Energetics & Thermochemistry. Ms. Kiely IB Chemistry SL Coral Gables Senior High School

Transcription

1 Energetics & Thermochemistry Ms. Kiely IB Chemistry SL Coral Gables Senior High School

2 Bell Ringer The following equation shows the formation of magnesium oxide from magnesium metal. 2Mg(s) + O2(g) 2MgO(s) ΔH = 1204kJ Which statement is correct for this reaction? A kj of energy are released for every mol of magnesium reacted. B. 602 kj of energy are absorbed for every mol of magnesium oxide formed. C. 602 kj of energy are released for every mol of oxygen gas reacted. D kj of energy are released for every two mol of magnesium oxide formed. QUIZ MOVED TO WEDNESDAY 10/24: I might be sent to a workshop on Monday. Just in case, the quiz has been moved to Wednesday. The quiz will include 5.1 and 5.2

3 ANSWER The following equation shows the formation of magnesium oxide from magnesium metal. 2Mg(s) + O2(g) 2MgO(s) ΔH = 1204kJ Which statement is correct for this reaction? D kj of energy are released for every two mol of magnesium oxide formed. ANSWER COULD HAVE BEEN option B if it said: B. 602 kj of energy are released for every mol of magnesium oxide formed.

4 Error in Calorimeter experiments Before we continue, let s talk about circumstances that can affect your data and cause error when using a calorimeter: -Open System (systematic error): If you were able to insulate your calorimeter more, then you would lose less heat to the surroundings. **Another way to help get more accurate results in an open system calorimeter: If you can get information not just on the water, but also on the can that is holding the water (its mass, its specific heat capacity), then you can add this into your calculations and get closer to the literature value.** -Incomplete combustion (systematic error): this can result for many reasons- but you can avoid it by making sure your fuel is burning in oxygen (making sure it is exposed to oxygen in some way, and not sealed with little oxygen to reach it. -Random errors: these will only affect your data and give you error if you do not 1) perform many trials, 2) apply your uncertainty values to your data based on the instruments being used, 3) propagate your uncertainties when averaging your data.

5 Incomplete combustion It happens, we ll talk about why in another chapter. For now, please note that IB will always ask you How can you better this experiment in the future? or How can you reduce error?. If the experiment is related to calorimetry, then please do not give a trivial answer such as Next time, we should measure more accurately. That is not detailed. Instead, you can say something like There may be error due to incomplete combustion, which can be reduced by burning the fuel in as much oxygen as possible. Also, heat is lost in this system due to it being primarily an open system. We can help trap in heat by insulating the apparatus/calorimeter.

6 HW Answers

7 Exercises 10 & 11

8 HW Answers

9

10 Calculating Changes in Enthalpy (ΔH) So far we have been focusing on Calorimetry, which is a method used to determine the amount of heat released by a chemical reaction (exothermic). For instance, when we are trying to calculate the heat released by an exothermic reaction, and we have access to the changes in temperature that have occurred to the system where the reaction is occurring, then we can use that information to help us calculate the ΔH of the reaction by assuming the heat changes of the system (and some other factors) are indicative of the heat released by the system. In a calorimetry experiment, the temperature changes (ΔT) occurring to the water as well as to the can, their respective heat capacities (c), and their respective masses (g) can all be used to calculate the heat released by the combustion of a fuel; (given assumptions).

11 OTHER WAYS of calculating Changes in Enthalpy (ΔH) When given a problem based on Calorimety you are supplied with data collected from an actual experiment. You are supplied with the increase in temperature that the system underwent, the amount in grams of the substance, and the specific heat capacity. They also give you how many moles of the substance are burned. It is not always possible, though, to design experiments that will help us get the data we re looking for in relation to the enthalpy changes of a reaction, especially if the reaction we want to solve for is potentially dangerous and releases too much energy. We can instead use theoretical data based off of other, similar reactions and what we know about their enthalpy data.this is when something called Hess Law comes into play.

12 Hess Law Hess Law states that the enthalpy change of a chemical reaction can be measured using the enthalpy changes of other reactions that have some of the the same reactants and products. Basically, you will be using the change in enthalpy values (ΔH) of reactions that can be measured in order to work out the change in enthalpy values for a particular reaction in which it cannot be measured directly. So let s say you are looking for ΔH of a reaction, but you only have information on two other similar reactions (which can also be theoretical reactions that do not actually exist) that contain some of those substances, ΔH₁ and ΔH₂. According to Hess Law: ΔH₁ + ΔH₂ = ΔH₃

13 Practice Using the equations below: C(s) + O2(g) CO2(g) Mn(s) + O2(g) MnO2(s) H = 390 kj H = 520 kj Determine what is H (in kj) for the following reaction? A. 910 B. 130 C. 130 D. 910 MnO2(s) + C(s) Mn(s) + CO2(g)

14 ANSWER Remember that Hess Law states ΔH₁ + ΔH₂ = ΔH₃ These are the rx s you have enthalpy information for: C(s) + O2(g) CO2(g) ΔH₁ = 390 kj Mn(s) + O2(g) MnO2(s) ΔH₂ = 520 kj This is the rx you want to solve for: ΔH₃= MnO2(s) + C(s) Mn(s) + CO2(g) ΔH₂ needs to be reversed so that Mn is on the product side, therefore the enthalpy value becomes positive. Cancel out like substances, and ensure that your resulting equation matches the one you are solving for. Then, use the formula: ΔH₁ + ΔH₂ = ΔH₃ ANSWER B. 130

15 Hess law is one way to calculate the enthalpy changes of reactions (rx) that we cannot measure directly in the lab; for example, theoretical rxs that may not necessarily occur in nature but are of interest to a scientist for the synthesis of a new drug or fuel. We can calculate the enthalpy change for a particular rx from the known enthalpy changes of other rxs that contain some of the same substances! Worked example pg. 180 This is the rx you want to solve ΔH for: S(s) + O2(g) SO2(g) ΔH =? These are the rxs you have ΔH values for: S(s) + 1½ O2(g) SO3(g) ΔH = -395 kj SO2(g) + ½ O2(g) SO3(g) ΔH = -98 kj Based on the original equation, your goal is to produce SO2. This means you should take a look at your given rxs, and make sure they are written in a manner so that SO2 is also a product. Therefore, let s change the second known one, meaning we must also reverse its enthalpy sign! SO3(g) SO2(g) + ½ O2(g) ΔH = +98 kj

16 H1 is the first known rx: S(s) + 1½ O2(g) SO3(g) ΔH1 = -395 kj H2 is the second known rx that we altered: SO3(g) SO2(g) + ½ O2(g) ΔH2 = 98 kj ΔH1 S(s) + 1½ O2(g) SO3(g) ΔH3 ΔH2 SO2(g) + ½ O2(g) According to Hess Law: ΔH1 + ΔH2 = ΔH = -297 kj/mol You can double check this by putting together the reactants and products of H1 and H2, respectively, into one large equation, and then simplifying it by cancelling out like substances. If you end up with the equation you are originally solving for, then you did it correctly! S(s) + 1½ O2(g) + SO3(g) SO3(g) + SO2(g) + ½ O2(g) Simplified: S(s) + O2(g) SO2(g) ΔH = -297 kj/mol

17 Let s practice using exercises from your book, Topic 5

18 Exercises A 17. ΔH = 394 kj + (+283) kj = 111 kj 18. ΔH = kj + (+66.4 kj) = kj 19. H = (2 ( 33.2 kj mol ¹)) + (+9.16 kj mol ¹) = kj mol ¹ 20. B) -3