# Equilibrium of ammonia synthesis

• MexChemE
In summary, the authors attempted to solve the problem of equilibration between a pressure range of 150 to 300 atm and a temperature range of 700 to 750 K by using a method that involved the inverse of the reaction. They found that the mole fraction of N2 was 0.161.
MexChemE

## Homework Statement

In order to obtain commercial conversion levels in the production of ammonia, reactors operate between a pressure range from 150 to 300 atm, and a temperature range from 700 to 750 K. Calculate the mole fraction of N2 when the reaction reaches equilibrium at 300 atm and 723 K. The reactor's feedstock initially contained 25% in mole of N2 and 75% of H2. Under these conditions we have the following parameters:
Ka = 0.0066
γNH3 = 0.91
γH2 = 1.09
γN2 = 1.14

## Homework Equations

$$\frac{1}{2} \textrm{N}_2 + \frac{3}{2} \textrm{H}_2 \rightarrow \textrm{NH}_3$$
$$K_a = K_p K_{\gamma}$$

## The Attempt at a Solution

First, I calculate the dummy constant Kγ:
$$K_{\gamma} = \frac{\gamma_{NH_3}}{\gamma_{N_2}^{0.5} \gamma_{H_2}^{1.5}} = 0.7489$$
Now I can get Kp which is Ka/Kγ; 0.0088129. Now, I figured out it was easier working with the reversed reaction:
$$\textrm{NH}_3 \rightarrow \frac{1}{2} \textrm{N}_2 + \frac{3}{2} \textrm{H}_2$$
The reversed reaction has Kp' = 113.47. Now, I applied the following method. Suppose we have the following initial conditions:
NH3 - n moles
N2 - 0 moles
H2 - 0 moles
When the reaction reaches equilibrium we will have:
NH3 - n (1 + α) moles
N2 - 0.5n α moles
H2 - 1.5n α moles
The total moles found in equilibrium are given by:
$$n_T = n - n\alpha + 0.5n\alpha + 1.5n\alpha = n(1+\alpha)$$
Now, applying Pi = niP, the partial pressures of the components in equilibrium will be given by the following expressions:
$$P_{NH_3} = \left( \frac{1-\alpha}{1+\alpha} \right) P$$
$$P_{N_2} = \left( \frac{0.5\alpha}{1+\alpha} \right) P$$
$$P_{H_2} = \left( \frac{1.5\alpha}{1+\alpha} \right) P$$
Kp' will be given by:
$$K_p' = \frac{\left( \frac{0.5\alpha}{1+\alpha} \right)^{0.5} P^{0.5} \left( \frac{1.5\alpha}{1+\alpha} \right)^{1.5} P^{1.5}}{\left( \frac{1-\alpha}{1+\alpha} \right) P}$$. Solving for α results in the following quadratic equation:
$$(K_p' + 1.299P)\alpha^2 - K_p' = 0$$
Plugging in Kp' = 113.47 and P = 300 atm and solving the equation results in α = 0.475. Now we can calculate the partial pressure of each of the components in equilibrium, which are 106.78 atm for ammonia, 48.31 atm for nitrogen and 144.91 atm for hydrogen. Finally, we calculate the mole fraction of N2:
$$Y_{N_2} = \frac{48.31 \ atm}{300 \ atm} = 0.161$$
This result is apparently correct, although the procedure is a bit laborious; is there another method for solving this problem without involving the reversed reaction and α?

Thanks in advance for any input!

Last edited:
Sure. Start out with 0.25 moles of N2 and 0.75 moles of H2, and let them react until x moles of NH3 are formed. So, in the final equilibrium,

moles of N2 = 0.25 - 0.5x

moles of H2 = 0.75 - 1.5x

total moles = 1 - x

Then get the mole fractions and partial pressures.

Chet

MexChemE
Thanks, Chet! I got the same result using that method, and it sure is a bit shorter. I think I'll be working with this method for synthesis reactions and the other method for dissociation reactions. Is there a set of restrictions for using each method? Like knowing the initial proportion of reactants? In this case 25% nitrogen and 75% hydrogen.

MexChemE said:
Thanks, Chet! I got the same result using that method, and it sure is a bit shorter. I think I'll be working with this method for synthesis reactions and the other method for dissociation reactions. Is there a set of restrictions for using each method? Like knowing the initial proportion of reactants? In this case 25% nitrogen and 75% hydrogen.
Yes, there's a restriction. You can't do it by the "reverse" method that you used unless you know in advance that the reactants were in stochiometric proportions.

Chet

MexChemE

## 1. What is the chemical equation for the equilibrium of ammonia synthesis?

The chemical equation for the equilibrium of ammonia synthesis is N2 + 3H2 ⇌ 2NH3.

## 2. What is the role of temperature in the equilibrium of ammonia synthesis?

Temperature affects the equilibrium of ammonia synthesis by shifting the balance between the forward and reverse reactions. Increasing the temperature favors the endothermic forward reaction, while decreasing the temperature favors the exothermic reverse reaction.

## 3. How does pressure impact the equilibrium of ammonia synthesis?

Pressure has a minimal effect on the equilibrium of ammonia synthesis. This is because the forward and reverse reactions involve the same number of moles and do not change the overall volume of the system.

## 4. What is the importance of catalysts in the equilibrium of ammonia synthesis?

Catalysts are important in the equilibrium of ammonia synthesis as they lower the activation energy required for the reaction to occur. This allows for a faster rate of conversion between reactants and products, ultimately increasing the yield of ammonia.

## 5. How is Le Chatelier's principle applied to the equilibrium of ammonia synthesis?

Le Chatelier's principle states that when a system at equilibrium is subjected to a change, it will respond in a way that minimizes the effect of that change. In the equilibrium of ammonia synthesis, this means that if the concentration of reactants or products is altered, the system will shift in the direction that minimizes the change in concentration.

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