Question

Consider the following reaction equilibrium: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) Initially, 1 mole of \(\mathrm{N}_{2}\) and 3 mole of \(\mathrm{H}_{2}\) are taken in a 2 litre flask. At equilibrium state, if the number of union of \(\mathrm{N}_{2}\) in \(0.6\), what is the total number of moles of all gases present in the flask: (a) \(0.8\) (b) \(1.6\) (c) \(3.2\) (d) \(6.4\)

Short answer

The total number of moles of all gases in the flask is 3.1, not matching any option directly; calculations must confirm total moles per the entire reaction setup.

Step-by-step solution

Identify Initial Moles and Setup Change

Initially, we start with 1 mole of \(\mathrm{N}_2\) and 3 moles of \(\mathrm{H}_2\). No \(\mathrm{NH}_3\) is present initially. Let the change in moles of \(\mathrm{N}_2\) forming \(\mathrm{NH}_3\) be \(x\). From the reaction equation, it's clear that to form \(2x\) moles of \(\mathrm{NH}_3\), \(x\) moles of \(\mathrm{N}_2\) and \(3x/2\) moles of \(\mathrm{H}_2\) are consumed.

Calculate Equilibrium Moles

At equilibrium, if \(0.6\) moles of \(\mathrm{N}_2\) have reacted, the number of moles of \(\mathrm{N}_2\) left is \(1 - 0.6 = 0.4\). For \(\mathrm{H}_2\), the moles left are \(3 - 3 \times 0.3 = 2.1\), since \(3 \times 0.3 = 0.9\) moles of \(\mathrm{H}_2\) have reacted. The number of moles of \(\mathrm{NH}_3\) formed is \(2 \times 0.3 = 0.6\).

Compute Total Moles at Equilibrium

Now, sum the moles of all gases present at equilibrium: \[0.4 \text{ moles of } \mathrm{N}_2 + 2.1 \text{ moles of } \mathrm{H}_2 + 0.6 \text{ moles of } \mathrm{NH}_3 = 3.1 \text{ moles.}\]

Adjust for Volume

Since the reaction occurs in a 2 litre flask, we have calculated the total moles in terms of moles per 2 litres. Hence, the total number of moles per litre is \(3.1/2 = 1.55\). The problem, however, asks for total moles present in the 2-litre flask, so the answer is \(3.1\) moles directly, without needing to adjust per litre for this question.

Key concepts

Equilibrium Constant

In chemical reactions, equilibrium is the point where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of the reactants and products remain constant over time. The equilibrium constant, denoted as \( K \), is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium. For a general reaction, \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant is given by:

• \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

In this formula:

• \([A]\), \([B]\), \([C]\), and \([D]\) are the molar concentrations of the components \(A\), \(B\), \(C\), and \(D\) respectively at equilibrium.
• \(a\), \(b\), \(c\), \(d\) are the stoichiometric coefficients from the balanced chemical equation.

The equilibrium constant helps predict the direction in which the reaction will proceed to reach equilibrium and indicates the extent to which reactants are converted to products.

Reaction Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. It provides a way of calculating the amounts of reactants and products involved in a chemical equation. Understanding reaction stoichiometry is key in solving problems involving equilibrium, like in the exercise above.In the reaction \(\mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), the stoichiometric coefficients are crucial:

• 1 molecule of \(\mathrm{N}_{2}\) combines with 3 molecules of \(\mathrm{H}_{2}\) to produce 2 molecules of \(\mathrm{NH}_{3}\).
• This means for every mole of \(\mathrm{N}_{2}\) reacting, 3 moles of \(\mathrm{H}_{2}\) are required, producing 2 moles of \(\mathrm{NH}_{3}\).

By using stoichiometry, we can determine how much reactant is needed or how much product can be formed. In equilibrium scenarios, stoichiometry helps calculate the changes in concentration or pressure of substances in a mixture. It's a fundamental aspect when predicting how shifts in concentrations affect the equilibrium position.

Mole Concept

The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. It serves as a bridge between the microscopic world of atoms and molecules, and the macroscopic quantities we can measure. One mole contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) particles.In chemical equations, moles are important for understanding the proportions of reactants and products. In the exercise, we began with specific moles of \(\mathrm{N}_2\) and \(\mathrm{H}_2\). Knowing the number of moles is essential for calculating the equilibrium composition.The mole concept allows you to:

• Calculate the moles of reactants and products involved in a reaction.
• Determine the yield of a reaction based on the initial moles of reactants.
• Use stoichiometry to find out how much of each substance will react or be produced.

Using the mole concept helps chemists quantify reactions and analyze equilibrium situations accurately, ensuring the proportions and relations between different elements and compounds are consistent and meaningful.