Gr.11 Chem PPT L.14.1 PDF

Chapter 14: Chemical Equilibrium Book Page 630 14.1-The concept of equilibrium and the equilibrium constant Objectives 1. Discuss how and why a chemical system comes to equilibrium 2. Identify the characteristics of this equilibrium 3. Calculate the concentrations of reactants and p...

Chapter 14: Chemical Equilibrium Book Page 630 14.1-The concept of equilibrium and the equilibrium constant Objectives 1. Discuss how and why a chemical system comes to equilibrium 2. Identify the characteristics of this equilibrium 3. Calculate the concentrations of reactants and products at equilibrium 3 Video 4 Do all reactions go to completion? Answer on Poll Everywhere https://PollEv.com/multiple_choice_polls/LdptBC4fMLBvjGZxZVixG/respond 5 Reactions proceeding to completion - In Stoichiometric calculations we assume that the reactions proceed to completion that is one of the reactants(LR) runs out. - In the complete combustion of methane the transformation of the entire amount of reactants into a stable product means that the reaction proceed to completion 6 Other reactions stop far short of completion Reaction stops before completion. The system reaches Chemical Equilibrium Example: Nitrogen dioxide NO2 (dark brown color) in a sealed glass vessel at 25°C, changes to dinitrogen tetroxide N2O4 (colorless) Expectation: What are your expectations “Cold Calling” 7 Other reactions stop far short of completion Reaction stops before completion. The system reaches Chemical Equilibrium Example: Nitrogen dioxide NO2 (dark brown color) in a sealed glass vessel at 25°C, changes to dinitrogen tetroxide N2O4 (colorless) Expectation: A complete reaction means the final solution is supposed to be colorless Observation: The intensity of brown color initially present decreases however the content of vessel do not become colorless. Instead the intensity of brown color(light-brown) becomes constant: this indicates that reaction stopped before completion and the system reached Chemical Equilibrium 8 What is chemical equilibrium? - George Zaidan and Charles Morton Reaction: 2NO2 (g) ⇆ N2O4 (g) Reaction: 2NO2 (g) ⇆ N2O4 (g) (Think-Pair-Share) The backward (reverse) The FORWARD reaction reaction is favored is favored 12 Characteristics of chemical equilibrium 13 14 VIDEO 15 Real life example: the Haber-Bosch Process 16 Show me technique 17 Think-Pair-Share 18 19 Think-Pair-Share 20 21 22 The equilibrium constant  From their observations of many chemical reactions, two Norwegian chemists, proposed in 1864 the law of mass action as a general description of the equilibrium condition. Law of mass action For a reversible reaction at equilibrium and at a constant temperature, a certain ratio of reactants and products concentration has a constant value, K (the equilibrium constant) As long as a given reaction is at equilibrium and temperature does not change, according to law of mass action, the value of K remains constant 25  Guldberg and Waage postulated that for a reaction of the type The law of mass action is represented by the following equilibrium expression: K is a constant called the equilibrium constant (given without units). The square brackets indicate the concentrations of the chemical species at equilibrium A, B, C, and D represent chemical species j, y, l, and m are their coefficients in the balanced equation If the equilibrium concentrations of the reaction components are known, the value of the equilibrium constant at a given temperature can be calculated The magnitude of equilibrium constant tells us whether an equilibrium reaction favors the products or reactants. If K is much greater than 1 → equilibrium will lie to the right and favors the products If K is much smaller than 1→ equilibrium will lie to the left and favors the reactants In this contexts, any number greater than 10 is considered to be much greater than 1 and any number less than 0.1 is much less than 1 27 28 Writing Equilibrium Expressions Write the equilibrium expression for the following reaction: Consider again the ammonia synthesis reaction: N2 (g) + 3H2 (g) → 2NH3(g) At 500 C, K = 6x10-2. The Table below gives three sets of data for the synthesis of ammonia. What can you infer? even though the individual sets of equilibrium concentrations are quite different for the different situations, the equilibrium constant remains the same (within Each set of equilibrium experimental error). concentrations is called an equilibrium position.  Equilibrium Positions The following results were collected for two experiments involving the reaction at 600 oC between gaseous sulfur dioxide and oxygen to form gaseous sulfur trioxide: Show that the equilibrium constant is the same in both cases.  Equilibrium Positions The following results were collected for two experiments involving the reaction at 600 oC between gaseous sulfur dioxide and oxygen to form gaseous sulfur trioxide: Show that the equilibrium constant is the same in both cases.  for a particular system at a particular temperature There is only one equilibrium constant but there are an infinite number of equilibrium positions. The equilibrium depends on the ratio of the remains the same (within constant concentrations experimental error) The equilibrium the equilibrium depends on the initial position adopted concentrations will not concentrations by a system always be the same. Calculating the Values of K Group work N2 (g) + 3H2 (g) ⇄ 2NH3 (g)  We can draw some important conclusions For a reaction of the form the equilibrium expression is If this reaction is reversed, then the new equilibrium expression is If the original reaction is multiplied by a factor n to give the equilibrium expression becomes  LET’S REVIEW The equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression raised to the nth power. Knew = (Koriginal)n For a reaction, at a given temperature, there are many equilibrium positions The equilibrium constant K values are customarily written without units. K always has the same value at a given temperature regardless of the amounts of reactants or products present initially. Find the expression of equilibrium constant Note: If pure solids or pure liquids are involved in a chemical reaction, their concentrations are not included in the equilibrium expression for the reaction. This simplification occurs only with pure solids or liquids, not with solutions or gases, since in these last two cases the concentrations can vary. 37 38 39 Individual work 40 Think-Pair-Share 41 Thank you Quizizz Time: Test your knowledge: https://quizizz.com/admin/quiz/65126454e4 6dc1f2b6cf296f?searchLocale= 42