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13 -- Part 2: Spontaneous Change: Entropy and Gibbs Energy

- Two moles of products are produced by three moles of reactants.
- The loss of one mole of gas is a sign of a loss of volume.
- The number of possible configurations for the molecule in the system is reduced by this loss.

- The number of forms in which the sucrose molecule can store their energy is reduced when they leave the solution.

- The four gases are likely to have different structures.

- The number of moles of gases is the same on 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 888-739-5110 We can't determine whether the entropy increases or decreases on the basis of just the generalizations.

- Predicting an increase or decrease in entropy will help us understand when a process will proceed spontaneously in the forward direction.

- Predict whether the outcome is uncertain in the following reactions.

- CONCEPT ASSESSMENT Figure 13-2 shows a process of expansion of an ideal gas into an evacuated bulb.
- After expansion into the vacuum, use a second particlein-a-box model to represent the system.

- These models can be used to explain why this expansion isSpontaneous.

- The term "entrapment" was introduced by a German physicist in 1854, before Boltzmann presented his equation.

- The ideal heat engine can achieve the maximum efficiency for converting heat into work.

- The process of changing the system variables by infinitesimal amounts is referred to as a reversible process.

- The subscript states that we are considering a reversible process.

- The method of integration used to get the summation of all these infinitesimal quantities is mathematical.
- See Are You Wondering.

- Section 13-1 has ideas that can be used to rationalize equation (13.2).
- The more energy added to a system, the greater the number of energy levels available to the particles.

- The increase in the number of energy levels at low temperatures is greater than the increase in the number of energy levels at high temperatures.

- Imagine the change from state 1 to state 2 is carried out in a series of infinitesimal steps.

- C/S can be related to other properties.
- The equation (13.6) describes C/S in terms of gas volumes for the isothermal expansion of an ideal gas.

- We must come up with a way to accomplish a given change in a different way.
- It can be done fairly easily for certain types of changes, even though it is not always easy.

- To carry out the vaporization in a vap reversible way, heat must be delivered in infinitesimal amounts.

- The amount of heat used to convert a small amount of liquid to an equivalent amount of Vapor at a constant temperature is infinitesimal.
- The heat is used to overcome the intermolecular forces of attraction, not to cause an increase in temperature.
- The total change for the vaporization is obtained by adding the values of dS and dq.

- The symbol tr represents a transition.
- For example, if we replace tr with fus for the melting of solid or vap for the vaporization of liquid, we know which transition is involved.

- If the phase transition occurs at the normal or normal transition temperature, the equation applies.
- A phase transition can only be carried out at the normal transition temperature.

- Consider the melting of ice at its normal melting point, which is 6.02 kJ mol-1.

- Entropy changes can be expressed on a per-mole basis.

- A useful generalization known as Trouton's rule states that for many liq and normal boiling point are uids at their normal boiling points.

- If the increased accessibility of energy levels produced in transferring one mole of molecule from liquid to vapor at 1 bar is the same for different liquids, then we should expect the same values.

- Instances in which Trouton's rule fails are understandable.
- The lower the entropy in the liquid state, the lower the hydrogen bonding among molecule will be.

- The increase in the vaporization process's entropy is greater than normal.

- Although a chemical equation is not necessary, writing one can help us see the process we use to find the value.

- We should check the sign of C/S when we solve this problem.
- Here, we expect an increase in the entropy because it is higher than that of a liquid.

- Its normal boiling point is -29.79.

- The transition from solid rhombic sulfur to solid monoclinic sulfur at 95.5 degC has an entropy change of 1.09 J mol-1 K-1.

- If you are familiar with the stant pressure heat capacity of the substance being heated.
- dS is the corre calculus technique of integrasing entropy change.

- In this chapter, we will discuss the nature of chemical reactions.
- When a reaction involving ideal gases is assumed to occur at a constant temperature, the amounts and partial pressures of the gases will change.
- We want to know how the ideal gas's entropy changes with pressure.

- We need to make sure that the external pressure is different from the gas pressure in order to expand or compress a gas.
- The following process shows the expansion of a gas.

- Any one of the infinite number of intermediate states is represented by this step.

- PdV L is 1 * 10 - 10 bar L.

- The infinite number of L - 1 * 10 - 10 bar L is PdV.

- Since the temperature is constant, we can write on it.

- If the volume increases or the pres are not sure, we can use the equation (13.8), f 6 Pi.
- As we saw earlier, the number of microstates and entropy increase as the space available we must make use of the first to the particles increases.

- C/H depends on C/T.

- The ideas of Ludwig Boltzmann are used to do this.
- Allow the gas to expand to a final volume Vf if you want it to be an ideal gas.

- Si and Wf are the number of microstates for the initial and final states of the gas, respectively.
- We need to get a value for the ratio.
- If there is only a single gas molecule in the container, you can do that.
- The volume of the container should affect the number of microstates available to this single molecule.

- The isothermal expansion of an ideal gas is given by equation (13.6).

- The following constant pressure process has an entropy change.

- Cp,m 75.3 J mol K.

- The enthalpy of fusion for ice is 6.01.

- We need to think of a way to carry out this process reversibly so that we can use the equations given in Table 13.1.

- Several changes occur during this process.

- The ice is heated from -10 to 0 and then melted.
- The water is heated from 0 to 10 degrees.
- The total entropy change can be calculated by using equations from Table 13.1 and adding them together.

- 100 g mol-1 is equal to 5.55 mol.

- The increase in temperature is expected to result in a positive overall entropy change.
- H O is converted from a solid state to a liquid state.

- The heating of water is much smaller.
- Even though the temperature increase is the same for both, the entropy change is more for the heating of water than it is for ice.

- One mole of neon gas, initially at 300 K and 1.00 bar, expands adiabatically against a constant external pressure of 0.50 bar until the gas pressure is also 0.50 bar.
- The heat capacity of Ne(g) is 20.8 J mol-1 K-1.

- One mole of neon gas expands against a constant external pressure of 0.50 bar until the gas pressure is also 0.50 bar.

- Spontaneous change is called Entropy and Gibbs Energy.
- The temperature of the system should be lowered to smaller and smaller values.
- The total energy available to the particles decreases when the temperature is lower.
- The particles are forced to occupy lower and lower energy levels until they are in the lowest energy level.

- The perfect crystal's entropy is zero.

- Scientists had already deduced the third law of thermodynamics by studying isothermal processes at very low temperatures before Boltzmann presented his equation.

- Clausius's approach to entropy provides equations for calculating the change in the entropy of substances that are heated or cooled.
- We can use these equations and the fact that S is 0 at T to assign a specific value to the entropy of any substance.

- The heat capacity of each phase changes.

- The assumption that Cp is constant is not valid.

- The temperature variation of the 4 is taken into account when calculating X(l, Tvap, 1 bar) C/Sdeg3 L Cp(liquid) ln.

- If the enthalpy changes and transition temperatures are known, the entropy changes can be calculated.
- If the heat capacity of each phase is known, the entropy can be calculated.
- Remembering that Sideg is a known quantity, we have Sfdeg as C/S.

- An extrapolation is required because experimental methods cannot be carried to that temperature.

- Consider the equation for a reaction.

- The following equation has a familiar form and is used for the reaction.

- Calculating the Standard Entropy of Reaction Use data from Appendix D is an example.

- The standard reaction entropy, C/rSdeg, is calculated from standard molar entropies.

- A check on this calculation can be done with some qualitative reasoning.
- If three moles of reactants are converted to two moles of products, the system's entropy should decrease.
- We should expect C/rSdeg 6 0.

- Data from Appendix D can be used to calculate the standard reaction entropy for the synthesis of ammonia from its elements.

- N2O3 is an unstable oxide that can easily be broken down.
- The standard reaction time for the decomposition of N2O3 to nitrogen monoxide and nitrogen dioxide is 25 minutes.

- At low tempera, 13-5 uses the standard molar entropies of NO and NO.

- When a gas absorbs energy, some of the energy goes up and the quanta of the molecule goes down.
- There are other ways to use energy.

- As the number of atoms per molecule increases, we can say that Standard molar entropy increases.

- The movement of atoms is suggested by arrows.
- The difference between NO21g2 and NO(g) helps account for the fact that NO21g2 is greater in entropy than NO(g).

- The answer to this question is yes, if we focus only on substances that are solid at 298.15 K, and we know the amount of heat required to raise the temperature of the solid from 0 K to 298.15 K at a constant.
- There is a plot of standard molar entropy for several monatomic solids.
- The enthalpy change per mole of solid when it is heated from 0 to 298.15 K at a constant pressure equal to 1 bar is represented by C/Hdeg.

- The result shows that the standard entropy value is propor tional to the heat absorbed by that solid to get it from 0 to 298.15 K. The heat that is absorbed by the solid in this process is dispersed through the various energy levels of the solid.

- The amount of energy stored in the solid is measured.

- In Section 13-1, we came to the tentative conclusion that processes in which the entropy increases should be spontaneously and that processes in which it decreases should be nonspontaneous.
- It can be difficult to explain the freezing of liquid water at -10 degrees.

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