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Chapter 12: Problem 27

At \(450^{\circ} \mathrm{C}\), the equilibrium constant \(K_{\mathrm{c}}\) for the HaberBosch synthesis of ammonia is 0.16 for the reaction written as $$ 3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ Calculate the value of \(K_{\mathrm{c}}\) for the same reaction written as $$ \frac{3}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g}) $$

Short Answer

Expert verified
The equilibrium constant for the second reaction is 0.4.

Step by step solution

01

Understanding the Problem

The given reaction for the synthesis of ammonia is \[3 \text{H}_2(g) + \text{N}_2(g) \rightleftharpoons 2 \text{NH}_3(g)\]and its equilibrium constant is \(K_c = 0.16\) at \(450^{\circ} \text{C}\). We are asked to find the equilibrium constant \(K'_{c}\) for the same reaction but written differently:\[\frac{3}{2} \text{H}_2(g) + \frac{1}{2} \text{N}_2(g) \rightleftharpoons \text{NH}_3(g)\]
02

Relationship Between Equilibrium Constants

Notice that the second reaction is just the stoichiometry of the first reaction divided by 2. For a reaction scaled by a factor \(n\), the new equilibrium constant \(K'_{c}\) is related to the original \(K_c\) by \[K'_{c} = K_{c}^{1/n}\]where \(n\) is the factor by which the stoichiometry is divided.
03

Calculate the New Equilibrium Constant

In this case, \(n = 2\) as the stoichiometry of all reactants and products is halved. Thus, \[K'_{c} = (K_{c})^{1/2} = (0.16)^{1/2}\]Calculate the square root of \(0.16\) to find \(K'_{c}\).
04

Perform the Calculation

Calculate \[(0.16)^{1/2} = 0.4\]So, the new equilibrium constant \(K'_{c}\) is 0.4.

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Key Concepts

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

Haber-Bosch process
The Haber-Bosch process is an essential method for synthesizing ammonia. It involves the direct combination of nitrogen from the air with hydrogen gas, typically derived from natural gas. This reaction occurs under high pressure and temperature, facilitated by a catalyst.
The primary reaction is:
  • 3H2(g) + N2(g) 鈫� 2NH3(g)
This process is vital for producing fertilizers, supporting global agriculture. Its industrial relevance stems from the balance between reaction conditions and efficiency. The Haber-Bosch process not only revolutionized agriculture but also deeply impacted chemical engineering and industrial practices.
Understanding this process helps grasp other key concepts in chemistry, such as equilibrium and reaction kinetics.
Chemical Equilibrium
In chemistry, equilibrium refers to the state where the rates of the forward and reverse reactions equalize. This means the concentrations of reactants and products remain steady over time.
In the context of the Haber-Bosch process:
  • The equilibrium is represented by the reversible reaction 3H2(g) + N2(g) 鈫� 2NH3(g).
At equilibrium, the concentration of ammonia (NH3) does not change unless external conditions such as pressure or temperature are altered.
Understanding chemical equilibrium involves recognizing how the equilibrium constant (Kc) indicates the position of equilibrium. A higher Kc value suggests more products relative to reactants at equilibrium.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It uses the coefficients in a balanced chemical equation to determine how much reactant is needed or how much product will form.
In our ammonia synthesis problem, stoichiometry helps adjust the reaction:
  • Original: 3H2 + N2 鈫� 2NH3
  • Modified: 1.5H2 + 0.5N2 鈫� NH3
This adjustment means changing the equilibrium constant. If a reaction鈥檚 stoichiometry is divided by a factor, the new equilibrium constant relates to the original by a power of this division factor.
Using stoichiometry correctly ensures accurate chemical reactions and calculations.
Ammonia Synthesis
Ammonia synthesis is the process of creating ammonia from hydrogen and nitrogen molecules. It's a key part of the Haber-Bosch process and fundamental in producing fertilizers.
The synthesis process:
  • Involves combining hydrogen (H2) and nitrogen (N2).
  • Occurs under high pressure and temperature with a catalyst to speed up the reaction.
Ammonia is crucial for multiple applications, primarily in fertilizers, which aid in food production globally.
The synthesis is a perfect example of applied chemical principles like equilibrium and kinetics, showing how theoretical chemistry translates to practical applications.
Reaction Quotient
The reaction quotient (Q) is a measure used to determine the direction a reaction will proceed to reach equilibrium. It鈥檚 calculated like the equilibrium constant (Kc) but at non-equilibrium conditions.
For the reaction:
  • 3H2 + N2 鈫� 2NH3
The reaction quotient is given by:
\[ Q = \frac{[NH_3]^2}{[H_2]^3[N_2]} \]If Q = Kc, the system is at equilibrium. If Q < Kc, the reaction will proceed forward, creating more products. If Q > Kc, the reaction will shift to produce more reactants.
Understanding Q helps predict reaction shifts and manage chemical processes effectively.

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Most popular questions from this chapter

At \(2300 \mathrm{~K}\) the equilibrium constant for the formation of \(\mathrm{NO}(\mathrm{g})\) is \(1.7 \times 10^{-3}\) $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) $$ (a) Analysis of the contents of a sealed flask at \(2300 \mathrm{~K}\) shows that the concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are both \(0.25 \mathrm{M}\) and that of \(\mathrm{NO}\) is \(0.0042 \mathrm{M}\). Determine if the system is at equilibrium. (b) If the system is not at equilibrium, in which direction does the reaction proceed? (c) Calculate all three equilibrium concentrations.

Predict whether the equilibrium for the photosynthesis reaction described by the equation $$ \begin{array}{r} 6 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{~s})+6 \mathrm{O}_{2}(\mathrm{~g}) \\ \Delta_{1} H^{\circ}=2801.69 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ would (i) shift to the right, (ii) shift to the left, or (iii) remain unchanged for each of these changes: (a) decrease the concentration of \(\mathrm{CO}_{2}\) at constant volume. (b) increase the partial pressure of \(\mathrm{O}_{2}\) at constant volume. (c) remove one half of the \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (d) decrease the total pressure by increasing the volume. (e) increase the temperature. (f) introduce a catalyst into a constant-volume system.

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) $$ has the value \(2.64 \times 10^{-3}\) at \(2300 . \mathrm{K}\). If a mixture of \(1.00 \mathrm{~mol} \mathrm{CO}\) and \(1.00 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) is allowed to come to equilibrium in a sealed, \(1.00-\mathrm{L}\) flask at \(2300 . \mathrm{K}\), (a) calculate the final concentrations of all four species: CO, \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2}\) (b) calculate the equilibrium concentrations after an additional 1.00 mol each of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) is added to the flask.

At \(627{ }^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.76\) for the reaction $$ 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) $$ Calculate \(K_{\mathrm{c}}\) at \(627{ }^{\circ} \mathrm{C}\) for (a) synthesis of 1 mol sulfur trioxide gas. (b) decomposition of \(2 \mathrm{~mol} \mathrm{SO}_{3}\)

Write a chemical equation for an equilibrium system that would lead to each expression \((\mathrm{a}-\mathrm{c})\) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2} \mathrm{~S}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{2}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{\left(P_{\mathrm{IF}}\right)}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{Cl}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\)
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