Problem 2 - Chemistry and Chemical Reactivity

Question

Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})\) (b) \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\)

Step-by-Step Solution

Answer

(a) \( -\frac{1}{2} \frac{d[\mathrm{NO}]}{dt} = -\frac{d[\mathrm{Br}_2]}{dt} = \frac{1}{2} \frac{d[\mathrm{NOBr}]}{dt} \); (b) \( -\frac{d[\mathrm{N}_2]}{dt} = -\frac{1}{3} \frac{d[\mathrm{H}_2]}{dt} = \frac{1}{2} \frac{d[\mathrm{NH}_3]}{dt} \).

Step 1: Understand Reaction (a)

In reaction (a), the equation is \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})\). The stoichiometry of NO and NOBr are both 2:1 with respect to the disappearance and formation of these substances.

Step 2: Calculate Relative Rate of Disappearance for NO

The rate of disappearance of \(\mathrm{NO}\) is \(-\frac{1}{2}\frac{d[\mathrm{NO}]}{dt}\). The negative sign indicates it is a reactant being consumed.

Step 3: Rate of Disappearance for Br2

For \(\mathrm{Br}_{2}\), the rate of disappearance is \(-\frac{d[\mathrm{Br}_{2}]}{dt}\) since it reacts in a one-to-one ratio.

Step 4: Rate of Formation for NOBr

The rate of formation for \(\mathrm{NOBr}\) is \(+\frac{1}{2}\frac{d[\mathrm{NOBr}]}{dt}\). The coefficient indicates that for every NOBr formed, two moles of NO are used, thus it relates through a factor of \(\frac{1}{2}\).

Step 5: Understand Reaction (b)

In reaction (b), the equation is \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\). The stoichiometry here will determine the rate relationships.

Step 6: Relative Rate of Disappearance for N2

The rate of disappearance for \(\mathrm{N}_{2}\) is \(-\frac{d[\mathrm{N}_{2}]}{dt}\) because nitrogen is consumed in a one-to-one ratio with ammonia production.

Step 7: Relative Rate of Disappearance for H2

For \(\mathrm{H}_{2}\), the rate of disappearance is \(-\frac{1}{3}\frac{d[\mathrm{H}_{2}]}{dt}\) since three moles of \(\mathrm{H}_{2}\) are consumed for every two moles of \(\mathrm{NH}_{3}\) produced.

Step 8: Relative Rate of Formation for NH3

The rate of formation for \(\mathrm{NH}_{3}\) is \(+\frac{1}{2}\frac{d[\mathrm{NH}_{3}]}{dt}\) because two moles of \(\mathrm{NH}_{3}\) are produced for every one mole of \(\mathrm{N}_{2}\) and three moles of \(\mathrm{H}_{2}\) consumed.

Key Concepts

Stoichiometry

Stoichiometry is the key to understanding chemical reactions and their rates. It refers to the relationship between the quantities of reactants and products in a chemical equation. By looking at the balanced chemical equation, we can determine these relationships.

The coefficients in the chemical equation reflect the proportions in which substances react and form. For example, in reaction (a) \(2 \mathrm{NO} + \mathrm{Br}_{2} \rightarrow 2 \mathrm{NOBr}\), the coefficients "2" and "1" indicate that two moles of \(\mathrm{NO}\) react with one mole of \(\mathrm{Br}_{2}\) to form two moles of \(\mathrm{NOBr}\).

This is how stoichiometry helps us calculate the rate of disappearance for reactants and the rate of formation for products. Knowing these relationships is vital for predicting the behavior of reactions under different conditions.

Rate of Disappearance

The rate at which a reactant disappears in a chemical reaction shows how quickly it is being consumed. This rate can be calculated from the change in concentration of the reactants over a given time period.

### Calculating Rate of Disappearance
For reactant \(\mathrm{NO}\) in reaction (a), the rate of disappearance is given by \(-\frac{1}{2}\frac{d[\mathrm{NO}]}{dt}\). The negative sign indicates that the concentration of a reactant decreases over time.

For \(\mathrm{Br}_{2}\), the rate of disappearance is \(-\frac{d[\mathrm{Br}_{2}]}{dt}\) because it has a one-to-one stoichiometric ratio to product formation. Similarly, in reaction (b), \(\mathrm{N}_{2}\) vanishes at \(-\frac{d[\mathrm{N}_{2}]}{dt}\), and \(\mathrm{H}_{2}\) at \(-\frac{1}{3}\frac{d[\mathrm{H}_{2}]}{dt}\) because it is consumed three times faster than \(\mathrm{NH}_{3}\) is formed.

This concept helps us understand the dynamics of how quickly a chemical reaction can proceed.

Rate of Formation

The rate of formation tells us how fast products are being generated from a reaction. This is determined by the stoichiometry of the balanced equation, which helps us track the increase in concentration over time.

### Calculating Rate of Formation
In reaction (a), \(\mathrm{NOBr}\) forms at a rate of \(+\frac{1}{2}\frac{d[\mathrm{NOBr}]}{dt}\). Here, the positive sign illustrates that the product's concentration is rising as the reaction progresses.

In reaction (b), the rate of formation for \(\mathrm{NH}_{3}\) is \(+\frac{1}{2}\frac{d[\mathrm{NH}_{3}]}{dt}\). This arises from the reaction's stoichiometry where two moles of \(\mathrm{NH}_{3}\) are produced for every mole of \(\mathrm{N}_{2}\) consumed.

Understanding this concept is crucial for calculating how much product can be obtained over time and optimizing reactions for industrial and laboratory processes.