Hydrogen gas reacts with nitrogen gas to form ammonia what type of chemical reaction is this

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The Haber Process is used in the manufacturing of ammonia from nitrogen and hydrogen, and then goes on to explain the reasons for the conditions used in the process. The process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic.

\[\ce{ N2(g) + 3H2(g) <=> 2NH3 (g)} \label{eq1}\]

with \(ΔH=-92.4 kJ/mol\).

A flow scheme for the Haber Process looks like this:

Figure \(\PageIndex{1}\): Scheme of the Haber Process

General Conditions of the Process

The catalyst: The catalyst is actually slightly more complicated than pure iron. It has potassium hydroxide added to it as a promoter - a substance that increases its efficiency.
The pressure: The pressure varies from one manufacturing plant to another, but is always high. You cannot go far wrong in an exam quoting 200 atmospheres.
Recycling: At each pass of the gases through the reactor, only about 15% of the nitrogen and hydrogen converts to ammonia. (This figure also varies from plant to plant.) By continual recycling of the unreacted nitrogen and hydrogen, the overall conversion is about 98%.

The proportions of nitrogen and hydrogen: The mixture of nitrogen and hydrogen going into the reactor is in the ratio of 1 volume of nitrogen to 3 volumes of hydrogen. Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. That means that the gases are going into the reactor in the ratio of 1 molecule of nitrogen to 3 of hydrogen. That is the proportion demanded by the equation.

In some reactions you might choose to use an excess of one of the reactants. You would do this if it is particularly important to use up as much as possible of the other reactant - if, for example, it was much more expensive. That does not apply in this case. There is always a down-side to using anything other than the equation proportions. If you have an excess of one reactant there will be molecules passing through the reactor which cannot possibly react because there is not anything for them to react with. This wastes reactor space - particularly space on the surface of the catalyst.

Temperature

Equilibrium considerations: You need to shift the position of the equilibrium (Equation \(\ref{eq1}\)) as far as possible to the right in order to produce the maximum possible amount of ammonia in the equilibrium mixture. The forward reaction is exothermic with \(ΔH=-92.4 kJ/mol\). According to Le Chatelier's Principle, this will be favored if you lower the temperature. The system will respond by moving the position of equilibrium to counteract this - in other words by producing more heat. To get as much ammonia as possible in the equilibrium mixture, you need as low a temperature as possible. However, 400 - 450° C is not a low temperature!
Rate considerations: The lower the temperature you use, the slower the reaction becomes. A manufacturer is trying to produce as much ammonia as possible per day. It makes no sense to try to achieve an equilibrium mixture which contains a very high proportion of ammonia if it takes several years for the reaction to reach that equilibrium. You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor.
The compromise: 400 - 450°C is a compromise temperature producing a reasonably high proportion of ammonia in the equilibrium mixture (even if it is only 15%), but in a very short time.

Notice that there are 4 molecules on the left-hand side of Equation \(\ref{eq1}\), but only 2 on the right. According to Le Chatelier's Principle, if you increase the pressure the system will respond by favoring the reaction which produces fewer molecules. That will cause the pressure to fall again. In order to get as much ammonia as possible in the equilibrium mixture, you need as high a pressure as possible. 200 atmospheres is a high pressure, but not amazingly high.

Rate considerations: Increasing the pressure brings the molecules closer together. In this particular instance, it will increase their chances of hitting and sticking to the surface of the catalyst where they can react. The higher the pressure the better in terms of the rate of a gas reaction.
Economic considerations: Very high pressures are very expensive to produce on two counts. You have to build extremely strong pipes and containment vessels to withstand the very high pressure. That increases your capital costs when the plant is built. High pressures cost a lot to produce and maintain. That means that the running costs of your plant are very high.
The compromise: 200 atmospheres is a compromise pressure chosen on economic grounds. If the pressure used is too high, the cost of generating it exceeds the price you can get for the extra ammonia produced.

Equilibrium considerations: The catalyst has no effect whatsoever on the position of the equilibrium. Adding a catalyst doesn't produce any greater percentage of ammonia in the equilibrium mixture. Its only function is to speed up the reaction.
Rate considerations: In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible time. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor.
Separating the ammonia: When the gases leave the reactor they are hot and at a very high pressure. Ammonia is easily liquefied under pressure as long as it is not too hot, and so the temperature of the mixture is lowered enough for the ammonia to turn to a liquid. The nitrogen and hydrogen remain as gases even under these high pressures, and can be recycled.

By mixing one part ammonia to nine parts air with the use of a catalyst, the ammonia will get oxidized to nitric acid.

\[\begin{align*} \ce{4 NH_3} + \ce{5 O_2} &\rightarrow \ce{4 NO} + \ce{6 H_2O} \\[4pt] \ce{2 NO} + \ce{O_2} &\rightarrow \ce{2 NO_2} \\[4pt] \ce{2 NO_2} + \ce{2 H_2O} &\rightarrow \ce{2 HNO_3} + \ce{H_2} \end{align*}\]

This page titled The Haber Process is shared under a not declared license and was authored, remixed, and/or curated by Jim Clark via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

You have learned that chemical equations provide us with information about the types of particles that react to form products. Chemical equations also provide us with the relative number of particles and moles that react to form products. In this section you will explore the quantitative relationships that exist between the quantities of reactants and products in a balanced equation. This is known as stoichiometry.

Stoichiometry, by definition, is the calculation of the quantities of reactants or products in a chemical reaction using the relationships found in the balanced chemical equation. The word stoichiometry is actually Greek from two words: \(\sigma \tau \omicron \iota \kappa \eta \iota \omicron \nu \), which means "element", and \(\mu \epsilon \tau \rho \omicron \nu), which means "measure".

The mole, as you remember, is a quantitative measure that is equivalent to Avogadro's number of particles. So how does this relate to the chemical equation? Look at the chemical equation below.

\[2 \ce{CuSO_4} + 4 \ce{KI} \rightarrow 2 \ce{CuI} + 2 \ce{K_2SO_4} + \ce{I_2} \nonumber \]

The coefficients used, as we have learned, tell us the relative amounts of each substance in the equation. So for every 2 units of copper (II) sulfate (\(\ce{CuSO_4}\)) we have, we need to have 4 units of potassium iodide (\(\ce{KI}\)). For every two dozen copper (II) sulfates, we need 4 dozen potassium iodides. Because the unit "mole" is also a counting unit, we can interpret this equation in terms of moles, as well: For every two moles of copper (II) sulfate, we need 4 moles potassium iodide.

The production of ammonia \(\left( \ce{NH_3} \right)\) from nitrogen and hydrogen gases is an important industrial reaction called the Haber process, after German chemist Fritz Haber.

\[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right) \nonumber \]

The balanced equation can be analyzed in several ways, as shown in the figure below.

Figure \(\PageIndex{1}\): This representation of the production of ammonia from nitrogen and hydrogen show several ways to interpret the quantitative information of a chemical reaction.

We see that 1 molecule of nitrogen reacts with 3 molecules of hydrogen to form 2 molecules of ammonia. This is the smallest possible relative amount of the reactants and products. To consider larger relative amounts, each coefficient can be multiplied by the same number. For example, 10 molecules of nitrogen would react with 30 molecules of hydrogen to produce 20 molecules of ammonia.

The most useful quantity for counting particles is the mole. So if each coefficient is multiplied by a mole, the balanced chemical equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This is the conventional way to interpret any balanced chemical equation.

Finally, if each mole quantity is converted to grams by using the molar mass, we can see that the law of conservation of mass is followed. \(1 \: \ce{mol}\) of nitrogen has a mass of \(28.02 \: \text{g}\), while \(3 \: \text{mol}\) of hydrogen has a mass of \(6.06 \: \text{g}\), and \(2 \: \text{mol}\) of ammonia has a mass of \(34.08 \: \text{g}\).

\[28.02 \: \text{g} \: \ce{N_2} + 6.06 \: \text{g} \: \ce{H_2} \rightarrow 34.08 \: \text{g} \: \ce{NH_3} \nonumber \]

Mass and the number of atoms must be conserved in any chemical reaction. The number of molecules is not necessarily conserved.

The equation for the combustion of ethane (\(\ce{C_2H_6}\)) is

\[2 \ce{C_2H_6} + 7 \ce{O_2} \rightarrow 4 \ce{CO_2} + 6 \ce{H_2O} \nonumber \]

Indicate the number of formula units or molecules in the balanced equation.
Indicate the number of moles present in the balanced equation.

Solution

Two molecules of \(\ce{C_2H_6}\) plus seven molecules of \(\ce{O_2}\) yields four molecules of \(\ce{CO_2}\) plus six molecules of \(\ce{H_2O}\).
Two moles of \(\ce{C_2H_6}\) plus seven moles of \(\ce{O_2}\) yields four moles of \(\ce{CO_2}\) plus six moles of \(\ce{H_2O}\).

For the following equation below, indicate the number of formula units or molecules, and the number of moles present in the balanced equation.

\[\ce{KBrO_3} + 6 \ce{KI} + 6 \ce{HBr} \rightarrow 7 \ce{KBr} + 3 \ce{H_2O} \nonumber \]

Answer

One molecules of \(\ce{KBrO_3}\) plus six molecules of \(\ce{KI}\) plus six molecules of \(\ce{HBr}\) yields seven molecules of \(\ce{KBr}\) plus three molecules of \(\ce{I_2}\) and three molecules of \(\ce{H_2O}\). One mole of \(\ce{KBrO_3}\) plus six moles of \(\ce{KI}\) plus six moles of \(\ce{HBr}\) yields seven moles of \(\ce{KBr}\) plus three moles of \(\ce{I_2}\) plus three moles of \(\ce{H_2O}\).