Unit 6 is all about the big idea Energy. Episode 59 discusses the questions 2021 - Question 4, 2017 - Question 5 and 2013 - Question 3. These are released FRQs from previous exams and copyright of the College Board.
Unit 6 is all about the big idea Energy (0:46). Episode 59 discusses the questions 2021 - Question 4, 2017 - Question 5 and 2013 - Question 3. These are released FRQs from previous exams and copyright of the College Board.
Question 4 of the 2021 exam starts with our favorite equation: mcAT (1:34). In part b) you calculate the mass of iron and in part (c) discuss the effect of doubling the mass of iron on the maximum temperature. Question 5 of the 2017 looks at similar concepts, but it is a combustion reaction (3:40). It starts again with calculating the magnitude of heat energy and using mcAT. In part (b) we calculate how much energy 1 mole of 2-propanol would release. Part (c) asks about the effect of having a water/propanol mixture on the final temperature. Question 3 of the 2013 exam starts with stoichiometry - identifying the limiting reactant (5:30) and in part (b) the inconsistent trial. In (c) we are using again mCAT for our calculations. (D) and (e) have us calculate the enthalpy - in d) using experimental data, in (e) as enthalpy of formation. In (f) we are provided with an explanation for the discrepancy between (d) and (e) and are asked to explain if that could be the reason.
Today’s Question of the day is about Enthalpy. How do we calculate the enthalpy using bond enthalpies?
A. bonds broken - bonds formed B. bonds formed - bonds broken C. bonds broken + bonds formed D. bonds formed + bonds broken
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Hi and welcome to the APsolute Recap: Chemistry Edition. Today’s episode will recap Unit 6 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 6: Thermodynamics. Unit 6 is an interesting unit. It focuses on only one big idea: Energy; and the heart of chemistry: breaking and forming bonds. It also lays the foundation for Unit 9: Applications of Thermodynamics and is often found in questions together with concepts from unit 9. 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 6. We will review 2021 - Question 4, 2017 - Question 5 and 2013 - Question 3. These are released FRQs from previous exams and copyright of the College Board.
Lets Zoom in:
Let’s start with 2021 - Q4. In part a) we are asked to calculate the amount of heat absorbed by the mixture given from minutes 0 to 4. Using mCAT and two significant figures, we calculate the amount of heat to be 190 J. In part b) we are now asked to calculate the mass of iron in grams that has reacted in the mixture. Looking at the equation given, we notice that the reaction releases 1650 kJ. The balanced coefficient for iron indicates that there are 4 moles of iron. Be careful before you start: units are friends! We calculated a value in J, but the enthalpy of reaction is in kJ. We need to convert! Let’s use dimensional analysis: Our 190 J is converted to kJ by multiplying it with 1 kJ over 1000 J. Then we use the enthalpy of reaction: times 4 moles of iron over 1650 kJ. In the last step we want grams, so we multiply by the molar mass of iron: times 55.85 g over 1 mol. If you set it up that way, all units except for grams should neatly cancel out and you’ll get 0.026g. Don’t be confused by the negative sign - this only indicates that energy is being released and doesn’t affect the mass. Mass is still a positive value. Part c) changes the experiment and doubles the amount of iron. Don’t be fooled, the sand isn’t part of the reaction, so it's not a limiting reactant problem! You are using twice as much iron in an open container where it reacts with oxygen from the air - which means that it is in excess. Doubling the iron will increase the maximum temperature, because you have more moles of iron that react, meaning you have more iron atoms that react and the released thermal energy increases, which increases the maximum temperature.
Switching now to 2017 - Question 5, which has similar concepts. However, in this case we have a combustion reaction and use the released energy to heat water. In part a) we are again asked to calculate the magnitude of heat energy and can use mCAT. Be careful: The tricky thing with the mcAT is the m. We do not have an aqueous solution where we add the mass of the compound to the water! We burn the propanol and heat the water only, assuming no energy is lost in between. So, the mass for this calculation is the mass of water, 125.00g. Another note: it asks you to calculate the energy in kJ! If you plug them in just as it is, you will get J, because the specific heat capacity is in J. But you will have to convert to kJ before you report your answer of 15.2 kJ. In part b) we are now asked to calculate the energy that one mole of 2-propanol would release. In our set up, we used 0.55g, which, using its molar mass, 60.1g/mol, is 0.0091 mol. Therefore one mole of 2-propanol would release 15.2 kJ over 0.0091 mol = 1689 kJ, or 1700 kJ with the correct number of significant figures. In part c) we again do the experiment with modifications - now there is some water mixed into the 2-propanol. And so, less propanol would be combusted, not as much energy would be transferred and the final water temperature would be less than in the first experiment.
The third question we are looking at is 2013 - Question 3. It starts with stoichiometry - a concept you just cannot escape! a) asks about the limiting reactant. There are two ways to solve this - You can either convert both reactants to moles - 0.1 moles HCl and 0.0124 moles of Mg (for the 0.50g) and then argue that the balanced chemical equation indicates a 2 H+ to 1 MgO relationship. Therefore you would need only 2 times 0.0124 moles of acid, which is 0.0246 moles, but you have 0.1 - more than enough! So, MgO is limiting. But, good news if you aren’t a fan of calculations - here is a second way to solve. You have four trials, in which the amount and molarity of acid is constant. The only thing that changes is the amount of magnesium oxide. You can also see that the change in temperature varies - which can only come from the change in mass of MgO. Therefore, MgO is limiting.
Part b) has you identify a trial that is inconsistent. The change in temperature should be proportional to the change in mass of our limiting reactant, MgO. Doubling mass should give you twice the change in temperature. In trial 2 and 4, 0.50g raised the temperature by approximately 4 degrees. In trial 3, half of the magnesium oxide, 0.25g raised the temperature approximately 2 degrees. In trial 1, however, 0.25g raised the temperature only 1 degree. Therefore, trial 1 is inconsistent. Part c) is our all-time favorite, calculating the thermal energy! You can use any of the three remaining trials. Just a few notes: you need to convert the mL of acid to g using the density given. In contrast to 2017 - Question 5, we now add the acid and the magnesium oxide together, so, the mass is the mass of acid PLUS mass of solid. If you choose trials 2 or 4, you will get an answer of 1700 J. For trial 3 you’ll get 880 J because we have a lower mass and lower change in temperature.
In part d) we are now asked to CALCULATE the change in enthalpy per molereaction. You can again either choose trials 2/4 or trial 3. And again, units are friends! If you have no idea what to do, look at the units! It asks you for kj/molreaction. We just calculated kJ, so we now have to divide it by the molreaction. So, we have our 1700 J or 880 J. First, we have to flip the sign, because the energy is being released: negative 1700 J or negative 880 Joules. We can calculate the moles by converting the mass of MgO to moles using the molar mass of 40.30g / mol. Lastly, in your calculation, be mindful of the questions asking for KILOjoules / molreaction: don’t forget to convert. You will notice that, no matter if you did the calculation for trials 2 and 4 or trial 3, you’ll always get the same answer: the standard enthalpy is negative 140 kJ/molreaction . That makes sense? Why? Because it is per mole of reaction and that is the same for all trials! That is the beauty of per mole of reaction!
Part e) shifts our focus towards a different concept: enthalpy of formation. You can use the tabulated values to calculate the accepted value for the enthalpy change by calculating the sum of the enthalpies of formation with their coefficients of the products minus the enthalpies of formation with their coefficients of the reactants. So we have: in brackets: [-467kJ/mol + (-286 kJ/mol)] minus, in brackets: [-602 kJ/mol + (2 x 0 kJ/mol)], which will be -151 kJ/molreaction. Last, but not least, in part f) we are made aware that the experimental value we calculated, -140 kJ/molreaction and the accepted value, -151 kJ/molreaction, do not agree. Could the leaking of heat energy from the calorimeter explain this? Yes! The experimental value is less negative, therefore more positive than the accepted value. If the calorimeter was leaking heat energy, the change in temperature would be smaller, the calculated heat energy is smaller and then, as it is the case here, the calculated enthalpy is less negative.
To recap…
Unit 6 is all about energy. The best advice: units are friends! Unit 6 involves a lot of calculations and units can show us the way and help us with the correct set up. But: if they are not respected, they might even trip us up! Be especially careful with joules and kilojoules. Unit 6 also often uses experimental data and experimental set ups. Make sure you read through the set up carefully. In preparation for this unit, go over labs and discuss potential experimental variations as well as experimental errors and how they affect your calculated values. Also be careful with signs - reflect on: when do you need to calculate a magnitude and when is it important to indicate if the reaction is endothermic or exothermic? The kJ/molereaction concept is unique to unit 6, too. Make sure you consider the coefficients!
Coming up next on the APsolute RecAP Chemistry Edition: Factors affecting equilibrium
Today’s Question of the day is about Enthalpy. How do we calculate the enthalpy using bond enthalpies?