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HSC Chemistry
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Module 1: Properties and Structure of Matter1.1 Properties of Matter
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1.2 Atomic Structure and Atomic Mass
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1.3 Periodicity
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1.4 Bonding
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Module 2: Introduction to Quantitative Chemistry2.1 Chemical Reactions and Stoichiometry
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2.2 Mole Concept
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2.3 Concentration and Molarity
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2.4 Gas Laws
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Module 3: Reactive Chemistry3.1 Chemical Reactions
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3.2 Predicting Reactions of Metals
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3.3 Rates of Reactions
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Module 4: Drivers of Reactions4.1 Energy Changes in Chemical Reactions
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4.2 Enthalpy and Hess's Law
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4.3 Entropy and Gibbs Free Energy
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Module 5: Equilibrium and Acid Reactions5.1 Static and Dynamic Equilibrium5 Topics
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5.2 Factors that Affect Equilibrium2 Topics
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5.3 Calculating the Equilibrium Constant2 Topics
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5.4 Solution Equilibria
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Module 6: Acid/Base Reactions6.1 Properties of Acids and Bases7 Topics
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6.2 Using Brønsted–Lowry Theory2 Topics
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6.3 Quantitative Analysis1 Topic
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Module 7: Organic Chemistry7.1 Nomenclature2 Topics
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7.2 Hydrocarbons2 Topics
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7.3 Products of Reactions Involving Hydrocarbons
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7.4 Alcohols1 Topic
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7.5 Reactions of Organic Acids and Bases
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7.6 Polymers2 Topics
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Module 8: Applying Chemical Ideas8.1 Analysis of Inorganic Substances3 Topics
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8.2 Analysis of Organic Substances
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8.3 Chemical Synthesis and Design
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Working ScientificallyWorking Scientifically Overview1 Topic
Lesson 17, Topic 2
In Progress
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) |
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1.42 | 0.71 | 2.14 |
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