EquilibriumConcentrationSE

Name: Date: Student Exploration: Equilibrium and Concentration Directions: Follow the instructions to go through the simulation. Respond to the questions and prompts in the orange boxes. Vocabulary: chemical equilibrium, concentration, equilibrium, equilibrium constant, reaction quotient, reversible reaction Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Gary has $5,000 in his bank account and earns a modest salary. Every month he pays for rent, food, utilities, and entertainment. A. How will Gary’s account change if he saves more than he spends? If he saves more and spends less the amount initially present in his account will increase. B. How will Gary’s account change if he spends more than he saves? His account will start decreasing from $5000 to less. C. What happens if Gary spends exactly as much as he saves? The amount in his account will remain the same. Gizmo Warm-up If Gary spends exactly as much as he earns, his savings will be in equilibrium . Equilibrium occurs when two opposing processes occur at the same rate, leading to no net change. In the Equilibrium and Concentration Gizmo, you will investigate how equilibrium can occur in chemical reactions. To begin, check that Reaction 1 is selected. Set Moles NO 2 to 8 and Moles N 2 O 4 to 0. 1. Click Play ( ) and observe the colliding molecules. What do you notice? The colliding molecules start to combine together and turn into N2O4. In the Gizmo, a blue flash appears every time two reactants combine to form a product. A red flash appears every time a product dissociates into reactants. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

2. Click Reset ( ), and set Moles NO 2 to 0 and Moles N 2 O 4 to 8. Click Play . What do you notice now? The molecules break up and turn into NO2 3. When a reaction can proceed in either direction, it is a reversible reaction . Based on what you have observed, is the synthesis of NO 2 into N 2 O 4 a reversible reaction? Explain. The synthesis of NO 2 into N 2 O 4 is a reversible reaction. Because both the forward and reverse reactions are observed. Activity A: Reversible reactions Get the Gizmo ready: ● Click Reset . Reaction 1 should be selected. ● Set Moles NO 2 to 8 and Moles N 2 O 4 to 0. ● Move the Sim. speed slider all the way to the right. Question: What are the characteristics of reversible reactions? 1. Predict: Suppose you began with 8 moles of NO 2 in the chamber. What do you think will happen if you let the reaction go for a long time? The rates of both reactions start to become the same and eventually equilibrium will bbe achieved. 2. Test: Click Play . Select the BAR CHART tab and check that Moles is selected. Observe the bar chart for about 30 seconds. As time goes by, what do you notice about the bars representing moles NO 2 and moles N 2 O 4 ? Tha bar representing moles NO2 decreases and the bar representing moles N2O4 increases. 3. Observe: Click Pause ( ). Select the GRAPH tab. Click the (–) zoom control on the horizontal axis until you can see the whole graph. What do you notice? The values are becoming similar and constent. This situation, in which the overall amounts of reactants and products does not change significantly over time, is called a chemical equilibrium . 4. Record: On the BAR CHART tab, turn on Show data values . How many moles of NO 2 and N 2 O 4 are there right now? Moles NO 2 4.32 Moles N 2 O 4 2.17 5. Calculate: Suppose all the NO 2 molecules were synthesized into N 2 O 4 . Given the equation 2NO 2 ⇄ N 2 O 4 , how many moles of N 2 O 4 would be produced? 4 moles of N2O4 Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

B. What happens to the rate of the reverse reaction as the products are produced? The rate increases. C. Why do reversible reactions always result in chemical equilibria? Because equilibrium usually occurs when the rates of the forward and reverse reaction are equal. If there is any imbalanced reactant or products, the reaction will shift to favor a side in order to reach equilibrium again,, Activity B: The equilibrium constant Get the Gizmo ready: ● Click Reset . Select Reaction 1 . ● Set Moles NO 2 to 2 and Moles N 2 O 4 to 7. Introduction: When investigating the rates of reactions, it often is useful to consider the concentrations of reactants rather than the total number of moles. Concentrations are often expressed in moles per liter, or mol/L. Brackets are used to signify concentration. For example, “[H 2 ] = 5.0 M” means the concentration of hydrogen gas in a chamber is 5.0 moles per liter. Question: What are the characteristics of reactions in equilibrium? 1. Record: On the BAR CHART tab, select Concentration . Check that Show data values is on. If necessary, use the arrows to adjust the scale of the chart. A. What are the current concentrations of each compound? [NO 2 ] 4.00M [N 2 O 4 ] 14.00M B. Click Play and wait for equilibrium to become established. Click Pause . What are the approximate equilibrium concentrations? [NO 2 ] 12.03M [N 2 O 4 ] 5.82M 2. Calculate: The value K c represents the ratio of products to reactants in a reaction at equilibrium. The greater the amount of products relative to reactants, the higher the resulting value of K c . For a general reaction between gases: a A( g ) + b B( g ) ⇌ cC( g ) + d D( g ), K c is calculated as follows: For the current reaction, 2NO 2 ⇌ N 2 O 4 , we have: Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

Based on the current concentrations of NO 2 and N 2 O 4 , what is K c ? 4.02 L mol^-2 Show your work here: 5.82 mol/L K c =1.75 L mol^-1 (12.03 mol/L)^2 3. Gather data: Experiment with a variety of initial concentrations of NO 2 and N 2 O 4 . For each set of initial concentrations, use the Gizmo to determine the equilibrium concentrations of each substance. In the last column, find K c for that trial. Run three trials for each set of initial conditions. Initial [NO 2 ] Initial [N 2 O 4 ] Equilibrium [NO 2 ] Equilibrium [N 2 O 4 ] K c 4 mol 8 mol 10.96 4.52 4.16*10^-2 4 mol 8 mol 11.13 4.43 3.98*10^-2 4 mol 8 mol 11.20 4.40 3.92*10^-2 1 mol 6 mol 6.99 3.01 4.30*10^-2 1 mol 6 mol 8.29 2.36 2.84*10^-2 1 mol 6 mol 7.55 2.73 3.61*10^-2 3 mol 5 mol 8.76 2.12 2.42*10^-2 3 mol 5 mol 8.52 2.24 2.62*10^-2 3 mol 5 mol 7.11 2.95 4.14*10^-2 4. Calculate: Find the average value of K c for each set of three trials. Trials 1-3: 4.01*10^-2 Trials 4-6: 3.58*10^-2 Trials 7-9: 2.99*10^-2 5. Analyze: What do you notice about the values of K c ? The values of Kc stay relatively constant with different amounts of initiated concentrations of the reactants. In general, the value of K c will be constant for a given reaction at a constant temperature, no matter the starting concentrations. That is why K c is known as the equilibrium constant . In this Gizmo, the values of K c will vary somewhat because there is a very limited number of molecules in the chamber. 6. On your own: Use the Gizmo to find K c for Reaction 4 : H 2 + I 2 ⇌ 2HI. Collect data at least 10 times and Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

Activity C: Reaction direction Get the Gizmo ready: ● Click Reset . Check that Reaction 4 is selected. ● Set Moles H 2 to 5, Moles I 2 to 5, and Moles HI to 3. Introduction: For a reversible reaction with equilibrium constant K c , it often is useful to know in which direction the reaction will proceed given the starting amounts of reactants A and B and products C and D. This is done by calculating the reaction quotient , Q c : Question: How can you predict the direction of a reversible reaction? 1. List: Select the BAR CHART tab. What are the initial concentrations of each substance? [H 2 ] 9.23 M [I 2 ] 9.23 M [HI] 5.54 M 2. Calculate: Use the equation above to find Q c for the current reaction. A. What is the current value of Q c ? 0.36 B. In activity B, what value of K c did you arrive at for this reaction? 32.1 C. How does Q c compare to K c ? Qc is much smaller than Kc 3. Analyze: Recall that Q c is equal to the ratio of product concentrations to reactant concentrations. A. If there is an excess of products, will Q c be greater than or less than K c ? Greater B. If there is an excess of reactants, will Q c be greater than or less than K c ? Less C. In the current situation, is there an excess of products or reactants? Reactants Explain: Qc is less than Kc, thus there is excess of reactants. D. When the reaction begins, do you expect [HI] to increase or decrease? Increase Explain: Excess of reactantrs would favor the forward reaction, thus more products produced. 4. Test: Click Play . What happens to [HI]? The value of HI eventually increases. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

Extension: Equilibrium calculations Get the Gizmo ready: ● Click Reset . Select Reaction 1 . ● Set Moles NO 2 to 0 and Moles N 2 O 4 to 6. Goal: Given K c and initial concentrations, calculate equilibrium concentrations. 1. List: Select the BAR CHART. What is the initial concentration N 2 O 4 ? [N 2 O 4 ] initial = 18.00 M 2. Experiment: Click Play and wait for a few seconds. Click Pause before equilibrium is reached. A. What is the current concentration of N 2 O 4 ? [N 2 O 4 ] = 12.19 M B. How much has the concentration of N 2 O 4 gone down? 6.03 M C. What is the current concentration of NO 2 ? [NO 2 ] = 6.03 M D. In general, if [N 2 O 4 ] is reduced by x , how much does [NO 2 ] increase? also by x This result may be surprising. It is true because at constant pressure, the overall density of particles in the container remains constant. So, if the concentration of one substance is reduced by x , the concentration of the other substance increases by x . 3. Manipulate: Begin with the general equation for K c : . A. What is the equation for K c for the reaction 2NO 2 ⇌ N 2 O 4 ? （ N2O4) K c = (NO2)^2 B. In this experiment, the initial concentration of NO 2 is zero. If the concentration of N 2 O 4 is reduced by x at equilibrium, the equilibrium concentration of NO 2 is equal to x . Substitute the following values into the equation you wrote in step A: [N 2 O 4 ] = ([N 2 O 4 ] initial – x ) [NO 2 ] = x K c = [N2O4]initial – x)/x^2 C. In activity A, you discovered that K c for this reaction was close to 0.042. Substitute this value and the initial concentration of N 2 O 4 into your equation. 0.042 = (18-x)/x^2 Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved