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The feed to an ammonia synthesis reactor contains 25 mole \(\%\) nitrogen and the balance hydrogen. The flow rate of the stream is \(3000 \mathrm{kg} / \mathrm{h}\). Calculate the rate of flow of nitrogen into the reactor in \(\mathrm{kg} / \mathrm{h}\). (Suggestion: First calculate the average molecular weight of the mixture.)

Short Answer

Expert verified
The rate of flow of nitrogen into the reactor is found to be \( 3000 \, \mathrm{kg/h} \)

Step by step solution

01

Determine the Average Molecular weight

The average molecular weight of the mixture can be obtained by summing up the products of the mole fraction and the molecular weight of each component. The mole fraction of Nitrogen (N2) and Hydrogen (H2) are 25 and 75 \(\%\) respectively. Molecular weights for Nitrogen and Hydrogen are \(28 \, \mathrm{g/mol}\) and \(2 \, \mathrm{g/mol}\) respectively. \( M_{avg} = 0.25*28 + 0.75*2 = 7 \, \mathrm{g/mol} \)
02

Convert Total Feed Rate from Mass to Moles

The total feed rate in mass (3000 \, \mathrm{kg/h}\) is converted into moles using the average molecular weight. \(n_{total} = \frac{3000000 \, \mathrm{g/h}}{7 \, \mathrm{g/mol}} \approx 428571.43 \, \mathrm{mol/h}\)
03

Calculate the Nitrogen Flow Rate

Finally, the mole flow rate of Nitrogen can be determined by multiplying the total feed rate in moles by the mole fraction of Nitrogen. Later, this value is converted to mass per hour by multiplying with the molecular weight of nitrogen. \(n_{N2} = 0.25 * 428571.43 \, \mathrm{mol/h} = 107142.86 \, \mathrm{mol/h} \). Converting moles to mass gives, \( m_{\mathrm{N2}} = 107142.86 \, \mathrm{mol/h} * 28 \, \mathrm{g/mol} \approx 3000000 \, \mathrm{g/h} = 3000 \, \mathrm{kg/h}\)
04

Provide the Final Answer

The nitrogen flow rate into the reactor is 3000 kg/h

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Key Concepts

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

Mole Fraction Explained
Imagine you have a mixture of two or more substances. The mole fraction is a way of expressing the concentration of a particular component in that mixture. It's simply the ratio of the number of moles of the component of interest to the total number of moles of all components in the mixture.

The formula for mole fraction (\(X\textsubscript{i}\)) is:
\[\begin{equation}X_i=\frac{n_i}{n_{total}}\end{equation}\]
where \begin{itemize}\item \(n_i\) is the number of moles of the component, and\item \(n_{total}\) is the total number of moles of all components in the mixture.\end{itemize}
For instance, if you have a container filled with nitrogen and hydrogen gases, and 25 moles out of a total of 100 moles are nitrogen, the mole fraction of nitrogen would be 0.25 or 25%. It's a simple but fundamental concept in chemical engineering calculations for understanding the composition of mixtures.
Average Molecular Weight
To comprehend the average molecular weight, consider it as the 'weighted average' of the molecular weights of all components in a mixture. Each component's molecular weight is multiplied by its respective mole fraction. By adding these products together, we get the average molecular weight of the entire mixture.

The equation looks like this:
\[\begin{equation}M_{avg}=\sum(X_i \times M_i)\end{equation}\]
where
  • \(X_i\) is the mole fraction,
  • \(M_i\) is the molecular weight of each component,
  • the sum is over all the components in the mixture.

This concept is particularly useful when dealing with mixtures, as it allows for the conversion between mass and moles of a mixture rather than having to do so for each individual component.
Molar Mass Conversion
One of the most frequent conversions in chemical engineering is from mass to moles or vice versa. This is where molar mass conversion comes into play. The molar mass (or molecular weight) is the mass of one mole of a substance.

The conversion formula is:
\[\begin{equation}n=\frac{m}{M}\end{equation}\]
where
  • \(n\) is the number of moles,
  • \(m\) is the mass of the substance (in grams or kilograms),
  • \(M\) is the molar mass of the substance (in grams per mole or kilograms per mole).

This equation comes in handy when there's a need to convert the given mass of a feed stream, like in the ammonia synthesis example, into moles before proceeding with composition-based calculations.
Ammonia Synthesis Process
Ammonia synthesis refers to the industrial process used to produce ammonia (\(NH_3\)) from nitrogen (\(N_2\)) and hydrogen (\(H_2\)) gases. The heart of this production is the Haber-Bosch process, which combines nitrogen and hydrogen under high temperatures and pressures in the presence of a catalyst. The reaction can be simplified as:
\[\begin{equation}N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)\end{equation}\]
Understanding the mole fraction and average molecular weight, as discussed previously, is crucial in this process. They determine the right proportions of nitrogen to hydrogen and help in calculating the quantities involved in the reaction. Chemical engineers utilize these calculations to ensure the optimum synthesis of ammonia, which is a key ingredient for fertilizers and essential for agriculture.

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Most popular questions from this chapter

A mixture of methanol (methyl alcohol) and water contains \(60.0 \%\) water by mass. (a) Assuming volume additivity of the components, estimate the specific gravity of the mixture at \(20^{\circ} \mathrm{C} .\) What volume (in liters) of this mixture is required to provide 150 mol of methanol? (b) Repeat Part (a) with the additional information that the specific gravity of the mixture at \(20^{\circ} \mathrm{C}\) is 0.9345 (making it unnecessary to assume volume additivity). What percentage error results from the volume- additivity assumption?

In April \(2010,\) the worst oil spill ever recorded occurred when an explosion and fire on the Deepwater Horizon offshore oil-drilling rig left 11 workers dead and began releasing oil into the Gulf of Mexico. One of the attempts to contain the spill involved pumping drilling mud into the well to balance the pressure of escaping oil against a column of fluid (the mud) having a density significantly higher than those of seawater and oil. In the following problems, you may assume that seawater has a specific gravity of 1.03 and that the subsea wellhead was 5053 ft below the surface of the Gulf. (a) Estimate the gauge pressure (psig) in the Gulf at a depth of \(5053 \mathrm{ft}\). (b) Measurements indicate that the pressure inside the wellhead is 4400 psig. Suppose a pipe between the surface of the Gulf and the wellhead is filled with drilling mud and balances that pressure. Estimate the specific gravity of the drilling mud. (c) The drilling mud is a stable slurry of seawater and barite (SG \(=4.37\) ). What is the mass fraction of barite in the slurry? (d) What would you expect to happen if the barite weight fraction were significantly less than that estimated in Part (c)? Explain your reasoning.

Certain solid substances, known as hydrated compounds, have well-defined molecular ratios of water to some other species. For example, calcium sulfate dihydrate (commonly known as gypsum, \(\left.\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right),\) has 2 moles of water per mole of calcium sulfate; alternatively, it may be said that 1 mole of gypsum consists of 1 mole of calcium sulfate and 2 moles of water. The water in such substances is called water of hydration. (More information about hydrated salts is given in Chapter 6 .) In order to eliminate the discharge of sulfuric acid into the environment, a process has been developed in which the acid is reacted with aragonite \(\left(\mathrm{CaCO}_{3}\right)\) to produce calcium sulfate. The calcium sulfate then comes out of solution in a crystallizer to form a slurry (a suspension of solid particles in a liquid) of solid gypsum particles suspended in an aqueous \(\mathrm{CaSO}_{4}\) solution. The slurry flows from the crystallizer to a filter in which the particles are collected as a filter cake. The filter cake, which is 95.0 wiff solid gypsum and the remainder CaSO_solution, is fed to a dryer in which all water (including the water of hydration in the crystals) is driven off to yield anhydrous (water-free) CaSO \(_{4}\) as product. A flowchart and relevant process data are given below. Solids content of slurry leaving crystallizer: \(0.35 \mathrm{kg} \mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} / \mathrm{L}\) slurry \(\mathrm{CaSO}_{4}\) content of slurry liquid: \(0.209 \mathrm{g} \mathrm{CaSO}_{4} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) Specific gravities: \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), 2.32 ;\) liquid solutions, 1.05 (a) Briefly explain in your own words the functions of the three units (crystallizer, filter, and dryer). (b) Takea basis of one liter of solution leaving the crystallizer and calculate the mass (kg) and volume (L) of solid gypsum, the mass of \(\mathrm{CaSO}_{4}\) in the gypsum, and the mass of \(\mathrm{CaSO}_{4}\) in the liquid solution. (c) Calculate the percentage recovery of \(\mathrm{CaSO}_{4}-\) that is, the percentage of the total \(\mathrm{CaSO}_{4}\) (precipitated plus dissolved) leaving the crystallizer recovered as solid anhydrous \(\mathrm{CaSO}_{4}\) (d) List five potential negative consequences of discharging \(\mathrm{H}_{2} \mathrm{SO}_{4}\) into the river passing the plant.

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