Question

Consider the chemical reaction: \(\mathrm{N}_{2}(\mathrm{~g})\) \(+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\). The rate of this reaction can be expressed in terms of time derivative of concentration of \(\mathrm{N}_{2}(\mathrm{~g})\), \(\mathrm{H}_{2}(\mathrm{~g})\) or \(\mathrm{NH}_{3}(\mathrm{~g})\). Identify the correct relationship amongst the rate expressions (a) rate \(\begin{aligned} \text { (b) rate } &=-\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{d} t}=-\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d} t} \\ &=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{d} t} \\ &=+2 \frac{\mathrm{d}\left[\mathrm{NH}_{2}\right]}{\mathrm{d} t}=-3 \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d} t} \\ \mathrm{~d} t \end{aligned}\) (c) rate \(\begin{aligned} &=-\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{d} t}=\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d} t} \\ &=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{d} t} \\ \text { (d) rate } &=-\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{d} t}=-\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{d} t} \\\ &=+\frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{d} t} \end{aligned}\)

Short answer

The correct relationship among the rate expressions is option (c): rate = -\frac{\mathrm{d}[\mathrm{N}_2]}{\mathrm{d} t} = \frac{1}{3} \frac{\mathrm{d}[\mathrm{H}_2]}{\mathrm{d} t} = +\frac{1}{2} \frac{\mathrm{d}[\mathrm{NH}_3]}{\mathrm{d} t}.

Step-by-step solution

Understand the Stoichiometry of the Chemical Reaction

The balanced chemical equation is given as \(\mathrm{N}_2(\mathrm{~g}) + 3\mathrm{H}_2(\mathrm{~g}) \rightarrow 2\mathrm{NH}_3(\mathrm{~g})\). The stoichiometry of the reaction indicates that one mole of \(\mathrm{N}_2\) reacts with three moles of \(\mathrm{H}_2\) to produce two moles of \(\mathrm{NH}_3\).

Relate the Rates of Consumption and Production

According to the stoichiometry, the rate at which \(\mathrm{N}_2\) is consumed is three times slower than the rate at which \(\mathrm{H}_2\) is consumed because it takes 3 moles of \(\mathrm{H}_2\) to react with 1 mole of \(\mathrm{N}_2\). Conversely, for every mole of \(\mathrm{N}_2\) consumed, two moles of \(\mathrm{NH}_3\) are produced. Thus, the rate of production of \(\mathrm{NH}_3\) is twice the rate of consumption of \(\mathrm{N}_2\).

Apply the Rate Relationship to Each Substance

The rate of reaction in terms of \(\mathrm{N}_2\) is \(\text{rate} = -\frac{\mathrm{d}[\mathrm{N}_2]}{\mathrm{d}t}\). For \(\mathrm{H}_2\), the rate is three times faster, reflected as \(\text{rate} = -\frac{1}{3}\frac{\mathrm{d}[\mathrm{H}_2]}{\mathrm{d}t}\). For \(\mathrm{NH}_3\), the rate of production is \(\text{rate} = +\frac{1}{2}\frac{\mathrm{d}[\mathrm{NH}_3]}{\mathrm{d}t}\). The negative sign indicates consumption, while the positive sign indicates production.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the quantitative relationship between reactants and products in a chemical reaction. When we refer to a chemical equation like \(\mathrm{N}_2(\mathrm{~g}) + 3\mathrm{H}_2(\mathrm{~g}) \rightarrow 2\mathrm{NH}_3(\mathrm{~g})\), we are talking about the stoichiometry of the reaction. Here, the coefficients (1 for \(\mathrm{N}_2\), 3 for \(\mathrm{H}_2\), and 2 for \(\mathrm{NH}_3\)) tell us in what ratio the molecules react and are formed.

Understanding stoichiometry is crucial for predicting the amounts of substances consumed and produced in a reaction. It also allows us to establish relationships between the rates at which reactants are used up and products are formed. For example, because one mole of \(\mathrm{N}_2\) reacts with three moles of \(\mathrm{H}_2\) to produce two moles of \(\mathrm{NH}_3\), we can deduce that the disappearance of \(\mathrm{N}_2\) and \(\mathrm{H}_2\) and the appearance of \(\mathrm{NH}_3\) are directly related by their stoichiometric coefficients.

This relationship is fundamental when it comes time to connect stoichiometry with the rate of reaction, which leads us to our next concept.

Rate of Reaction

The rate of reaction is a measure of how quickly a reactant is consumed or a product is formed over time. It is often expressed as the change in concentration of a substance per unit time. Reflecting on our previous example, the stoichiometry tells us that one mole of \(\mathrm{N}_2\) is used for every three moles of \(\mathrm{H}_2\) consumed and this results in two moles of \(\mathrm{NH}_3\) produced.

We can express these changes quantitatively using the concept of rate. For instance, the rate of disappearance of \(\mathrm{N}_2\) is equivalent to the rate of disappearance of \(\mathrm{H}_2\) divided by 3, as it takes three times more \(\mathrm{H}_2\) molecules to react with \(\mathrm{N}_2\). In contrast, the rate of formation of \(\mathrm{NH}_3\) is twice the rate of disappearance of \(\mathrm{N}_2\), due to the 2:1 ratio indicated in the balanced equation.

Thus, the stoichiometric coefficients play a critical role in connecting the changes in concentration of reactants and products to the overall rate of reaction. By observing the rates of change for each substance involved in the reaction, we can understand the dynamics of the chemical process and calculate how long it will take for a reaction to go to completion or reach a certain point.

Concentration Time Derivative

The concentration time derivative is a mathematical way of expressing the rate of reaction. It is given by the symbol \(\frac{\mathrm{d}[\mathrm{Substance}]}{\mathrm{d}t}\), where \(\mathrm{d}[\mathrm{Substance}]\) represents the infinitesimally small change in concentration of the substance, and \(\mathrm{d}t\) represents the infinitesimally small change in time.

In our example reaction, the concentration time derivative for \(\mathrm{N}_2\) is \(\frac{\mathrm{d}[\mathrm{N}_2]}{\mathrm{d}t}\), which would be negative since \(\mathrm{N}_2\) is a reactant and its concentration decreases over time. Conversely, the concentration time derivative for \(\mathrm{NH}_3\) is \(\frac{\mathrm{d}[\mathrm{NH}_3]}{\mathrm{d}t}\), which would be positive, reflecting its formation. The stoichiometry of the reaction lets us relate these derivatives to each other. The rate at which \(\mathrm{N}_2\) is consumed is \(\frac{-1}{3}\) times the rate at which \(\mathrm{H}_2\) is consumed, and the rate at which \(\mathrm{NH}_3\) is produced is twice the rate at which \(\mathrm{N}_2\) is consumed.

This is more than just academic; quantifying the rate of change in concentration over time allows chemists to predict and control conditions in reactions, such as temperature and pressure, to optimize product yield and minimize waste. Understanding these concepts is vital for chemists and industry professionals who work with chemical reactions on a daily basis.