SLG Chem2 LG 3.1 The Equilibrium Condition PDF

Learning Guide Module Subject Code CHEM 2 Chemistry 2 Module Code 3.0 Chemical Equilibrium Lesson Code 3.1 The Equilibrium Condition Time Frame 30 minutes Components Tasks...

Learning Guide Module Subject Code CHEM 2 Chemistry 2 Module Code 3.0 Chemical Equilibrium Lesson Code 3.1 The Equilibrium Condition Time Frame 30 minutes Components Tasks TA ATA (min)a (min)b Target By the end of this learning guide, the students should be able to: 1 1. describe the condition of equilibrium in a reversible reaction in terms of macroscopic changes in the system and on the molecular level; and 2. define equilibrium as applied to a chemical system. Hook Are you familiar with the game called tug-of-war which others 2 refer to as rope-pulling? This game involves two teams at the end of each rope with the goal of pulling the other team over a specified line (Figure 3.1). Figure 3.1 Two teams pulling at opposite ends of a rope. (Image retrieved from https://dekkak.com/international-tug-of-war-day/) Have you tried playing this game or at least observed other kids playing it? As an observer, have you noticed that at certain times during the game, the rope seems to stop moving? Why would there be no apparent movement of the rope? Ignite In a tug-of-war game, we can see that at times there is no 20 Chemistry 2 The Equilibrium Condition |Page 1 of 6  apparent movement in the rope, although there is a great tension on the rope as it is being pulled because the two teams are pulling it with equal force or we can say that the force exerted by the two opposing teams is balanced. The word balance is often associated with the term equilibrium. We may find these two words as synonyms in most references. To be in equilibrium is to be in a state of balance. In tug-of-war where two teams exert equal forces and that the rope does not move, we can say that the rope is in static equilibrium - one in which the object is at rest. Equilibria (plural) can also be dynamic. Imagine that you are digging a hole. While doing so, your friend is also refilling it. If you dig the hole twice as fast as your friend refills it, the hole gets larger. If your friend fills the hole twice as fast than you dig, the hole gets smaller. But, if both work at the same speed, the size of the hole does not change even if you and your friend keep on digging and refilling it. This scenario is analogous to Chemical Equilibrium - a state of dynamic equilibrium that involves chemical reactions. It occurs when opposing reactions are proceeding simultaneously at equal rates. The rate at which products in a reversible reaction are formed from the reactant equals the rate at which the reactants are formed from the products. As a result, the concentrations of all reactants and products remain constant with time. Since concentration ceases to change, it may appear that everything has stopped, though in reality, both the forward and backward reactions continue to occur indefinitely. Let us examine the decomposition of a colorless N 2O4 gas to brown NO2 gas and see how this reaction illustrates chemical equilibrium. As shown in Figure 3.2, we begin with a sample of colorless N2O4 inside a sealed tube resting in a beaker (leftmost image). When the solid N2O4 is warmed, it vaporizes (bp = 21.2oC) and turns pale brown (middle image). The pale brown color slowly darkens, but after a few moments, the color stops changing (rightmost image). Chemistry 2 The Equilibrium Condition |Page 2 of 6  Figure 3.2. A macroscopic and molecular view of the N2O4-NO2 system. (Image from Brown, T. L., Bursten, B., Lemay, H. E., Murphy, C. J., & Woodward, P. M. (2012). Chemistry: The Central Science (12th ed.). Glenview, Illinois: Pearson Education, Inc.) What is happening here? At the molecular level, N2O4 molecules collide and a few begin to split into NO2 molecules. As time passes, more N2O4 molecules decompose, and the concentration of NO 2 increases. As more NO2 molecules are formed, it can be viewed in the macroscopic level that the tube turns pale brown and eventually darkens because NO2 is reddish-brown. As the reaction progresses, the decomposition of N2O4 slows down as the number of the molecules decreases. At the same time, the increased number of NO2 molecules that collide and combine, speed up the re-formation of N2O4. Eventually, N2O4 molecules decompose into NO2 molecules as fast as the NO2 molecules recombine to form N2O4. At this point the system has reached equilibrium: the concentrations of the reactant and product stop changing because the forward (fwd) and reverse (rev) rates have become equal: At equilibrium: ratefwd = raterev [3. 1] Thus, a system at equilibrium continues to be dynamic at the molecular level (the reaction does not stop), but we observe no further net change in the amounts of substances because changes in the forward direction are balanced by changes in the reverse direction. (Note: If you have access to the internet, you may view the reaction of N2O4 and NO2 through the link Chemistry 2 The Equilibrium Condition |Page 3 of 6  https://www.youtube.com/watch?v=zrei1gnHXAY) This equilibrium reaction is represented by writing the equation for the reaction with two half-headed arrows pointing in opposite directions (⇌), indicating that the reaction is reversible. Remember that a reversible reaction occurs in two directions ― forward (towards the formation of the products) and reverse/backward (towards the formation of the reactants). Thus, the above reaction is written as N2O4(g) ⇌ 2NO2(g) [3.2] Colorless Brown At a given temperature, when the system reaches equilibrium, product and reactant concentrations are constant. Therefore, their ratio must be a constant. We will use the N2O4-NO2 system to derive this constant. At equilibrium as shown in Equation 3.1, Ratefwd = Raterev We learned in kinetics that both forward and reverse reactions are elementary steps, so we can write their rate laws directly from the balanced equation: kfwd [N2O4]eq = krev[NO2]2eq [3.3] Where kfwd and krev are the forward and reverse rates constants, respectively, and the subscript “eq” refers to concentrations at equilibrium. Rearranging, k fwd = NO2  eq 2 k rev N 2 O4 eq [3.4] The ratio of constants gives rise to a new overall constant called the equilibrium constant (K): Chemistry 2 The Equilibrium Condition |Page 4 of 6  K= k fwd = NO2 2 eq k rev N 2 O4 eq [3.5] The equilibrium constant K is a number equal to a particular ratio of equilibrium concentrations of products and reactants at a particular temperature. What is the unit of the equilibrium constant K? Equilibrium constants are reported without units. This is because each term in the equilibrium constant expression represents the ratio of the measured quantity of the substance (molar concentrations or pressure) to the thermodynamic standard-state quantity of the substance. Recall that these standard state conditions are 1 M for solution and1 atm for 2.0𝑀 gases. Thus, a concentration of 2.0 M becomes = 2.0 while 1𝑀 0.5𝑎𝑡𝑚 a pressure of 0.5 atm becomes = 0.5. With these quantity 1𝑎𝑡𝑚 terms unitless, the ratio of terms we use to find the value of K is also unitless. In the next module, we will examine how the magnitude of K provides important information about the composition of an equilibrium mixture. Navigate Questions to ponder on (non-graded): 5 You are encouraged to answer these items to enhance your understanding of chemical equilibrium. 1. Describe how the graph of concentration (y-axis) versus time (x-axis) for reactants and products would look like when equilibrium is already established. 2. Can we establish a dynamic equilibrium in an open system? Why do you say so? 3. At equilibrium, do the concentrations of reactants have to be equal to the concentrations of products? Knot In summary, 2 When the forward and reverse reactions occur at the same rate, the system has reached dynamic equilibrium and concentrations no longer change. For dynamic equilibrium to occur, the reaction should take place in a closed system. The equilibrium constant (K) is a number based on a particular ratio of product and reactant concentrations or partial pressures. The quantity of the equilibrium constant is unitless. Chemistry 2 The Equilibrium Condition |Page 5 of 6 References: 1. Albarico, J.M. (2013). THINK Framework. Based on Ramos, E.G. and N. Apolinario. (n.d.) Science LINKS. Quezon City: Rex Bookstore Inc. 2. Brown, T. L., Bursten, B., Lemay, H. E., Murphy, C. J., & Woodward, P. M. (2012). Chemistry: The Central Science (12th ed.). Glenview, Illinois: Pearson Education, Inc. 3. Burdge, J., & Overby, J. (2012). Chemistry: Atoms First. New York, New York: McGraw-Hill Companies, Inc. 4. Ebbing, D. & Gammon, S. (2009). General Chemistry (9th ed.). Boston, New York: Houghton Mifflin Company. 5. Silberberg, M. S. (2010). Chemistry: The Molecular Nature of Matter and Change (5th ed.). New York, New York: McGraw-Hill Companies, Inc. Online Source: 1. Topic 4: Chemical Equilibrium (n.d.) Retrieved from https://www.edu.gov.mb.ca. Prepared by: Monaliza M. Callueng Reviewed by: Genalyn Alice R. Villoria Position: Special Science Teacher IV Position: Special Science Teacher V Campus: PSHS-IRC Campus:PSHS-CVC © 2020 Philippine Science High School System. All rights reserved. This document may contain proprietary information and may only be released to third parties with approval of management. Document is uncontrolled unless otherwise marked; uncontrolled documents are not subject to update notification. Chemistry 2 The Equilibrium Condition |Page 6 of 6