Problem 16 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

For the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) \(\rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) under certain conditions of temperature and partial pressure of the reactants, the rate of formation of \(\mathrm{NH}_{3}\) is \(10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\). The rate of consumption of \(\mathrm{H}_{2}\) under same condition is (a) \(1.5 \times 10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\) (b) \(1.76 \times 10^{-4} \mathrm{~kg} \mathrm{~h}^{-1}\) (c) \(6.67 \times 10^{-4} \mathrm{~kg} \mathrm{~h}^{-1}\) (d) \(3 \times 10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\)

Step-by-Step Solution

Answer

The rate of consumption of \(\mathrm{H}_{2}\) is \(1.5 \times 10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\).

Step 1: Understanding the Stoichiometry

Look at the balanced chemical equation: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\). It indicates that 1 mole of \(\mathrm{N}_{2}\) reacts with 3 moles of \(\mathrm{H}_{2}\) to produce 2 moles of \(\mathrm{NH}_{3}\).

Step 2: Relating the Rates of Reaction

The rate of formation of \(\mathrm{NH}_{3}\) is given as \(10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\). Using the stoichiometric coefficients from the balanced equation, since 2 moles of \(NH_{3}\) are produced for every 3 moles of \(H_{2}\) consumed, the rate of consumption of \(H_{2}\) will be \(\frac{3}{2}\) times the rate of formation of \(NH_{3}\).

Step 3: Calculating the Rate of Consumption of \(H_{2}\)

Multiply the rate of formation of \(NH_{3}\) by \(\frac{3}{2}\) to obtain the rate of consumption of \(H_{2}\): \(10^{-3} \mathrm{~kg} \mathrm{~h}^{-1} \times \frac{3}{2} = 1.5 \times 10^{-3} \mathrm{~kg} \mathrm{~h}^{-1}\).

Key Concepts

Chemical Reaction Rates

Understanding chemical reaction rates is crucial when studying how quickly reactants are transformed into products in a chemical process. It reflects how fast a reaction proceeds and has practical implications in fields such as chemical engineering, environmental science, and pharmacology.

For example, in the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\), the rate at which \(\mathrm{NH}_{3}\) is formed can be measured in kilograms per hour. To draw meaningful conclusions from this information, it's necessary to relate the rate to the balanced chemical equation, paying close attention to the stoichiometry involving moles of reactants and products. This allows us to calculate reaction rates for all reactants and products from any one rate that is known.

When tackling stoichiometry problems involving reaction rates, keep in mind:

• The rate of formation of products and the rate of consumption of reactants are interconnected through stoichiometric coefficients.
• Stoichiometry helps in predicting the rate of consumption of reactants like \(\mathrm{H}_{2}\) if the rate of formation of a product like \(\mathrm{NH}_{3}\) is known, as illustrated in the exercise.
• The reaction rate can vary based on environmental conditions such as temperature, pressure, and the presence of a catalyst.

Balancing Chemical Equations

Balancing chemical equations is a foundational skill in chemistry, vitally important for understanding and correctly solving stoichiometry problems. When a chemical equation is balanced, it obeys the Law of Conservation of Mass, meaning that the number of atoms for each element are the same on both sides of the equation.

The equation for the synthesis of ammonia, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\), serves as a classic example. Here's what to keep in mind:

• Each side of the equation must represent the same quantity of each element; in this case, we have 2 nitrogen atoms and 6 hydrogen atoms on each side.
• Stoichiometric coefficients, like the '3' in front of \(\mathrm{H}_{2}\), indicate the relative amounts of reactants and products involved and are essential for determining reaction rates.
• Mastering the balance of equations is fundamental, as it allows for the precise calculation of reactants needed or products formed in a chemical reaction.

A balanced equation is a roadmap guiding the stoichiometric conversion between moles and other measurable quantities like mass or volume.

Mole Concept

The mole concept is one of the pillars of stoichiometry, allowing chemists to quantify substances based on a common measure: the number of particles or entities in a substance. One mole corresponds to approximately \(6.022 \times 10^{23}\) entities, known as Avogadro's number, which is akin to using 'dozen' for counting eggs.

Understanding the mole concept enables us to link mass (in kilograms or grams) with an actual number of molecules or atoms. Hence, in the given exercise, when we talk about the reaction rates in terms of kilograms per hour, we could, if needed, translate this to moles per hour for a more molecular-level understanding.

Some key takeaways include:

• Stoichiometry problems often require converting mass to moles to apply the balanced equation correctly.
• The molar mass of a substance (the mass of one mole of that substance) is a vital link between the mass of a sample and the number of moles present.
• The mole concept is the bridge between the macroscopic world we observe and the microscopic world of atoms and molecules.

It's important to be familiar with these conversions to effectively solve stoichiometry problems involving reactions, as they provide clarity on the quantitative relations between reactants and products.