Problem 103 - Chemistry The Central Science In Si Units

Question

A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(g)\). The reaction ceases before either reactant has been totally consumed. At this stage \(3.0 \mathrm{~mol} \mathrm{~N}_{2}, 3.0 \mathrm{~mol} \mathrm{H}_{2}\), and \(3.0 \mathrm{~mol} \mathrm{NH}_{3}\) are present. Howmany moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

Step-by-Step Solution

Answer

Originally, there were 4.5 moles of \( \mathrm{N}_2 \) and 7.5 moles of \( \mathrm{H}_2 \).

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the formation of ammonia from nitrogen and hydrogen is given by:\[ \mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightarrow 2 \mathrm{NH}_3(g) \]This equation tells us that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.

Step 2: Define Variables for Initial Moles

Let the initial moles of nitrogen be \( x \) and hydrogen be \( y \). Since the reaction ceases before complete consumption, we know the remaining moles of each gas are non-zero.

Step 3: Consider Moles at Equilibrium

At equilibrium, we are given the composition: - \( 3.0 \) moles of \( \mathrm{N}_2 \)- \( 3.0 \) moles of \( \mathrm{H}_2 \)- \( 3.0 \) moles of \( \mathrm{NH}_3 \)

Step 4: Set Up Stoichiometric Relationships

According to the stoichiometry of the balanced equation, for every 2 moles of \( \mathrm{NH}_3 \) formed, 1 mole of \( \mathrm{N}_2 \) and 3 moles of \( \mathrm{H}_2 \) are consumed. Let \( z \) be the moles of \( \mathrm{NH}_3 \) formed (which is 3 moles at equilibrium):- \( \mathrm{N}_2 \) reacted = \( z / 2 \)- \( \mathrm{H}_2 \) reacted = \( 3z / 2 \)Therefore:- Remaining \( \mathrm{N}_2 \) = \( x - z/2 = 3 \)- Remaining \( \mathrm{H}_2 \) = \( y - 3z/2 = 3 \) Substitute the value of \( z = 3 \) into these equations:

Step 5: Solve the Equations

Substitute \( z = 3 \) into the remaining nitrogen and hydrogen amounts:\[ x - \frac{3}{2} = 3 \] which simplifies to: \[ x = 4.5 \]For hydrogen:\[ y - \frac{9}{2} = 3 \]which simplifies to: \[ y = 7.5 \] So the original moles of \( \mathrm{N}_2 \) and \( \mathrm{H}_2 \) were 4.5 and 7.5 moles respectively.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that focuses on the quantitative relationships between the reactants and products in a chemical reaction. This concept is crucial when dealing with reactions like ammonia synthesis, where precise proportions of nitrogen and hydrogen gases create ammonia.

In a balanced reaction, stoichiometry helps us determine how much of each reactant we need and how much product will form. For example, if we know the equation for forming ammonia is \( \mathrm{N}_2 + 3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3 \), stoichiometry tells us that one mole of nitrogen reacts with three moles of hydrogen, thereafter producing two moles of ammonia.

This proportional relationship is vital for calculating the amounts of reactants and products present at any stage of the reaction. Understanding stoichiometry can also help us understand why sometimes not all reactants are used up at the end of a reaction, such as in the given exercise where some nitrogen and hydrogen remain unreacted.

Balanced Chemical Equation

A balanced chemical equation is essential in demonstrating the stoichiometry of a reaction. It precisely shows the conservation of mass, meaning the same number of each type of atom appears on both sides of the equation.

When we write the balanced equation for ammonia synthesis, we start with: \( \mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightarrow 2 \mathrm{NH}_3(g) \). This equation tells us that one molecule of nitrogen needs three molecules of hydrogen to produce two molecules of ammonia. Each element is accounted for on both reactant and product sides, ensuring they are equal.

Understanding how to balance equations is critical. It helps us predict the amounts of products that will form and prevents costly errors in industrial applications, such as ammonia production.

• The equation must balance in terms of atoms for each element.
• We use coefficients to ensure equal amounts of elements on both sides.

This understanding lays the groundwork for applying stoichiometric calculations correctly.

Ammonia Synthesis

Ammonia synthesis is an industrial reaction crucial for producing fertilizers and various chemicals. It involves the combination of nitrogen and hydrogen gases to form ammonia, \( \mathrm{NH}_3 \). This process typically occurs under high pressure and temperature to increase reaction efficiency.

The key factors in ammonia synthesis include:

• Reactants: Nitrogen and hydrogen gases.
• Balanced Equation: \( \mathrm{N}_2 + 3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3 \).
• Conditions: High pressure, high temperature, and the use of a catalyst.

In the exercise, we evaluate the amount of reactants that transform into ammonia before the reaction ceases. Understanding ammonia synthesis helps illustrate how chemical equilibrium limits production and requires adjustments in conditions to optimize yields.

This synthesis is not only pivotal for enhancing agricultural productivity but also portrays the importance of balancing economic and environmental considerations in chemical production.