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HSC Chemistry

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  1. Module 5: Equilibrium and Acid Reactions
    5.1 Static and Dynamic Equilibrium
    5 Topics
  2. 5.2 Factors that Affect Equilibrium
    2 Topics
  3. 5.3 Calculating the Equilibrium Constant
    2 Topics
  4. 5.4 Solution Equilibria
  5. Module 6: Acid/Base Reactions
    6.1 Properties of Acids and Bases
    7 Topics
  6. 6.2 Using Brønsted–Lowry Theory
    2 Topics
  7. 6.3 Quantitative Analysis
    1 Topic
  8. Module 7: Organic Chemistry
    7.1 Nomenclature
    2 Topics
  9. 7.2 Hydrocarbons
    2 Topics
  10. 7.3 Products of Reactions Involving Hydrocarbons
  11. 7.4 Alcohols
    1 Topic
  12. 7.5 Reactions of Organic Acids and Bases
  13. 7.6 Polymers
    2 Topics
  14. Module 8: Applying Chemical Ideas
    8.1 Analysis of Inorganic Substances
    3 Topics
  15. 8.2 Analysis of Organic Substances
  16. 8.3 Chemical Synthesis and Design
  17. Working Scientifically
    Working Scientifically Overview
    1 Topic

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Lesson 3, Topic 2
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Calculating the Equilibrium Constant

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Example

Question: Calculate the equilibrium constant for the decomposition of ammonia, given the following information.

[NH3] (mol L-1)[N2] (mol L-1)[H2] (mol L-1)
1.420.712.14

The decomposition equilibrium for ammonia is:

2\text{NH}_3 (g) ⇌ \text N_2 (g) + 3\text H_2 (g)

Now, the equilibrium constant expression for this reaction may be written as follows, substituting in the substances and molar ratios:

K_{\text{eq}} = \frac{[\text N_2][\text H_2]^3}{[\text{NH}_3]^2}

Finally, the values from the table should be substituted into this expression to calculate Keq:

K_{\text {eq}} \frac{[0.71][2.14]^3}{[1.42]^2} = 3.5