If a mixture of 3 moles of \(\mathrm{H}_{2}\) and one mole of \(\mathrm{N}_{2}\) is completely converted into \(\mathrm{NH}_{3}\). What would be the ratio of the initial and final volume at same temperature and pressure?

Short Answer

Expert verified
The ratio of the initial volume to the final volume is 2:1.

Step by step solution

01

Write the Balanced Chemical Equation

The balanced chemical equation for the synthesis of ammonia (NH3) from hydrogen (H2) and nitrogen (N2) is: \[3\mathrm{H}_{2} + \mathrm{N}_{2} \rightarrow 2\mathrm{NH}_{3}\]This reaction depicts that three molecules of hydrogen gas react with one molecule of nitrogen gas to form two molecules of ammonia.
02

Calculate Initial and Final Moles of Gases

Initially, there are 3 moles of \(\mathrm{H}_{2}\) and 1 mole of \(\mathrm{N}_{2}\), making a total of 4 moles of gas. After the reaction, according to the balanced equation, there will be 2 moles of \(\mathrm{NH}_{3}\) produced.
03

Apply Avogadro's Law

According to Avogadro's law, equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. Since the volume is directly proportional to the number of moles at constant temperature and pressure, the ratio of the initial volume (\(V_i\)) to the final volume (\(V_f\)) can be given by the ratio of the initial moles to the final moles.\[\frac{V_i}{V_f} = \frac{\text{Initial moles}}{\text{Final moles}}\]\[\frac{V_i}{V_f} = \frac{4}{2}\]
04

Calculate the Volume Ratio

Substituting the values from the previous step, we get:\[\frac{V_i}{V_f} = \frac{4 \, \text{moles}}{2 \, \text{moles}} = 2\]The ratio of the initial volume to the final volume is 2:1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equation Balancing
Balancing chemical equations is fundamental in understanding chemical reactions. It is crucial because the Law of Conservation of Mass dictates that mass is neither created nor destroyed in an isolated system. Consequently, we must have the same number of each type of atom on both sides of the equation.

In the synthesis of ammonia, the balanced equation \[3\mathrm{H}_{2} + \mathrm{N}_{2} \rightarrow 2\mathrm{NH}_{3}\] shows a stoichiometrically correct reaction, where the hydrogen and nitrogen atoms are conserved. Balancing involves making sure that the number of atoms for each element is equal on both sides. In this example, we start with six hydrogen atoms and two nitrogen atoms, resulting in two NH3 molecules, which also contain six hydrogen atoms and two nitrogen atoms in total.
Stoichiometry
Stoichiometry lies at the heart of chemical reactions, allowing us to calculate relative quantities of reactants and products. Using the balanced equation and the mole concept, stoichiometry enables us to understand the ratios in which chemicals combine and react.

For instance, the ammonia synthesis reaction requires a specific ratio of hydrogen to nitrogen. Our balanced equation shows that for every one mole of nitrogen, we need three moles of hydrogen to make two moles of ammonia. This 1:3 ratio is essential for stoichiometric calculations and helps identify how much reactant is required and how much product can be yielded from a reaction.
Gas Volume Ratios
Avogadro's Law plays a central role in understanding gas volume ratios in chemical reactions. It tells us that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules. Hence, the volume of gas is proportional to the number of moles of that gas under these conditions.

In applying this law to our exercise, we observe that the initial volume of gas (from 3 moles of H2 and 1 mole of N2) is twice as much as the final volume of gas (2 moles of NH3), given that the temperature and pressure remain constant. This demonstrates that the volumes of gases are directly related to their molar amounts, a fact that can greatly simplify calculations involving gaseous reactants or products in chemical reactions.
Mole Concept
The mole concept is a fundamental bridge between the atomic scale and the macroscopic world. It provides a means to quantify amounts of substances based on the number of particles. One mole of any substance contains Avogadro's number of particles, approximately \(6.022 \times 10^{23}\) entities.

In the context of our ammonia production exercise, we equate moles of substance with Avogadro's concept to find out that the number of moles initially present is the sum of the moles of hydrogen and nitrogen. This enables us to calculate not just the amounts of substances in moles but also to relate those amounts to observable quantities like volume in the case of gases.

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Most popular questions from this chapter

The equilibrium constant of the reaction; \(\mathrm{SO}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{SO}_{2(\mathrm{~g})}+1 / 2 \mathrm{O}_{2(\mathrm{~g})}\); is \(0.20 \mathrm{~mole}^{1 / 2}\) litre \(^{-1 / 2}\) at \(1000 \mathrm{~K}\). Calculate equilibrium constant for $$ 2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}=2 \mathrm{SO}_{3(\mathrm{~g})} $$

State which one is homogeneous or heterogeneous? (a) \(\quad \overline{\text { C }_{\text {Diamond }}} \rightleftharpoons \bar{C}_{\text {graphite }}\) (b) \(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})}=\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\) (c) \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\) (d) \(\quad \mathrm{MgCO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{MgO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}\) (e) \(\mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}=\mathrm{PCl}_{5(\mathrm{~g})}\)

The first order diffraction of \(X\) -rays from a certain set of crystal planes oceurs at an angle of \(11.8^{\circ}\) from the planes. If the planes are \(0.281 \mathrm{~nm}\) apart, what is the wavelength of \(X\) -rays?

Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is \(125 \mathrm{pm}\). (a) What is the length of the side of the unit cell? (b) How many unit cells are there in \(1.00 \mathrm{~cm}^{3}\) of aluminium?

\(\Delta G^{\circ}\) for \(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3}\) is \(-16.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). Find out \(K_{\mathrm{p}}\) for the reaction. Also report \(K_{\mathrm{p}}\) and \(\Delta G^{\circ}\) for: $$ \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \quad \text { at } 25^{\circ} \mathrm{C} $$

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