Gizmo Answer Key: Equilibrium and Concentration

Name: Rishi Nelson Date: 07-09-2022 Student Exploration: Equilibrium and Concentration Directions: Follow the instructions to go through the simulation. Respond to the questions andprompts 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? He will have more money B. How will Gary’s account change if he spends more than he saves? He will have less money C. What happens if Gary spends exactly as much as he saves? It will remainunchanged. 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 molecules are merging 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. 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 are disintegrating. 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. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

It is a reversible reaction because both the forward reaction: blue flashes, and the reverse reaction; redflashes can be seen Activity A: Reversiblereactions 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 amount of moles present in NO2 and N2O4 will remain constant and the same if you allow thereaction to run for a sufficiently enough time. 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 ? While moles of NO2 decline, moles of N2O4 rise. 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? I see that the NO2 and N2O4 moles level off and remain constant. This situation, in which the overall amounts of reactants and products does not change significantly overtime, 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 4.45 Moles N 2 O 4 1.78 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 6. Experiment: Click Reset . On the INITIAL SETTINGS tab, set Moles NO 2 to 0 and Moles N 2 O 4 to 4. Click Play . Click Pause when the bars of the bar chart stop moving very much. A. List the current amounts of each substance: Moles NO 2 4.0 Moles N 2 O 4 2.0 B. How do these results compare to starting with 8 moles of NO 2 ? Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

The quantity from this experiment is comparable to what we obtained with 8 moles of NO2 and 0moles of N2O. 7. Summarize: In each trial, you started with the same amounts of nitrogen and oxygen. In this situation, did the equilibrium amounts change depending on the direction of the reaction? The equilibrium amounts were similar in both cases. 8. Set up the Gizmo: Click Reset and select the EXPERIMENT tab on the left. On the INITIAL SETTINGS tab on the right, select Reaction 2 . Set Moles NO to 5, Moles NO 2 to 5, and Moles N 2 O 3 to 0. What are the reactants and product of this reaction? Reactants: NO NO2 Product: N2O3 (Note: In this reaction, some of the NO 2 reactants combine to form N 2 O 4 , as in reaction 1.) 9. Observe: Recall that a blue flash appears every time two reactants combine to form a product. A red flash appears every time a product dissociates into reactants. Click Play . A. At first, do you notice more blue flashes or red flashes? Blue flashes B. What do you notice about the frequency of blue and red flashes as time goes by? After some time, the number of blue and red flashes is equal. C. Click Reset . This time, start the experiment with 0 moles of NO and NO 2 and 5 moles of N 2 O 3 . Click Play . What do you notice about the red and blue flashes now? At first, there is a higher frequency of red flashes, but over time the frequency is even 10. Explain: Think about how the numbers of blue and red flashes reflect the rates of the forward (reactants → products) and reverse (products → reactants) reactions. A. What happens to the rate of the forward reaction as the reactants are consumed? As reactants are used up, the forward reaction rate falls. B. What happens to the rate of the reverse reaction as the products are produced? With the production of products, the rate of reverse increases. C. Why do reversible reactions always result in chemical equilibria? Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

When the rates of forward and backward reactions are equal, chemical equilibrium occurs. Upuntil there is no longer an excess of reactants, the forward reaction will be faster than thereverse reaction, and the opposite is true for an excess of products. Activity B: The equilibriumconstant 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, ormol/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 ] 4.00 M [N 2 O 4 ] 14.00 M B. Click Play and wait for equilibrium to become established. Click Pause . What are the approximate equilibrium concentrations? [NO 2 ] 12 M [N 2 O 4 ] 6 M 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: Based on the current concentrations of NO 2 and N 2 O 4 , what is K c ? 0.042 Show your work here: Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

6.0 K c = 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 lastcolumn, 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 2 7 12 6 0.042 2 7 11.69 6.31 0.046 2 7 12.78 5.21 0.031 3 5 13.12 4.88 0.028 3 5 13.47 4.53 0.025 3 5 13.36 4.64 0.026 4 3 12.61 5.39 0.034 4 3 13.05 4.94 0.029 4 3 13.10 4.90 0.028 4. Calculate: Find the average value of K c for each set of three trials. Trials 1-3: 0.039 Trials 4-6: 0.026 Trials 7-9: 0.030 5. Analyze: What do you notice about the values of K c ? They essentially remain the same. 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 average your results to get the best approximation of K c . Show your data and work on a separate sheet of paper. (Hint: Because of the coefficient “2” in front of HI, you will have to square the concentration of HI to find K c .) K c = 14.76 7. 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 doneby 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? 14.36 C. How does Q c compare to K c ? Qc is less 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: Q is lower than K, hence there are more reactants. D. When the reaction begins, do you expect [HI] to increase or decrease? increase Explain: Since there are too many reactants and too few products, the amount of HI will rise as the reaction proceeds. 4. Test: Click Play . What happens to [HI]? [HI] increases Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

Extension: Equilibriumcalculations 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 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 ] = 11.94 B. How much has the concentration of N 2 O 4 gone down? 6.07 C. What is the current concentration of NO 2 ? [NO 2 ] = 6.07 D. In general, if [N 2 O 4 ] is reduced by x , how much does [NO 2 ] increase? 5.9 This result may be surprising. It is true because at constant pressure, the overall density of particles in thecontainer 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 = 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 = N2O4initial-x/x 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. Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

0.042 = 18-x/x D. Rearrange the terms of your equation to form a quadratic equation in the form ax 2 + bx + c = 0. = 0 4. Solve: Because the equation is in the form ax 2 + bx + c = 0, you can use the quadratic formula (shown below) to solve for x . Ignore negative solutions because the concentrations cannot be negative. Show your work. 5. Predict: Based on the value for x , what do you expect the equilibrium concentrations of NO 2 and N 2 O 4 to be? [NO 2 ] 11.98 [N 2 O 4 ] 6 Check your work by solving for K c using K c = 0.4 If you don’t get the correct value of K c , recheck your work. 6. Test: Click Play and wait for equilibrium to be established. What are the actual equilibrium values of each substance? [NO 2 ] 12.4 [N 2 O 4 ] 5.82 How close were these results to your predicted results? within 0.2 7. Challenge: Suppose you begin with 6 moles of NO 2 and 5 moles of N 2 O 4 . Assuming a value for K c of 0.042, Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved

predict the equilibrium concentrations of NO 2 and N 2 O 4 . (Use the Gizmo to determine the initial concentrations.) Show your work on a separate sheet of paper. After you have made your predictions, click Play and record the experimental results. Predicted: [NO 2 ] 11.98 [N 2 O 4 ] 6.02 Experimental: [NO 2 ] 11.93 [N 2 O 4 ] 6.07 Reproduction for educational use only. Public sharing or posting prohibited. © 2020 ExploreLearning™ All rights reserved