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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

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

Step by step solution

01

- 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 \( \mathrm{O}_{2} \).
02

- 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 \( \mathrm{O}_{2} \).
03

- Write the balanced equation

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Define reaction rate. Assuming constant temperature and a closed reaction vessel, why does the rate change with time?

What is the central idea of collision theory? How does this model explain the effect of concentration on reaction rate?

How are integrated rate laws used to determine reaction order? What is the reaction order in each of these cases? (a) A plot of the natural logarithm of [reactant] vs. time is linear. (b) A plot of the inverse of [reactant] vs. time is linear. (c) [reactant] vs. time is linear.

The effect of substrate concentration on the first-order growth rate of a microbial population follows the Monod equation: \(\mu=\frac{\mu_{\max } S}{K_{\mathrm{s}}+S}\) where \(\mu\) is the first-order growth rate \(\left(\mathrm{s}^{-1}\right), \mu_{\max }\) is the maximum growth rate \(\left(\mathrm{s}^{-1}\right), S\) is the substrate concentration \(\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) and \(K_{\mathrm{s}}\) is the value of \(S\) that gives one-half of the maximum growth rate (in \(\mathrm{kg} / \mathrm{m}^{3}\) ). For \(\mu_{\max }=1.5 \times 10^{-4} \mathrm{~s}^{-1}\) and \(K_{\mathrm{s}}=0.03 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Plot \(\mu\) vs. \(S\) for \(S\) between 0.0 and \(1.0 \mathrm{~kg} / \mathrm{m}^{3}\). (b) The initial population density is \(5.0 \times 10^{3}\) cells \(/ \mathrm{m}^{3}\). What is the density after \(1.0 \mathrm{~h}\), if the initial \(S\) is \(0.30 \mathrm{~kg} / \mathrm{m}^{3} ?\) (c) What is it if the initial \(S\) is \(0.70 \mathrm{~kg} / \mathrm{m}^{3}\) ?

Like any catalyst, palladium, platinum, or nickel catalyzes both directions of a reaction: addition of hydrogen to (hydrogenation) and its elimination from (dehydrogenation) carbon double bonds. (a) Which variable determines whether an alkene will be hydrogenated or dehydrogenated? (b) Which reaction requires a higher temperature? (c) How can all-trans fats arise during hydrogenation of fats that contain some double bonds with a cis orientation?

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