Question

At \(573 \mathrm{K},\) gaseous \(\mathrm{NO}_{2}(\mathrm{g})\) decomposes, forming \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) If a vessel containing \(\mathrm{NO}_{2}(\mathrm{g})\) has an initial concentration of \(2.8 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) how long will it take for \(75 \%\) of the \(\mathrm{NO}_{2}(\mathrm{g})\) to decompose? The decomposition of \(\mathrm{NO}_{2}(\mathrm{g})\) is second order in the reactant and the rate constant for this reaction, at \(573 \mathrm{K},\) is \(1.1 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}.\)

Short answer

About 97.4 seconds.

Step-by-step solution

Set Up the Problem

We are given a second-order decomposition reaction where NO2(g) decomposes into NO(g) and O2(g). The initial concentration of NO2(g) is given as \(2.8 \times 10^{-2} \text{ mol/L}\), and the rate constant \(k\) is \(1.1 \text{ L/mol} \cdot \text{s}\). We want to find out how long it takes for 75% of the NO2(g) to decompose.

Calculate the Final Concentration

If 75% of NO2(g) decomposes, then 25% of it remains. The final concentration of NO2(g), \([NO_2]_f\), is 25% of the initial concentration. Calculate it as follows:\[[NO_2]_f = 0.25 \times 2.8 \times 10^{-2} = 0.7 \times 10^{-2} \text{ mol/L}\]

Use Second Order Kinetics Equation

For a second-order reaction, the integrated rate law is given by:\[\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt\]where \([A]_t\) is the final concentration, \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is time. Substitute the known values into the equation.

Substitute Values into the Equation

We know the initial concentration \([A]_0\) is \(2.8 \times 10^{-2} \text{ mol/L}\), the final concentration \([A]_t\) is \(0.7 \times 10^{-2} \text{ mol/L}\), and the rate constant \(k\) is \(1.1 \text{ L/mol} \cdot \text{s}\). Substitute these values:\[\frac{1}{0.7 \times 10^{-2}} = \frac{1}{2.8 \times 10^{-2}} + 1.1 \times t\]

Solve for Time \(t\)

Calculate the inverse of the concentrations and rearrange to solve for \(t\):\[\frac{1}{0.7 \times 10^{-2}} = 142.86 \\frac{1}{2.8 \times 10^{-2}} = 35.71\]Substitute back:\[142.86 = 35.71 + 1.1t\]Subtract 35.71 from both sides:\[107.15 = 1.1t\]Finally, divide by 1.1:\[t \approx 97.41 \text{ seconds}\]

Provide Final Answer

The time required for 75% of NO2(g) to decompose is approximately 97.4 seconds.

Key concepts

Second-Order Reaction

In chemical kinetics, a second-order reaction implies that the rate of reaction depends on the concentration of one reactant raised to the second power or the concentration of two reactants each raised to the first power. For our exercise, the decomposition of \(\mathrm{NO}_2 ext{(g)}\) is a second-order reaction. This means that the rate at which \(\mathrm{NO}_2 ext{(g)}\) decomposes increases with the square of its concentration.

If you double the concentration of \(\mathrm{NO}_2 ext{(g)}\), the rate of decomposition will increase by a factor of four. This type of relationship is crucial in predicting how fast reactions proceed, especially when the time factor is involved. Understanding the order of a reaction helps chemists to determine the most effective conditions for a reaction to occur quickly and efficiently.

Decomposition Reaction

A decomposition reaction involves a single compound breaking down into two or more simpler substances. For the given reaction of \(\mathrm{NO}_2 ext{(g)}\), it decomposes into \(\mathrm{NO} ext{(g)}\) and \(\mathrm{O}_2 ext{(g)}\). Decomposition reactions are commonly endothermic, meaning they consume heat. This is why the reaction in our exercise is studied at a high temperature of 573 K.

Key characteristics of decomposition reactions include:

• Formation of simpler products.
• Often require energy input (heat, light, etc.) to proceed.
• Can be reversible under certain conditions.

The knowledge of how decomposition reactions work is crucial in predicting by-products and controlling processes in industrial applications, such as the synthesis of chemicals, energy production, and waste management.

Rate Constant

The rate constant, denoted by \(k\), is a crucial factor in a reaction's rate law and is unique for every reaction at a given temperature. In our case, the rate constant for the decomposition of \(\mathrm{NO}_2 ext{(g)}\) at 573 K is \(1.1 ext{ L/mol} \cdot \text{s}\). It acts as a proportionality factor in the rate law, linking the concentration of the reactants to the reaction rate.

It provides insights into how fast or slow a reaction occurs under specific conditions. Some essential points about the rate constant include:

• It is dependent on temperature; increasing temperature usually increases \(k\).
• It has units that vary based on reaction order (\(\text{L/mol} \cdot \text{s}\) for second-order here).
• A larger \(k\) value means a faster reaction.

Understanding the rate constant helps chemists to customize reactions for desired yields and efficiency.

Integrated Rate Law

The integrated rate law for a second-order reaction is expressed as: \[\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt\] where \([A]_t\) is the concentration of the reactant at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant. This equation helps to relate the concentrations of reactants to time, allowing chemists to calculate how long it will take for a certain amount of reactant to decompose or transform.

Key points for integrated rate laws include:

• They provide a mathematical way to determine the time needed for a specific reaction conversion.
• The form differs for zero, first, and second-order reactions.
• It is derived by integrating the differential rate law.

Knowing how to use the integrated rate law is invaluable for understanding reaction kinetics, especially when designing experiments and predicting reaction timeframes.