Question

An amount of 3 moles of \(\mathrm{N}_{2}\) and some \(\mathrm{H}_{2}\) is introduced into an evacuated vessel. The reaction starts at \(t=0\) and equilibrium is attained at \(t=t_{1} .\) The amount of ammonia at \(t=2 t_{1}\) is found to be \(34 \mathrm{~g} .\) It is observed that \(\frac{w\left(\mathrm{~N}_{2}\right)}{w\left(\mathrm{H}_{2}\right)}=\frac{14}{3}\) at \(t=\frac{t_{1}}{3}\) and \(t=\frac{t_{1}}{2} .\) The only correct statement is (a) \(w\left(\mathrm{~N}_{2}\right)+w\left(\mathrm{H}_{2}\right)+w\left(\mathrm{NH}_{3}\right)=118 \mathrm{~g}\) at \(t=t_{1}\) (b) \(w\left(\mathrm{~N}_{2}\right)+w\left(\mathrm{H}_{2}\right)+w\left(\mathrm{NH}_{3}\right)=102 \mathrm{~g}\) at \(t=2 t_{1}\) (c) \(w\left(\mathrm{~N}_{2}\right)+w\left(\mathrm{H}_{2}\right)+w\left(\mathrm{NH}_{3}\right)=50 \mathrm{~g}\) at \(t=t_{1} / 3\) (d) \(w\left(\mathrm{~N}_{2}\right)+w\left(\mathrm{H}_{2}\right)+w\left(\mathrm{NH}_{3}\right)\) cannot be predicted

Short answer

The correct statement is (a) since the mass of NH3 at t=2t1 is 34 g, which implies 2 moles of NH3. Hence at equilibrium, t1, there must be 1 mole of NH3 made from 0.5 mole of N2 and 1.5 moles of H2. As the molar ratio N2:H2 is 1:3, the initial moles of H2 must be 4.5. The total weight at t1 is thus w(N2) + w(H2) + w(NH3) = (3-0.5) * 28 + (4.5-1.5) * 2 + 1 * 17 = 70 + 6 + 17 = 93 g. However, the given statements do not include this value, which indicates there is a mismatch in the provided exercise and the options or that additional information is needed to determine the exact masses at given times.

Step-by-step solution

Calculation of mass of NH3

First, calculate the moles of ammonia (NH3) using its mass at time t = 2t1. The molar mass of NH3 is 17 g/mol. Using the given mass of 34 g, we find the number of moles using the formula: moles = mass / molar mass.

Determine initial moles of N2 and H2

We use the stoichiometry of the reaction N2 + 3H2 -> 2NH3. For every 1 mole of N2 reacted, 3 moles of H2 are required, and 2 moles of NH3 are produced. Given 3 moles of N2 at the start and using the ratio of masses at t1/3 and t1/2, find the moles of H2 initially present.

Use the mass ratio at t1/3 and t1/2

According to the given mass ratio of N2 to H2, we have w(N2)/w(H2) = 14/3. This ratio is constant at t1/3 and t1/2, which indicates that the change in mass of N2 and H2 is due to the reaction progressing, but the ratio must remain the same.

Apply the mass ratio to find the mass of H2

From the mass ratio w(N2)/w(H2) = 14/3 and knowing the molar masses of N2 and H2 (28 g/mol and 2 g/mol respectively), we can establish the relationship between the moles of N2 and H2.

Find the correct statement

By using the calculated masses and moles, check each statement (a) to (d) by adding the masses of N2, H2, and NH3. The correct statement will be the one that matches the observed masses at the specified time.

Key concepts

Stoichiometry

Understanding stoichiometry is essential for solving chemical equilibrium problems, as it connects the quantities of reactants and products in a chemical reaction. In the context of our exercise, the stoichiometry of the given reaction, \( \mathrm{N}_{2} + 3\mathrm{H}_{2} \rightarrow 2\mathrm{NH}_{3} \), outlines a precise link between the nitrogen, hydrogen, and ammonia molecules.

Consider this like a recipe, where for every one part of nitrogen (\( \mathrm{N}_{2} \)), you need three parts of hydrogen (\( \mathrm{H}_{2} \)) to cook up two parts of ammonia (\( \mathrm{NH}_{3} \)). When 3 moles of \( \mathrm{N}_{2} \) are used up, naturally, this means we expect 3 times as many moles of \( \mathrm{H}_{2} \) to be consumed if the reaction went to completion.

Therefore, stoichiometry allows us to formulate a direct relationship, not only between the amounts of \( \mathrm{N}_{2} \) and \( \mathrm{H}_{2} \) but also between the reactants and the products at different points in the reaction's progress. Through this, we not only discern the necessary proportions but can also evaluate which statements about the system's mass at different times are accurate.

Molar Mass Calculation

To make sense of the quantities involved in chemical reactions, converting grams to moles and vice versa is vital, which is where the concept of molar mass comes into play. In our exercise, the molar mass of ammonia (\( \mathrm{NH}_{3} \)) is 17 g/mol, which means that each mole of ammonia weighs 17 grams.

Through the expression \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \), we decipher the number of ammonia moles from its given mass. For instance, if we have 34 g of \( \mathrm{NH}_{3} \), we can calculate the number of moles as \( \frac{34 \text{ g}}{17 \text{ g/mol}} \), which gives us 2 moles of ammonia. This calculation is a cornerstone for further analysis in chemical equilibrium problems, especially when combined with the stoichiometry of the reaction to find the correspondence between the reactants' masses at any time during the reaction.

Reaction Progress and Mass Ratio

Monitoring the reaction progress and understanding the mass ratio of reactants are crucial steps in solving equilibrium problems. At different times, such as \( t=\frac{t_{1}}{3} \) and \( t=\frac{t_{1}}{2} \), the problem statement specifies the mass ratio \( \frac{w(\mathrm{N}_{2})}{w(\mathrm{H}_{2})}=\frac{14}{3} \), which remains constant regardless of how far the reaction has progressed.

This implies that for every 14 g of nitrogen consumed, 3 g of hydrogen are used. It is an aspect of the reaction mechanism that, despite the consumption of reactants, their mass ratio does not change. Using the molar masses of \( \mathrm{N}_{2} \) and \( \mathrm{H}_{2} \), we can deduce the relationship between their consumed moles. This invariant mass ratio is fundamental to identifying the correct statement among the options provided in our exercise, helping us to determine the combined mass of all substances at different reaction times.