Question

Reaction rate is expressed in terms of changes in concentration of reactants and products. Write a balanced equation for the reaction with this rate expression: $$ \text { Rate }=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\Delta t}=\frac{1}{4} \frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t}=\frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t} $$

Short answer

The balanced equation is: \[ 2 \mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2} + \mathrm{O}_{2} \]

Step-by-step solution

- Understand the rate expression

The rate expression given is: \[ \text{Rate}=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\Delta t}=\frac{1}{4} \frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t}=\frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t} \]This expression relates the change in concentrations of different species over time. It indicates that the decomposition of \( \mathrm{N}_{2} \mathrm{O}_{5} \) produces \( \mathrm{NO}_{2} \) and \( \mathrm{O}_{2} \).

- Derive the stoichiometric coefficients

The rates are given in terms of the concentrations of \( \mathrm{N}_{2} \mathrm{O}_{5} \), \( \mathrm{NO}_{2} \), and \( \mathrm{O}_{2} \). The denominator represents the stoichiometric coefficients: \( -\frac{1}{2} \) for \( \mathrm{N}_{2} \mathrm{O}_{5} \), \( \frac{1}{4} \) for \( \mathrm{NO}_{2} \), and \( 1 \) for \( \mathrm{O}_{2} \). We can interpret this as: 2 moles of \( \mathrm{N}_{2} \mathrm{O}_{5} \) decompose to produce 4 moles of \( \mathrm{NO}_{2} \) and 1 mole of \( \mathrm{O}_{2} \).

- Write the balanced equation

Using the stoichiometric coefficients derived, write the balanced chemical equation: \[ 2 \mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2} + \mathrm{O}_{2} \]

Key concepts

stoichiometric coefficients

Stoichiometric coefficients are numbers that appear before the chemical formulas in a balanced chemical equation. They indicate the proportion of reactants and products in a chemical reaction.
These coefficients tell us the relative number of moles of each substance involved.
In our example:
2 moles of \( \text{N}_2\text{O}_5 \) decompose to form 4 moles of \( \text{NO}_2 \) and 1 mole of \( \text{O}_2 \).
Understanding these numbers helps us balance the equation correctly.
This balance ensures that the same number of each type of atom exists on both the reactant and product sides.
To derive these coefficients, you can look at how each substance's concentration changes over time, as given by the reaction rate expression.
From the step-by-step solution, we obtained that:

• 2 moles of \( \text{N}_2\text{O}_5 \) decompose.
• This produces 4 moles of \( \text{NO}_2 \).
• And 1 mole of \( \text{O}_2 \).

These proportions are key to writing a balanced chemical equation.

balanced chemical equation

A balanced chemical equation represents a chemical reaction with equal numbers of each type of atom on both sides.
This balance is crucial because it follows the Law of Conservation of Mass.
In our example, we started with the reaction rate expression:
\( \text{Rate} = -\frac{1}{2} \frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = \frac{1}{4} \frac{\Delta[\text{NO}_2]}{\Delta t} = \frac{\Delta [\text{O}_2]}{\Delta t} \)
From here, we derived the stoichiometric coefficients.
This helped us write the balanced chemical equation:
\[ 2 \text{N}_2\text{O}_5 \rightarrow 4 \text{NO}_2 + \text{O}_2 \]
This equation tells us that 2 moles of \( \text{N}_2\text{O}_5 \) react to produce 4 moles of \( \text{NO}_2 \) and 1 mole of \( \text{O}_2 \).
It's good practice to double-check your work to ensure that all atoms are conserved in the equation.
Balancing helps you understand the mole relationships and how reactants turn into products.

decomposition reaction

A decomposition reaction occurs when a single compound breaks down into two or more simpler substances.
This type of reaction is common in chemistry.
In our example:
\( 2 \text{N}_2\text{O}_5 \rightarrow 4 \text{NO}_2 + \text{O}_2 \)
This equation is a classic decomposition reaction where one molecule of \( \text{N}_2\text{O}_5 \) breaks down into molecules of \( \text{NO}_2 \) and \( \text{O}_2 \).
Decomposition reactions usually require energy from heat, light, or electricity to break the bonds in the compound.
Understanding decomposition reactions is essential for studying chemical stability and reactions that release energy.
The rate at which a decomposition reaction occurs can be measured and expressed using reaction rates, which was addressed in the original exercise.
In summary:

• Decomposition reactions break a single compound into simpler substances.
• They typically require energy input.
• They help us understand the behavior of chemical compounds.

Recognizing a decomposition reaction can help identify products and reactants more easily.