The FRQ questions on the AP exam often combine content from two or three different units. In today’s episode, we are emphasizing questions that require knowledge and skills from Unit 7: Equilibrium.
The FRQ questions on the AP exam often combine content from two or three different units. In today’s episode, we are emphasizing questions that require knowledge and skills from Unit 7: Equilibrium. You will rarely find an FRQ that solely focuses on Unit 7, since it lays the foundation for Unit 8, acids and bases as well as parts of Unit 9, Application of Thermodynamics. It also links back to concepts from earlier units, like gas laws. The questions we are using today are online accessible. Our suggestion: Answer the questions yourself and then listen to this episode to hear the explanations, as well as do’s and don'ts for answering questions of Unit 7. We will review 2016 - Question 6, 2015 - Question 4 and 2014 - Question 4. These are released FRQs from previous exams and copyright of the College Board.
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Hi and welcome to the APsolute Recap: Chemistry Edition. Today’s episode will recap Unit 7 Free-response questions.
Let’s Zoom Out
The FRQ questions on the AP exam often combine content from two or three different units. In today’s episode, we are emphasizing questions that require knowledge and skills from Unit 7: Equilibrium. You will rarely find an FRQ that solely focuses on Unit 7, since it lays the foundation for Unit 8, acids and bases as well as parts of Unit 9, Application of Thermodynamics. It also links back to concepts from earlier units, like gas laws. The questions we are using today are online accessible. Our suggestion: Answer the questions yourself and then listen to this episode to hear the explanations, as well as do’s and don'ts for answering questions of Unit 7. We will review 2016 - Question 6, 2015 - Question 4 and 2014 - Question 4. These are released FRQs from previous exams and copyright of the College Board.
Lets Zoom in:
And start with discussing 2016 - Question 6. This is a short answer question which has 4 points. It focuses on two aspects: Understanding the magnitude of K as well as explaining how a change affects the equilibrium. In part (a) we are asked to determine the concentration of the barium complex ion, taking into account the equilibrium constant. K in this case is very large, 7.7 times 107. This indicates that the reaction will pretty much go to completion. We are starting with 50 mL of 0.20M barium nitrate and 50 mL of 0.30M EDTA. Since we are mixing 50 mL of both reactants, but with barium nitrate having the lower concentration, the barium ion will be the limiting reactant. We can calculate the 0.01 mol of barium by multiplying 0.05 L times 0.2 mol/L. Our new volume is 0.1 L since we combined the two solutions. And so, the concentration will be 0.01 mol divided by 0.1L, which is 0.1 molar.
In part (b) we are diluting the solution to 1.00L and we are asked to determine and justify if the number of moles of barium 2+ is greater, less than or equal to the barium 2+ ions in the original solution. Don’t forget: even though the K value is very large, it is still an equilibrium concentration! So, how does diluting it change the equilibrium concentration? When we dilute our solution from 0.1L to 1.00L, the original concentrations of the barium -EDTA complex, the barium 2+ and the EDTA itself are divided by 10. We can set this up as a calculation for Q: 1/10th of the concentration of the barium-EDTA complex, divided by 1/10th of the concentration of barium 2+ which is multiplied by 1/10th of the concentration of EDTA. Even without knowing the actual concentration, we can see that we have: 1/10 divided by the product of 1/10 times 1/10. That will give us 10. Therefore, Q is 10 times bigger than K. And because Q is larger than K, to move to equilibrium, the reverse reaction will increase and produce more reactants - leading to more barium 2+ ions. And so, the concentration of barium 2+ will be larger than in the original solution.
Let’s move to 2014 - Question 4. This question combined gas laws with equilibrium. In part (a) we can use everyone's favorite gas law equation: PVnRT, to calculate the moles of carbon dioxide present after 20 minutes. Plugging in our numbers and choosing the right R-value of 0.0821 L atm/mol K, we know that there are 0.0115 mol of carbon dioxide in the container. In part (b) we are presented with the claim that all of the calcium carbonate in BOTH setups had completely decomposed. Using the data in the experiment, we are asked to explain if we agree. AND WE DO NOT! You can approach this from two perspectives and either one will earn you credit: Option 1) Stoichiometry. We are starting with 50.0g of calcium carbonate. Using stoichiometry and the 1:1 relationship between calcium carbonate and carbon dioxide, we can see that we would have produced 0.5 mol of carbon dioxide, but we just calculate it to be “only” 0.0115 mol. Option 2) Both experimental set ups, the one with 50.0g and the one with 100.0g of calcium carbonate produced a pressure of 1.04 atm. This indicates that not all of the calcium carbonate can have decomposed - because then we would have different final pressure. As a result, we can derive that an equilibrium was reached at 1.04 atm and solid calcium carbonate remains in the container - obviously in one set up more than the other, but that wasn’t even the question!
In part (c) we are disturbing the equilibrium by adding CO2. Initially, as the prompt indicates, the pressure increases to 1.5atm. But what would the pressure be after equilibrium has been reached? Keep two things in mind: first, equilibrium can be reached from either side: from the reactants decomposing or from the products reacting to form the reactant. The key is the second part: the value of the equilibrium constant K, which tells us the ratio of products over reactants - and in this case the pressure from the gas - doesn’t change, unless you change the temperature. So, the position of the equilibrium remains the same: at a pressure of 1.04 atm! The reaction will increase the reverse reaction and the CO2 will react until the pressure reaches 1.04atm. Now, could we determine the VALUE of the equilibrium constant Kp for this experiment? The answer to this prompt in part (d) is YES! How so? When writing the equilibrium constant Kp we only include the pressure of gasses. Calcium carbonate and calcium oxide are both solids - which, together with pure liquids are never part of the equilibrium expression and therefore aren’t part of the calculated K value. So the only component that makes up K is the pressure of carbon dioxide at equilibrium. Which is, as discussed several times, 1.04 atm. And tada, the Kp value is 1.04. Don’t forget - the equilibrium constant is unitless!
Last, but not least, 2015, Question 4. This question focuses on solubility equilibria. In part (a) are you prompted to write the equilibrium equation for the dissolution of calcium hydroxide. Calcium hydroxide dissolves to form calcium 2+ and 2 hydroxide anions. We are using an equilibrium arrow, since the small Ksp indicates that we do have an equilibrium that actually lies on the side of the reactants. What is the molar solubility of calcium hydroxide in 0.10M calcium nitrate? If you need a recap, listen to episode 34! We can use an ICE table for it, but keep in mind: our initial concentration of calcium is NOT 0. Since we are dissolving in 0.1 molar calcium nitrate, our initial concentration is 0.1. Setting up the ICE table and making the assumption that the x is much smaller than 0.1 we can reduce the equation to: 1.3 x 10-6 = 0.10 times 4x2 instead of using 0.10 plus x times 4x2. Solving for x we get a value of 0.0018M, which is our molar solubility.
In part (c) we are adding four water molecules around the calcium ion. The key point here is the correct orientation: the partially negative oxygen is facing the calcium cation, whereas the partially positive hydrogens are pointing away.
To recap…
Unit 7, Equilibrium, lays the foundation for Units 8, acids and bases as well as parts of Unit 9, Application of Thermodynamics. It also ties in concepts from earlier units. Key concepts in unit 7 are understanding the magnitude of K, the equilibrium constant. If K is very large, the reaction goes to completion. In solubility equilibria, the K is often very small, indicating that only a very small amount will dissolve. K can either refer to concentration of aqueous solutions or pressures. Equilibrium is a dynamic state that can be reached “from both ends”. The equilibrium quotient Q describes the concentrations or pressure at a certain time. When comparing Q to K we can determine the direction in which the reaction will shift to reach equilibrium.
Coming up next on the APsolute RecAP Chemistry Edition: Molecular Structure and Acid Strength
Today’s Question of the day is about acid/base equilibria.
Would a weak acid have a very large or a very small equilibrium constant, Ka?