Problem 42 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Forty per cent of a mixture of \(0.2 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) and \(0.6 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) react to give \(\mathrm{NH}_{3}\) according to the equation: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at constant temperature and pressure. Then, the ratio of the final volume to the initial volumes of gases is (a) \(4: 5\) (b) \(5: 4\) (c) \(7: 10\) (d) \(8: 5\)

Step-by-Step Solution

Answer

The ratio of the final volume to the initial volumes of gases is 4:5 which corresponds to option (a).

Step 1: Determine the Mole Ratio from the Balanced Equation

First, analyze the balanced chemical equation def alongside information on the proportion that reacted. According to the balanced chemical equation, 1 mole of def reacts with 3 moles of def to produce 2 moles of def . The initial moles of def and def are 0.2 and 0.6, respectively.

Step 2: Calculate the Amount of moles that Reacted

Forty percent of the initial moles react, so 40% of 0.2 moles of def is 0.08 moles def , and 40% of 0.6 moles of def is 0.24 moles def . However, def is limiting reactant; thus, 0.08 moles of def pand 0.24 moles of def will actually react.

Step 3: Determine the Change in Moles after the Reaction

Subtract the moles that reacted from the initial moles to find the amount remaining. def reduces by 0.08 moles to 0.12 moles, and def reduces by 0.24 moles to 0.36 moles. According to the balanced equation, 0.08 moles def will produce 0.16 moles of def .

Step 4: Calculate the Total Moles of Gas Before and After the Reaction

The total initial moles are the sum of def and def moles, which is 0.2 + 0.6 = 0.8 moles. The final total moles are the sum of remaining def pand def moles and new def moles, which is 0.12 + 0.36 + 0.16 = 0.64 moles.

Step 5: Determine the Ratio of the Final Volume to the Initial Volume

Assuming constant temperature and pressure, the volume is directly proportional to the number of moles. Therefore, the final volume to initial volume ratio equals the final moles to initial moles ratio, which is 0.64:0.8. On simplifying this ratio by dividing both numbers by 0.16, we get 4:5.

Key Concepts

Stoichiometry

Stoichiometry is essentially the math behind chemistry. It is the study of the quantitative relationships or ratios between reactants and products in chemical reactions. Stoichiometry hinges on the law of conservation of mass, where the total mass of the reactants equals the total mass of the products.
In our example, stoichiometry is used to predict the outcome of the reaction when nitrogen (N₂) and hydrogen (H₂) combine to form ammonia (NH₃). The balanced equation provides the mole ratio required for the substances to react completely; in this case, one mole of N₂ reacts with three moles of H₂ to produce two moles of NH₃. To apply stoichiometry, we start by looking at the moles provided in the question and the percentage that reacted. From there, the solution is a matter of using these ratios to calculate the amounts of reactants consumed and products formed, retaining a balanced state throughout the process.
When solving stoichiometry problems, it's important to:

• Begin with a balanced chemical equation.
• Convert quantities to moles using the mole concept.
• Use mole ratios to find the relation between reactants and products.
• Convert moles back to desired units, if necessary (like volume or mass).

The correct stoichiometric calculations are critical in making sure that chemical reactions are understood quantitatively and can be used effectively in applications like chemical manufacturing and predicting reaction yields.

Chemical Reaction Balancing

Balancing a chemical reaction is a fundamental step in understanding stoichiometry. It refers to the process of ensuring that the number of atoms for each element is the same on both the reactant and product sides of a chemical equation. This is essential because the Law of Conservation of Mass dictates that no atoms are lost or gained in a chemical reaction; they are simply rearranged.
In the given problem, the reaction between N₂ and H₂ to form NH₃ is already balanced for us: N₂(g) + 3H₂(g) 2NH₃(g). It's important to note that balancing a chemical equation requires one to adjust only the coefficients (the numbers in front of molecules), not the subscripts within chemical formulas, as altering subscripts changes the substances involved. Here's a simple outline for balancing reactions:

• Write the unbalanced equation with all reactants and products.
• Count the number of atoms of each element on both sides of the equation.
• Add coefficients in front of compounds to balance the atoms, starting with the element that appears in the least number of compounds.
• Proceed systematically until all elements are balanced.
• Check your work by ensuring all atoms balance and that the smallest whole number coefficients are used.

Remember, a balanced chemical equation is crucial for correctly applying stoichiometry and for understanding the mole concept.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe. One mole is defined as the amount of a substance that contains the same number of entities (atoms, molecules, ions, etc.) as there are atoms in 12 grams of carbon-12. This number, known as Avogadro's number, is approximately 6.022 x 10²³.
Using the mole concept allows chemists to count particles by weighing them. In our chemical equilibrium problem, we use moles to quantify amounts of nitrogen, hydrogen, and ammonia. Knowing the initial moles of nitrogen (0.2 mol) and hydrogen (0.6 mol), and understanding that 40% of these moles react according to the stoichiometric ratios provided by the balanced equation, leads us to the number of moles that take part in the reaction and the number of moles of products formed.
Here are some key points when working with the mole concept:

• The mole allows us to relate masses of different chemical substances involved in a reaction.
• Molar masses, taken from the periodic table, help convert grams to moles and vice versa.
• In chemical equations, coefficients represent mole ratios which are used in stoichiometric calculations.
• Understanding moles is essential for calculating concentrations, determining reaction yields, and converting between mass and gaseous volume (using the ideal gas law).

By mastering the mole concept, students can tackle a wide range of chemical calculations and develop a deeper understanding of the quantitative aspects of chemistry.