Revision Quiz Entropy and Free Energy Attempt review PDF

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Quiz Entropy and Free Energy with full questions and answers...

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17/02/2020

Revision Quiz: Entropy and Free Energy: Attempt review

Started on Tuesday, 17 December 2019, 1:10 PM State Finished Completed on Monday, 17 February 2020, 11:15 PM Time taken 62 days 10 hours Marks 8.05/20.00 Grade 40.25 out of 100.00

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 1 Correct Mark 1.00 out of 1.00

What is the sign of ΔS in the following reaction? CaCO3(s) → CaO(s) + CO2(g) Assume that the reaction is carried out at 520 K. Select one: 1. Zero 2. Negative 3. The entropy can only be determined exactly at absolute zero 4. It is impossible to determine without experimental data.

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5. Positive One solid species is changed into one solid species and one gaseous species. Gas molecules have many more ways of distributing energy than solids (in technical parlance, there are more "degrees of freedom"). So there is a large entropy increase in going from a solid to a gas as well as an entropy increase from forming 2 species from 1.So it must be positive. The temperature is a bit of a red herring, but of course the higher the temperature the more important the inuence of the gas will be.

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A key point to keep in mind is the idea that as molecules become more mobile their entropy increases because they have so many more ways of taking up the energy. If you want to read more about this, you might want to look up information about the equipartition theorem which will give you an insight into why it is that pV = nRT. The correct answer is: Positive

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 2 Incorrect Mark 0.00 out of 1.00

An enzyme denatures at 45.7 degrees Celsius. The enthalpy of unfolding is 856 kJ mol –1. What is the entropy change, ΔS, for the transition? Give your answer to 2 signicant gures. You don't need to include the units but it might be worth thinking about what they should be. Answer: 

No. No. You are out by a factor of a thousand because you forgot that the units were kJ mol–1. But well done for converting to absolute temperature. As always, we seem to be talking about the Gibbs energy relation. At the denaturation temperature what we're saying is that the system is at equilibrium, so the Gibbs energy is zero - we can equate ΔH with TΔS. And as the pressure is constant ΔH ≈ q p.

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So we can calculate the entropy simply by the relation: ΔS = q/T = ΔH/T Entropy changes gives us a measure how widely the energy is being spread out and re-distributed when we carry out a process. Because as temperature rises the available thermal energy increases, our calculation has to take account of this. So our calculation can be thought of as the energy transfer, in relation to the temperature at which that change occurs.  The correct answer is: 2700

856kJ = 856 000 J 856 000 /( 275 + 45.7 = 320.7) = 856 000 / 320.7 = 2669 = 2700 JK-1 

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 3 Incorrect Mark 0.00 out of 1.00

Which of the following substances is likely to have the greatest molar entropy? Select one: 1. A diatomic gas 2. A monoatomic gas  Nearly but not quite.... A gas has a high entropy by comparison with solids or liquids. However, there are fewer modes of motion and therefore fewer degrees of freedom in a monatomic than in a diatomic gas. This means that the molar entropy is less for a monatomic gas. 3. A solid 4. A liquid 5. They are all the same.

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Entropy is all about being able to distribute energy as widely as possible. Gases will have more entropy per mole than anything else. And among gases, the more complex you make the molecule the great the entropy. If you want to learn more about this, you should go and look up the equipartition theorem, which shows how molecular energy distributes itself evenly between the various kinds of motion of a molecule. So the more mobile the molecule the more ways to spread the energy there will be. It is through this equipartition theorem that that basic equation pV = nRT emerges. The correct answer is: A diatomic gas

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 4 Incorrect Mark 0.00 out of 1.00

Which of the following statements is FALSE?

Select one: a. It is not possible to calculate the absolute value of ΔG at a given temperature. b. For a reaction to occur the ΔS of the system must be greater than or equal to 0 c. At very high temperatures the value of ΔS dominates that of ΔH in the calculation of ΔG d. S has an absolute value at a given temperature and pressure e. When we calculate H (rather than ΔH) it is always based on an arbitrary zero  Yes. This is indeed true. This means that we are unable to compare values of H between substances (which we can do for entropy). We therefore always use values of ΔH when comparing substances as these do not rely on the arbitrary zero point.

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The Gibbs equation: ΔG = ΔH - TΔS. is among the most important in all of science. The reason is that, although it applies only under conditions of constant temperature and pressure, it allows us to make predictions about spontaneity (remember that all of the terms are state functions, so we don't actually need to observe the process - it may be maddeningly or perhaps thankfully slow). It's worth,therefore, dwelling on each term to think through it aects the overall picture. The correct answer is: For a reaction to occur the ΔS of the system must be greater than or equal to 0

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 5 Incorrect Mark 0.00 out of 1.00

When cooking aubergines, especially when frying them, it is customary to cut them into pieces, places them in a sieve and sprinkle them with lots of salt. A weight is placed on top and after a couple of hours there is a large puddle of liquid underneath the sieve. Which of the following "explanations" is incomplete or actually false? Select one: a. The presence of the salt increases the osmotic pressure of the salt water. b. The chemical potential of the water on the outside of the aubergine is lower so water moves outwards. c. There is an imbalance in the ionic strength leading to dierence in vapour pressure between the inside and the outside. d. The high ionic strength of the salt draws the water out of the aubergine. e. "Ah, it's the power of Entropy!"  This is true, actually. But in choosing this you mark yourself out as being impressed by the pretentious and pompous statement of would-be Zen masters.

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It's astonishing how much deep science is involved in even the most basic operation in the kitchen. If this is the kind of thing that interests you there are several excellent books on the market. Take a look at Harold McGee's On Food and Cooking which focuses on a culture and chemistry. Peter Barham's Science of Cooking is rather more physical in approach. Herve This has a chattily written book called Les Secrets de la Casserole - The Secrets of the Saucepan. All worth looking at. The correct answer is: The high ionic strength of the salt draws the water out of the aubergine.

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 6 Partially correct Mark 0.55 out of 1.00

Which of the following statements about vapour pressure is true? Select one or more:

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a. The vapour pressure can be changed by dissolved solutes (e.g. salt or a molecular substance) This is absolutely true! Dissolved substances lower the vapour pressure. One way to think about this is that they reduce the chemical potential (see Thermodynamics lectures). Alternatively you can imagine the solute as "getting in the way" of solvent molecules that are trying to escape from the liquid. The more stu is dissolved the fewer the solvent molecules can escape - hence the vapour pressure drops.

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b. At the boiling point the vapour pressure of a liquid is equal to the applied pressure Yes. That is exactly the point. This is the reason for using a pressure cooker. By increasing the pressure, you can raise the boiling point and cook food faster (remember the Arrhenius equation?). On the other hand the boiling point drops with altitude. This is why the Clausius-Clapeyron equation is so important in determining where and when you get clouds and when it's likely to rain! c. Solids have a vapour pressure d. The vapour pressure increases with temperature. on a cold one.

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Yes it does. That's why clothes dry faster on a drying line on a warm day than

e. Vapour pressure dierences are what causes osmosis in plants.

Your answer is partially correct. You have correctly selected 3. Vapour pressure is an incredibly important property of a liquid that has huge impacts on biology and on our everyday lives. The correct answers are: The vapour pressure increases with temperature., Solids have a vapour pressure, At the boiling point the vapour pressure of a liquid is equal to the applied pressure, The vapour pressure can be changed by dissolved solutes (e.g. salt or a molecular substance), Vapour pressure dierences are what causes osmosis in plants.

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 7 Incorrect Mark 0.00 out of 1.00

For the following combustion reaction determine ΔGº, given the data below. C2 H5 OH(l) + 3O2(g) → 2CO2(g) + 3H2O (g) ΔH f(C 2H 5OH) = -277.69 kJmol -1 ΔH f(H 2O) = -241.82 kJmol -1 ΔH f(CO 2) = -393.51 kJmol -1 S(O 2) = 205.1 JK -1mol -1 S(CO 2) = 213.7 JK -1mol -1 S(H 2O) = 69.9 JK -1mol -1 S(C 2H 5OH) = 160.7 JK -1mol -1 Needless to say, you should assume standard conditions. Select one:

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1. There is insucient information to do the calculations. 2. -333.14 kJmol -1 3. -1748.78 kJmol-1 4. -1193.41 kJmol -1 5. 40157.41 kJmol -1  No....You don't appear to have converted JK -1mol -1 to kJK -1mol -1when working out the entropy change of the reaction. This is essential as the values of ΔH are given in kJK-1mol-1 and you must be consistent throughout your calculations. Think through this and try again.

This is a more involved calculation than sum as it takes a few steps. But what it does is bring together several ideas into one place. It's a  proper test of your understanding and ability to work with these quantities. If, after trying again, you're struggling with this you should post a question to the forum and discuss things with your colleagues.

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Revision Quiz: Entropy and Free Energy: Attempt review

I think if you asked a scientist which is the most important quantity in thermodynamics then the Gibbs energy would come pretty close to the top of the list because what it does is to provide a precise limit on "the possible". Chemical, physical, and therefore biological processes will only be feasible, or, if you prefer, spontaneous, if ΔG < 0. Anything else will need to be coupled in some other way in order to happen. This is why we spend so much time getting you to do such calculations. The correct answer is: -1193.41 kJmol-1

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 8 Incorrect Mark 0.00 out of 1.00

In which direction would you expect the equilibrium to shift in the following reaction as the temperature is raised? C 6H 12O 6 + 6O 2 → 6CO 2 + 6H 2O Select one: 1. It depends on the value of ΔG o 2. To the right 3. It depends on the value of ΔH 4. There is no change in equilibrium position 5. To the left  As temperature increases the entropy becomes more signicant in determining the favourability of a reaction, according to the 2nd law of thermodynamics, ΔG = ΔH - TΔS. Hence, the equilibrium will start to shift towards the reaction favoured on entropy grounds. In this case this is the forwards reaction. 7 moles of reactants go to 12 moles of products hence the products have more entropy than the reactants and the ΔS of the forwards reaction is positive (favourable)

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Many students struggle with this question. The key thing is to cut through the detail and think about what determines the position of an equilibrium. It is determined by the value of the overall ΔG which in turn is related to the equilibrium constantK by: ΔG = -RTlnK While it looks as if T is nothing more than a multiplier you have to consider that G has both enthalpic and entropic components. And the entropic component is strongly temperature dependent. When you have gases involved in the reaction, the entropy will be a strong driver to shift equilibria. The result is that the position of equilibrium is usually quite temperature sensitive. The correct answer is: To the right

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 9 Correct Mark 1.00 out of 1.00

Which of the following statements is the correct denition of the Gibbs energy? Select one: a. G = q + w b. G = S – TH c. G = U – TS d. G = H – TS

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Correct. Well done. This is entirely correct.

e. G = UH - T

This is probably the most important equation in thermodynamics. You must get this right. No pressure....... (or change in temperature, come to think of it).

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The correct answer is: G = H – TS

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 10 Partially correct Mark 0.50 out of 1.00

A packet of salt is dissolved into water. Which of the following statements is true? Select one or more: a. ΔG= 0 b. ΔG > 0

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c. ΔG < 0 Yes. This is true. For whatever detailed molecular reasons (that we will worry about elsewhere) the free energy is negative. We could also say that the reaction is exergonic. d. Dissolution is spontaneous

Your answer is partially correct.

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You have correctly selected 1. The free energy, also known as the Gibbs energy, is the crucial quantity that determines whether processes in nature take place or not. It is a quantity that combines the First and Second Law and which also tells us what is the maximum amount of useful work we can derive from a particular process. Even without being quantitative, we can often draw useful conclusions simply by making observations of behaviour in simple systems. The correct answers are: ΔG < 0, Dissolution is spontaneous

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 11 Incorrect Mark 0.00 out of 1.00

Which of the following statements is FALSE? Select one: 1. ΔG can be determined at particular conditions if the values of ΔGº, T and the proportions of reactants and products are known 2. A large value of Keq means that the reaction is close to completion at equilibrium  This is true. K eq is the ratio of the products to the reactants at the equilibrium position. The larger it is the more the products dominate, and hence the closer the reaction is to going to completion. 3. Entropy cannot be calculated for non-standard conditions 4. ΔGo allows Keq to be calculated. 5. At equilibrium ΔG is zero.

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The Gibbs energy is a measure of the driving force of a reaction - i.e. how are away it is from equilibrium. It is a quantity that is valid provided you keep the temperature and pressure constant. It is one of the most fundamental ideas in thermodynamics of relevance to biology since it allows us to understand how biological systems are powered. The correct answer is: Entropy cannot be calculated for non-standard conditions

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 12 Correct Mark 1.00 out of 1.00

Which of the following statements is FALSE? Select one: 1. At equilibrium ΔG is zero. 2. ΔGo allows Keq to be calculated.

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3. Entropy cannot be calculated for non-standard conditions This is false. The entropy can be calculated under non-standard conditions using a similar and related equation to that for the calculation of ΔG at non-standard conditions. The equation used is S = Sº RTlnX, where X is the mole fraction (amount of component/total amount). in other words, you need to factor in the entropy of mixing in order to correct the standard entropy. 4. ΔG can be determined at particular conditions if the values of ΔGº, T and the proportions of reactants and products are known 5. A large value of Keq means that the reaction is close to completion at equilibrium

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The Gibbs energy is a measure of the driving force of a reaction - i.e. how are away it is from equilibrium. It is a quantity that is valid provided you keep the temperature and pressure constant. It is one of the most fundamental ideas in thermodynamics of relevance to biology since it allows us to understand how biological systems are powered. The correct answer is: Entropy cannot be calculated for non-standard conditions

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Revision Quiz: Entropy and Free Energy: Attempt review

Question 13 Correct Mark 1.00 out of 1.00

Which of the following statements is TRUE? Select one: a. If ΔG of a reaction is negative the reaction is endergonic. b. If ΔG of a reaction is positive the reaction is fast. c. If ΔG of a reaction is negative the reaction cannot proceed.

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d. If ΔG of a reaction is 0 the reaction is at equilibrium. Correct. When the value of ΔG is 0, there is no thermodynamic reason for the reaction to proceed either forwards or backwards . The reaction is therefore at equilibrium. (Note, this is ΔG = 0 not ΔGº = 0). e. ΔS contributes more to the value of ΔG at low temperatures than at high temperatures.

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ΔGis the ultimate arbiter of whether a process will proceed spontaneously or not. So it's very important to be condent about the sign convention. It is also important to remember that for almost all processes, knowing whether or not a process is...