Problem 34 - Chemistry: An Atoms-Focused Approach

Question

Nitric oxide (NO) can be removed from gas-fired power-plant emissions by reaction with methane as follows: \(\mathrm{CH}_{4}(g)+4 \mathrm{NO}(g) \rightarrow 2 \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) Write an equation relating each of the following pairs of rates: a. The rate of formation of \(\mathrm{N}_{2}\) to the rate of formation of \(\mathrm{CO}_{2}\) b. The rate of formation of \(\mathrm{CO}_{2}\) to the rate of consumption of NO c. The rate of consumption of \(\mathrm{CH}_{4}\) to the rate of formation of \(\mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

Based on the given reaction equation: a. The rate of formation of \(\mathrm{N}_{2}\) is 2 times the rate of formation of \(\mathrm{CO}_{2}\). b. The rate of consumption of NO is 4 times the rate of formation of \(\mathrm{CO}_{2}\). c. The rate of consumption of \(\mathrm{CH}_{4}\) is 1/2 times the rate of formation of \(\mathrm{H}_{2} \mathrm{O}\).

Step 1: a. The rate of formation of \(\mathrm{N}_{2}\) to the rate of formation of \(\mathrm{CO}_{2}\)

To find the relationship between the rate of formation of \(\mathrm{N}_{2}\) to the rate of formation of \(\mathrm{CO}_{2}\), consult the balanced chemical equation: \(\mathrm{CH}_{4}(g)+4 \mathrm{NO}(g) \rightarrow 2 \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) From the balanced chemical equation, it can be seen that 2 moles of \(\mathrm{N}_{2}\) are formed as 1 mole of \(\mathrm{CO}_{2}\) is formed. Therefore, the relationship between the rate of formation of \(\mathrm{N}_{2}\) to the rate of formation of \(\mathrm{CO}_{2}\) is: Rate of formation of \(\mathrm{N}_{2}\) = 2 × Rate of formation of \(\mathrm{CO}_{2}\)

Step 2: b. The rate of formation of \(\mathrm{CO}_{2}\) to the rate of consumption of NO

To find the relationship between the rate of formation of \(\mathrm{CO}_{2}\) to the rate of consumption of NO, consult the balanced chemical equation: \(\mathrm{CH}_{4}(g)+4 \mathrm{NO}(g) \rightarrow 2 \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) From the balanced chemical equation, it can be seen that 1 mole of \(\mathrm{CO}_{2}\) is formed as 4 moles of NO are consumed. Therefore, the relationship between the rate of formation of \(\mathrm{CO}_{2}\) to the rate of consumption of NO is: Rate of consumption of NO = 4 × Rate of formation of \(\mathrm{CO}_{2}\)

Step 3: c. The rate of consumption of \(\mathrm{CH}_{4}\) to the rate of formation of \(\mathrm{H}_{2} \mathrm{O}\)

To find the relationship between the rate of consumption of \(\mathrm{CH}_{4}\) to the rate of formation of \(\mathrm{H}_{2} \mathrm{O}\), consult the balanced chemical equation: \(\mathrm{CH}_{4}(g)+4 \mathrm{NO}(g) \rightarrow 2 \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) From the balanced chemical equation, it can be seen that 1 mole of \(\mathrm{CH}_{4}\) is consumed as 2 moles of \(\mathrm{H}_{2} \mathrm{O}\) are formed. Therefore, the relationship between the rate of consumption of \(\mathrm{CH}_{4}\) to the rate of formation of \(\mathrm{H}_{2} \mathrm{O}\) is: Rate of consumption of \(\mathrm{CH}_{4}\) = 1/2 × Rate of formation of \(\mathrm{H}_{2} \mathrm{O}\)

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that allows us to understand the quantitative relationships in chemical reactions. When we talk about stoichiometry, we're essentially discussing the ratio in which reactants and products participate in a reaction.
This concept is crucial because it helps predict how much of a product will be formed or how much of a reactant will be needed. Consider the chemical equation from our original exercise:

• \( \mathrm{CH}_{4}(g) + 4 \mathrm{NO}(g) \rightarrow 2 \mathrm{N}_{2}(g) + \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g) \)

The stoichiometry of this equation tells us that:

• 1 mole of methane (\( \mathrm{CH}_{4} \)) reacts with 4 moles of nitric oxide (\( \mathrm{NO} \))
• This forms 2 moles of nitrogen gas (\( \mathrm{N}_{2} \)), 1 mole of carbon dioxide (\( \mathrm{CO}_{2} \)), and 2 moles of water vapor (\( \mathrm{H}_{2} \mathrm{O} \))

Understanding these ratios is key to calculating reaction rates and determining the exact quantities of reactants and products involved.

Balanced Chemical Equations

A balanced chemical equation is one where the number of atoms for each element is the same on both sides of the equation. This represents the law of conservation of mass: mass is neither created nor destroyed in a chemical reaction. When balancing an equation, coefficients are used in front of each compound to ensure that the number of atoms for each element matches on both sides.

• In the equation \( \mathrm{CH}_{4} + 4 \mathrm{NO} \rightarrow 2 \mathrm{N}_{2} + \mathrm{CO}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \), each element's atoms are balanced:

• For Carbon, we have 1 mole on both sides
• For Hydrogen, 4 atoms on both sides
• For Nitrogen, 4 atoms on both sides
• For Oxygen, 6 atoms on both sides

Balancing ensures that the equation has physical meaning and allows us to use it to predict reactants and products accurately.

Rate of Reaction

The rate of reaction is a measure of how quickly a reaction takes place. It can be thought of as how fast the reactants are turned into products. The rate can be expressed in terms of the concentration change of a reactant or product over time.
Consider our reaction:

• Rate of formation of \( \mathrm{N}_{2} \): Shows how quickly \( \mathrm{N}_{2} \) is produced
• Rate of consumption of \( \mathrm{NO} \): Indicates how fast \( \mathrm{NO} \) disappears
  

For instance, from the given reaction, we find:

• The rate of \( \mathrm{NO} \) consumption is 4 times the rate of formation of \( \mathrm{CO}_{2} \)
• The rate at which \( \mathrm{CH}_{4} \) is consumed is half the rate of formation of \( \mathrm{H}_{2} \mathrm{O} \)

Understanding these rates and their relationships helps in controlling industrial reactions and predicting how the concentration of substances changes with time.

Gas-phase Reactions

Gas-phase reactions involve substances in the gaseous state and are often influenced by various factors such as temperature and pressure. Since gases are highly energetic, their reactions can be quite rapid and are pivotal in processes like combustion, the atmosphere's chemistry, and industrial applications.

• In our reaction, both the reactants (\( \mathrm{CH}_{4} \) and \( \mathrm{NO} \)) and products (like \( \mathrm{N}_{2}, \mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O} \)) exist in the gas phase.

Gas-phase reactions are significant owing to:

• Their ease of mixing due to gaseous diffusion
• The large volume changes that can occur, often used in engines and power generation

Controlling conditions such as temperature and pressure can drastically alter the reaction rate and yield, making understanding gas-phase reactions vital for chemical engineering applications.