Chapter 14: Problem 72
The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ at \(450^{\circ} \mathrm{C}\) is \(0.159 .\) Calculate the equilibrium composition when \(1.00 \mathrm{~mol} \mathrm{~N}_{2}\) is mixed with \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) in a \(5.00-\mathrm{L}\) vessel.
Short Answer
Step by step solution
Write the Expression for Equilibrium Constant
Define Initial Concentrations
Establish Changes in Concentration
Substitute into Equilibrium Expression
Solve for x
Calculate Equilibrium Concentrations
Verify with Equilibrium Constant
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Constant
- \( K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \).
The value of \( K_c \) is specific to a particular reaction at a given temperature. In this exercise, \( K_c = 0.159 \) at \( 450^{\circ} C \), indicating that the reaction reaches equilibrium with a specific ratio of ammonia to nitrogen and hydrogen concentrations under these conditions.
Reaction Quotient
- \( Q_c = K_c \), the system is at equilibrium.
- \( Q_c < K_c \), the reaction will proceed forward to produce more products.
- \( Q_c > K_c \), the reaction will proceed in reverse to produce more reactants.
Le Chatelier's Principle
For instance, if additional \( H_2 \) were added to our reaction, Le Chatelier's Principle predicts that the equilibrium would shift right, forming more \( NH_3 \) to reduce the increased concentration of \( H_2 \). This principle is a powerful tool for predicting the behavior of chemical systems in terms of balancing changes in concentration and conditions.
Chemical Concentrations
- A concentration decrease in a reactant typically results in less product being formed.
- Conversely, an increase in product concentration can cause the system to shift left, forming more reactants.
- \([\text{N}_2] = \frac{1.00 \text{ mol}}{5.00 \text{ L}} = 0.20 \text{ M}\)
- \([\text{H}_2] = \frac{3.00 \text{ mol}}{5.00 \text{ L}} = 0.60 \text{ M}\)
- \([\text{NH}_3] = 0 \text{ M}\) initially