Question

A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(\mathrm{~g})\). 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. How many moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

Short answer

The initial moles of N₂ and H₂ were \( 4.5 \: \mathrm{mol} \) and \( 7.5 \: \mathrm{mol} \), respectively.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for this reaction is: \( \mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g) \rightleftharpoons 2\mathrm{NH}_{3}(g) \)

Find the change in moles of each substance

Since nitrogen and hydrogen are reacting to form ammonia, we can find the change in moles of each substance at equilibrium using the balanced chemical equation and the mole ratios. Let \( \Delta n_{\mathrm{N}_{2}} \) be the change in moles of nitrogen, \( \Delta n_{\mathrm{H}_{2}} \) be the change in moles of hydrogen, and \( \Delta n_{\mathrm{NH}_{3}} \) be the change in moles of ammonia. From the balanced chemical equation, we see that the mole ratio of N₂:H₂:NH₃ is 1:3:2. In other words, for every mole of nitrogen consumed as the reactants, 3 moles of hydrogen are consumed, and 2 moles of ammonia are produced. Now, we are given that at equilibrium, there are 3.0 moles of NH₃ and 3.0 moles of each of N₂ and H₂. Since 2 moles of ammonia are produced by 1 mole of nitrogen, we can express the change in moles of nitrogen as: \( \Delta n_{\mathrm{N}_{2}} = -\cfrac{1}{2}\Delta n_{\mathrm{NH}_{3}} \) Similarly, for hydrogen, we can express the change in moles as: \( \Delta n_{\mathrm{H}_{2}} = -3\Delta n_{\mathrm{N}_{2}} \)

Calculate the initial moles of N₂ and H₂

Now that we have the expressions for the changes in moles, we can calculate the initial moles of N₂ and H₂. Before the reaction goes to equilibrium, the change in moles of ammonia is the amount of ammonia present at equilibrium, which is 3.0 moles: \( \Delta n_{\mathrm{NH}_{3}} = 3.0 \: \mathrm{mol} \) Using the first equation we obtained in Step 2: \( \Delta n_{\mathrm{N}_{2}} = -\cfrac{1}{2} \times 3.0 \: \mathrm{mol} = -1.5 \: \mathrm{mol} \) We know that at equilibrium, there are 3.0 moles of N₂ left. So the initial moles of N₂ can be calculated using the following equation: \( n_{\mathrm{N}_{2}}^{\mathrm{initial}} = n_{\mathrm{N}_{2}}^{\mathrm{equilibrium}} - \Delta n_{\mathrm{N}_{2}} \) \( n_{\mathrm{N}_{2}}^{\mathrm{initial}} = 3.0 \: \mathrm{mol} - (-1.5 \: \mathrm{mol}) = 4.5 \: \mathrm{mol} \) Now, using the second equation we obtained in Step 2: \( \Delta n_{\mathrm{H}_{2}} = -3\Delta n_{\mathrm{N}_{2}} = -3 (-1.5 \: \mathrm{mol}) = 4.5 \: \mathrm{mol} \) We know that at equilibrium, there are 3.0 moles of H₂ left. So the initial moles of H₂ can be calculated using the following equation: \( n_{\mathrm{H}_{2}}^{\mathrm{initial}} = n_{\mathrm{H}_{2}}^{\mathrm{equilibrium}} - \Delta n_{\mathrm{H}_{2}} \) \( n_{\mathrm{H}_{2}}^{\mathrm{initial}} = 3.0 \: \mathrm{mol} - 4.5 \: \mathrm{mol} = 7.5 \: \mathrm{mol} \) The initial moles of N₂ and H₂ were \( 4.5 \: \mathrm{mol} \) and \( 7.5 \: \mathrm{mol} \), respectively.

Key concepts

The Mole Concept

Understanding the mole concept is essential for grasping the base of all stoichiometric calculations in chemistry. A mole represents Avogadro's number, which is approximately 6.022 x 10²³ entities, be it atoms, molecules, or ions. Essentially, the mole serves as a bridge between the microscopic world of atoms and the macroscopic world we observe.

For example, when we state that a substance has a molar mass of 1 gram per mole, this means that one mole of this substance weighs 1 gram. When dealing with chemical reactions, the mole concept allows us to count particles by weighing them, thus simplifying the process of measuring out exact amounts of substances for reactions.

Applying the Mole Concept

To solve the textbook problem, one must first translate the given masses or volumes of reactants and products into moles using their molar masses or molar volumes respectively. This step is essential before using the balanced chemical equation to understand the stoichiometry of the reaction.

Balanced Chemical Equations

A balanced chemical equation provides the 'recipe' for a chemical reaction, indicating the relative amounts of reactants and products. The law of conservation of mass dictates that matter cannot be created or destroyed in a chemical reaction. Hence, a balanced equation ensures that the number of atoms for each element is the same on both sides of the equation.

Returning to our example, the balanced equation given was:
N2(g) + 3H2(g) ⇌ 2NH3(g)
This indicates that one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia gas. When solving problems, it's important to always start with a balanced chemical equation as it sets the stage for any subsequent stoichiometric calculations.

Reaction Stoichiometry

Reaction stoichiometry involves using the coefficients of a balanced equation to determine the proportions of reactants and products involved in a chemical reaction. It tells us how many moles of one substance are needed to react with another, and how many moles of a product are formed as a result.

In the exercise problem, reaction stoichiometry allowed us to determine the initial moles of nitrogen and hydrogen gas before the reaction reached equilibrium. By understanding the mole ratio from the balanced equation, we could relate the amount of ammonia produced back to the amounts of the original reactants consumed.

Calculations in Stoichiometry

Stoichiometric calculations typically involve setting up ratios using the coefficients from the chemical equation. This step is demonstrated in the solution through the given moles of ammonia produced, relating it to nitrogen and hydrogen through the balanced equation. Knowing this relationship, you can calculate initial amounts of reactants and strategize how much of each substance will be needed for a given chemical reaction to occur.