Problem 27
Question
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 $$ \frac{3}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g}) $$
Step-by-Step Solution
Verified Answer
The equilibrium constant for the second reaction is 0.4.
1Step 1: 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:\[\frac{3}{2} \text{H}_2(g) + \frac{1}{2} \text{N}_2(g) \rightleftharpoons \text{NH}_3(g)\]
2Step 2: 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.
3Step 3: 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 \(K'_{c}\).
4Step 4: Perform the Calculation
Calculate \[(0.16)^{1/2} = 0.4\]So, the new equilibrium constant \(K'_{c}\) is 0.4.
Key Concepts
Haber-Bosch processChemical EquilibriumStoichiometryAmmonia SynthesisReaction Quotient
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:
Understanding this process helps grasp other key concepts in chemistry, such as equilibrium and reaction kinetics.
The primary reaction is:
- 3H2(g) + N2(g) ↔ 2NH3(g)
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:
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.
In the context of the Haber-Bosch process:
- The equilibrium is represented by the reversible reaction 3H2(g) + N2(g) ↔ 2NH3(g).
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:
Using stoichiometry correctly ensures accurate chemical reactions and calculations.
In our ammonia synthesis problem, stoichiometry helps adjust the reaction:
- Original: 3H2 + N2 ↔ 2NH3
- Modified: 1.5H2 + 0.5N2 ↔ NH3
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:
The synthesis is a perfect example of applied chemical principles like equilibrium and kinetics, showing how theoretical chemistry translates to practical applications.
The synthesis process:
- Involves combining hydrogen (H2) and nitrogen (N2).
- Occurs under high pressure and temperature with a catalyst to speed up the reaction.
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’s calculated like the equilibrium constant (Kc) but at non-equilibrium conditions.
For the reaction:
\[ Q = \frac{[NH_3]^2}{[H_2]^3[N_2]} \]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.
For the reaction:
- 3H2 + N2 ↔ 2NH3
\[ Q = \frac{[NH_3]^2}{[H_2]^3[N_2]} \]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.
Other exercises in this chapter
Problem 25
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View solution Problem 30
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