Reversible Reactions And Equilibrium
Overview
Most reactions seem to run one way and stop, but a surprising number can go forwards and backwards. Heat blue hydrated copper(II) sulfate and it turns white; add water and the blue returns. That single demonstration is the gateway to one of chemistry's most powerful ideas: dynamic equilibrium, and the industrial processes that depend on it.
In this lesson you will meet reversible reactions and the special arrow that marks them, learn what it means for a closed system to reach equilibrium, and see how changing temperature, pressure and concentration shift the balance. You will then apply all of this to the Haber process for ammonia and the Contact process for sulfuric acid.
Objectives
- State that some chemical reactions are reversible as shown by the symbol ⇌.
- Describe how changing the conditions can change the direction of a reversible reaction for: (a) the effect of heat on hydrated compounds (b) the addition of water to anhydrous compounds limited to copper(II) sulfate and cobalt(II) chloride.
- [Supplement] State that a reversible reaction in a closed system is at equilibrium when: (a) the rate of the forward reaction is equal to the rate of the reverse reaction (b) the concentrations of reactants and products are no longer changing.
- [Supplement] Predict and explain, for a reversible reaction, how the position of equilibrium is affected by: (a) changing temperature (b) changing pressure (c) changing concentration (d) using a catalyst using information provided.
- [Supplement] State the symbol equation for the production of ammonia in the Haber process, N (g) + 3H (g) 2NH (g) 2 2 3.
- [Supplement] State the sources of the hydrogen (methane) ⇌ and nitrogen (air) in the Haber process.
- [Supplement] State the typical conditions in the Haber process as 450 °C, 20 000 kPa / 200 atm and an iron catalyst.
- [Supplement] State the symbol equation for the conversion of sulfur dioxide to sulfur trioxide in the Contact process, 2SO (g) + O (g) 2SO (g) 2 2 3.
- [Supplement] State the sources of the sulfur dioxide (burning ⇌ sulfur or roasting sulfide ores) and oxygen (air) in the Contact process.
- [Supplement] State the typical conditions for the conversion of sulfur dioxide to sulfur trioxide in the Contact process as 450 °C, 200 kPa / 2 atm and a vanadium(V) oxide catalyst.
- [Supplement] Explain, in terms of rate of reaction and position of equilibrium, why the typical conditions stated are used in the Haber process and in the Contact process, including safety considerations and economics.
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Lesson Evaluation
Congratulations on completing the lesson on Reversible Reactions And Equilibrium. Now that youve explored the key concepts and ideas, its time to put your knowledge to the test. This section offers a variety of practice questions designed to reinforce your understanding and help you gauge your grasp of the material.
You will encounter a mix of question types, including multiple-choice questions, short answer questions, and essay questions. Each question is thoughtfully crafted to assess different aspects of your knowledge and critical thinking skills.
Use this evaluation section as an opportunity to reinforce your understanding of the topic and to identify any areas where you may need additional study. Don't be discouraged by any challenges you encounter; instead, view them as opportunities for growth and improvement.
- Which symbol shows that a reaction is reversible? A. -> B. = C. the double arrow (reversible) symbol D. + Answer: C
- Hydrated copper(II) sulfate is: A. white B. blue C. pink D. green Answer: B
- At dynamic equilibrium in a closed system: A. the reaction has stopped B. only the forward reaction occurs C. forward and reverse rates are equal D. concentrations keep rising Answer: C
- The typical conditions for the Haber process are: A. 450 C, 200 kPa, vanadium(V) oxide B. 450 C, 20 000 kPa, iron C. 25 C, 100 kPa, platinum D. 1000 C, 20 000 kPa, nickel Answer: B
- Adding a catalyst to a reversible reaction at equilibrium: A. increases the yield of product B. shifts equilibrium to the right C. does not change the position of equilibrium D. stops the reverse reaction Answer: C
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