Problem 32 - Chemistry: An Atoms-Focused Approach

Question

\(\mathrm{N}_{2} \mathrm{O}_{5},\) when dissolved in water, decomposes to produce \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) $$2 \mathrm{N}_{2} \mathrm{O}_{5}(a q) \rightarrow 4 \mathrm{NO}_{2}(a q)+\mathrm{O}_{2}(a q)$$ The rate of formation of \(\mathrm{O}_{2}\) is \(4.0 \times 10^{-3} M \cdot \mathrm{min}^{-1}\) a. What is the rate of formation of \(\mathrm{NO}_{2}(a q) ?\) b. What is the rate of disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}(a q) ?\)

Step-by-Step Solution

Answer

Answer: (a) The rate of formation of NO2(aq) is \(16.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}\). (b) The rate of disappearance of N2O5(aq) is \(-8.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}\).

Step 1: Rate Law Relationship for Reactant and Product

We are given the balanced chemical equation: $$2 \mathrm{N}_{2} \mathrm{O}_{5}(a q) \rightarrow 4 \mathrm{NO}_{2}(a q)+\mathrm{O}_{2}(a q)$$ We can see from the stoichiometry that for every 2 moles of N2O5 consumed, 4 moles of NO2 are produced, and 1 mole of O2 is produced. We can write the rate law relationships as follows: $$\frac{\Delta[\mathrm{N}_{2} \mathrm{O}_{5}]}{\Delta t} \times \frac{1}{-2} = \frac{\Delta[\mathrm{NO}_{2}]}{\Delta t} \times \frac{1}{4} = \frac{\Delta[\mathrm{O}_{2}]}{\Delta t} \times \frac{1}{1}$$

Step 2: Calculate the Rate of Formation of NO2

We are given the rate of formation of O2: $$\frac{\Delta[\mathrm{O}_{2}]}{\Delta t} = 4.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}$$ We can use the rate law relationship to find the rate of formation of NO2: $$\frac{\Delta[\mathrm{NO}_{2}]}{\Delta t} = \frac{-\Delta[\mathrm{N}_{2} \mathrm{O}_{5}]}{2\Delta t}$$ $$\frac{\Delta[\mathrm{NO}_{2}]}{\Delta t} = 4 \times \frac{\Delta[\mathrm{O}_{2}]}{\Delta t}$$ $$\frac{\Delta[\mathrm{NO}_{2}]}{\Delta t} = 4 \times 4.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}$$ $$\frac{\Delta[\mathrm{NO}_{2}]}{\Delta t} = 16.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}$$ So, the rate of formation of NO2(aq) is \(16.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}\).

Step 3: Calculate the Rate of Disappearance of N2O5

We can now use the rate law relationship to find the rate of disappearance of N2O5: $$\frac{\Delta[\mathrm{N}_{2} \mathrm{O}_{5}]}{\Delta t} = -2 \times \frac{\Delta[\mathrm{O}_{2}]}{\Delta t} = -2 \times 4.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}$$ $$\frac{\Delta[\mathrm{N}_{2} \mathrm{O}_{5}]}{\Delta t} = -8.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}$$ So, the rate of disappearance of N2O5(aq) is \(-8.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}\). The negative sign indicates that the concentration of N2O5(aq) is decreasing over time. a. The rate of formation of NO2(aq) is \(16.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}\). b. The rate of disappearance of N2O5(aq) is \(-8.0 \times 10^{-3}\,\mathrm{M\cdot min}^{-1}\).

Key Concepts

Rate of Reaction

In chemical kinetics, the **rate of reaction** is a measure of how fast a chemical reaction occurs. It is defined as the change in the concentration of a reactant or product per unit time. In our exercise, we are interested in how quickly \[ \mathrm{N}_{2}\mathrm{O}_{5} \] decomposes and how fast \[ \mathrm{NO}_{2} \] and \[ \mathrm{O}_{2} \] are formed. Scientists often use units of concentration such as molarity (M, moles per liter) and time in minutes or seconds.

• The given rate of \( \mathrm{O}_{2} \) formation helps determine the reaction speed and allows calculations of other rates in the reaction.
• Writing the rate as \( 4.0 \times 10^{-3} \, \mathrm{M\cdot min}^{-1} \) indicates that the concentration of \( \mathrm{O}_{2} \) increases by \( 4.0 \times 10^{-3} \) moles per liter every minute.

Understanding the rate helps predict how long it takes for reactants to turn into products. This is crucial for industrial applications where reaction speed determines production efficiency.

Stoichiometry

**Stoichiometry** is a section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In the given reaction:\[ 2 \mathrm{N}_{2} \mathrm{O}_{5}(a q) \rightarrow 4 \mathrm{NO}_{2}(a q)+\mathrm{O}_{2}(a q) \]stoichiometry allows us to see how many moles of a product will be formed from a given amount of reactants, and vice versa. It uses a balanced chemical equation to relate different substances in the reaction. Key points include:

• For every 2 moles of \( \mathrm{N}_{2} \mathrm{O}_{5} \) destroyed, 4 moles of \( \mathrm{NO}_{2} \) are created and 1 mole of \( \mathrm{O}_{2} \) is released.
• This relationship enables conversion between the rates of different substances by using ratios.

Stoichiometry is the **bridge** between the qualitative and quantitative aspects of chemical reactions, allowing the calculation of how much of a substance is needed or produced.

Reaction Rates

**Reaction rates** quantify how quickly or slowly a reaction takes place. Unlike the rate of reaction, which focuses on a single substance, reaction rates can describe the speed of the entire process involving multiple substances. In our example, the equation explains how fast each component of the reaction is consumed or produced:

• The rate of disappearance of \( \mathrm{N}_{2} \mathrm{O}_{5} \) is negative, showing a decrease in its concentration over time, calculated as \(-8.0 \times 10^{-3} \, \mathrm{M\cdot min}^{-1}\).
• Conversely, the rates of formation for \( \mathrm{NO}_{2} \) and \( \mathrm{O}_{2} \) are positive, indicating increasing concentrations.

Using the given rate of \( \mathrm{O}_{2} \) formation, we determine the \( \mathrm{NO}_{2} \) formation rate as \( 16.0 \times 10^{-3} \, \mathrm{M\cdot min}^{-1}\).Understanding reaction rates is essential for optimizing chemical processes, ensuring products are made efficiently, and reactions are safe to conduct in controlled environments.