict chemical kinetics under various conditions, enabling finely-tuned experiments and deeper understanding of reaction mechanisms.