## Search found 62 matches

- Sat Mar 17, 2018 1:53 pm
- Forum: *Enzyme Kinetics
- Topic: Final-Collision Theory
- Replies:
**3** - Views:
**236**

### Re: Final-Collision Theory

Can someone please explain the difference between the collision theory and the transition state theory? Collision theory states that molecules need to collide with the correct orientation and with enough kinetic energy for a reaction to occur. Transition state theory builds on the collision theory....

- Sat Mar 17, 2018 1:41 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: S and delta S
- Replies:
**4** - Views:
**237**

### Re: S and delta S

You can find the absolute entropy by using the Boltzmann equation S=k

_{B}*lnW. k_{B}is Boltzmann's constant (1.38x10^{-23}J/K).- Sat Mar 17, 2018 11:27 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Catalysts
- Replies:
**2** - Views:
**89**

### Re: Catalysts

Catalysts are not included in the overall balanced reaction because they are present in the beginning and regenerated at the end. If you think about it, the catalyst would cancel out. Also, catalysts participate in the reaction by lowering the activation energy, but they are not necessarily reactant...

- Fri Mar 16, 2018 12:11 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Negative Ea
- Replies:
**3** - Views:
**106**

### Re: Negative Ea

The activation energy cannot be negative, but it can be decreased with the addition of a catalyst as it provides an alternative pathway between reactants and products.

- Fri Mar 16, 2018 12:08 am
- Forum: *Organic Reaction Mechanisms in General
- Topic: Functional Groups Overlapping
- Replies:
**2** - Views:
**472**

### Functional Groups Overlapping

Is it possible for functional groups to overlap? For example, can an aldehyde contain part of an amine?

- Sun Mar 11, 2018 12:43 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.63
- Replies:
**2** - Views:
**92**

### Re: 15.63

For the problem, you would just plug in the values given into the equation ln(k

_{2}/k_{1})=E_{a}/R((1/T_{1})-(1-T_{2})). The activation energy isn't negative, so it would just be ln(k_{2}/(1.5x10^{10})=38/.008314((1/298)-(1/310)).- Sun Mar 11, 2018 9:25 am
- Forum: *Organic Reaction Mechanisms in General
- Topic: Activation Energy and Temperature
- Replies:
**3** - Views:
**169**

### Re: Activation Energy and Temperature

Yes, these conditions are possible. Having a high temperature just means you're more likely to reach the activation energy, though it may be higher. The effect these two conditions would have on the rate constant, in terms of increasing or decreasing it, would depend on the magnitude of each condit...

- Sun Mar 11, 2018 9:21 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: reversible/irreversible
- Replies:
**2** - Views:
**78**

### Re: reversible/irreversible

There are a couple formulas for work that differ based on if a reaction is undergoing reversible or irreversible expansion. If the reaction is reversible, then use w=-nRTln(V final /V initial ), where T is the absolute temperature. When a reaction is irreversible and the external pressure is constan...

- Sun Mar 11, 2018 9:10 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: figure 15.27 interpretations
- Replies:
**1** - Views:
**48**

### Re: figure 15.27 interpretations

For figure 15.26, the reason that there is a peak followed by a sudden drop is that at low temperatures, there are not many molecules that have a high enough kinetic energy to collide and react. Therefore, instead, there is a higher fraction of molecules towards the beginning of the plot of kinetic ...

- Mon Mar 05, 2018 9:59 pm
- Forum: General Rate Laws
- Topic: rate constants
- Replies:
**7** - Views:
**188**

### Re: rate constants

Kayla Tchorz-Dis 1F wrote:I'm also a little confused on the concept of K. In our notes it says K is dependent on temperature and activation energy. So if it's dependent, how is it a constant?

The rate constant is constant in that it won't change under those specified conditions.

- Mon Mar 05, 2018 9:57 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Determining overall rate order
- Replies:
**3** - Views:
**105**

### Re: Determining overall rate order

Sometimes the presence of products does affect the reaction order, but I believe that we aren't really going to deal too much with that in this class. An example of this can be found on page 621 of the textbook in case you wanted to know more!

- Wed Feb 28, 2018 8:19 pm
- Forum: Second Order Reactions
- Topic: Half Life
- Replies:
**2** - Views:
**104**

### Half Life

I have a general question about the half life of reactions. Why does the half life of a zero order and second order reaction depend on the initial concentration of the reactant, but the concentration of a first order reaction does not? Can someone explain this conceptually?

- Wed Feb 28, 2018 8:27 am
- Forum: First Order Reactions
- Topic: Order of a reaction
- Replies:
**3** - Views:
**112**

### Re: Order of a reaction

To add on, the order of a reaction reveals to us what power of the concentration the rate is proportional to. For example, in a first-order reaction, its rate is proportional to the first power of the concentration. Also changing the concentration of different reactants has different effects on the ...

- Sun Feb 25, 2018 10:55 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Order
- Replies:
**5** - Views:
**163**

### Re: Order

The textbook briefly talks about fractional order reactions (page 621). I believe that if the fractional order is in the denominator, it means that the reaction slows down as the concentration of it increases. On the other hand, if the fractional order is in the numerator, the reaction rate increase...

- Sun Feb 25, 2018 10:48 pm
- Forum: General Rate Laws
- Topic: rate constants
- Replies:
**7** - Views:
**188**

### Re: rate constants

In addition to what is stated above, the rate constant k is independent of the concentrations of the reactants. You would multiply k by the concentrations to find the rate, but the concentrations themselves do not impact k.

- Sat Feb 24, 2018 9:17 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Order
- Replies:
**5** - Views:
**163**

### Re: Order

To add on, the order also reveals something about the rate of reaction. For example, a first-order reaction is one in which the rate of it is directly proportional to the concentration of the reactant concentration. A second-order reaction's rate is proportional to the second power of the concentrat...

- Wed Feb 21, 2018 11:44 pm
- Forum: General Rate Laws
- Topic: NO2 Example from Lecture
- Replies:
**2** - Views:
**94**

### Re: NO2 Example from Lecture

The equation is -d[NO2]/dt = d[NO]/dt = 2 * d[O2]/dt because as you stated, the rate at which NO2 concentration decreases is equal to the rate at which NO concentration increases. In addition, I believe that because O2 concentration increases half as fast, 2 x the rate of oxygen concentration increa...

- Wed Feb 21, 2018 11:23 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: mistake in Ch 14. #35.b?
- Replies:
**1** - Views:
**69**

### Re: mistake in Ch 14. #35.b?

Yes, this error is included in the Errors in Solutions Manual document posted on Lavelle's website.

- Tue Feb 20, 2018 6:44 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: HW 14.9
- Replies:
**2** - Views:
**63**

### Re: HW 14.9

To get the number of mols of electrons, you write a half reaction for both the oxidation and reduction. In 14.9a, for example, 2Ce 4+ (aq) is reduced while 3I - (aq) is oxidized. The two half reactions: 2Ce 4+ (aq) + 2e - --> 2Ce 3+ (aq) 3I - (aq) --> I 3 - (aq) + 2e- From this, you can see that 2 e...

- Mon Feb 19, 2018 3:36 pm
- Forum: Balancing Redox Reactions
- Topic: 14.1 chromium
- Replies:
**1** - Views:
**53**

### Re: 14.1 chromium

Each O is usually 2-, and there are 7 of them. 7 x 2- = 14-. The overall charge of the molecule is 2-. The difference between the overall charge and the oxygens is 12+. Each Cr is therefore 6+.

- Mon Feb 19, 2018 3:31 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15b [ENDORSED]
- Replies:
**1** - Views:
**82**

### 14.15b [ENDORSED]

This problem asks for the half-reactions and the cell diagram of H

^{+}(aq) + OH^{-}(aq) --> H_{2}O(l), the Bronsted neutralization reaction. Where did the O_{2}in both the half reactions come from (in the solutions manual)?- Sun Feb 18, 2018 8:48 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Nernst Equation
- Replies:
**4** - Views:
**160**

### Re: Nernst Equation

In addition to what was stated above, the Nernst equation allows us to estimate the potentials of cells under nonstandard conditions.

- Sun Feb 18, 2018 8:40 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: inert metals
- Replies:
**1** - Views:
**70**

### Re: inert metals

I believe that the inert metal used is usually Pt(s) or C(gr). Inert metals are only necessary if the reactants do not act as electrodes.

- Sat Feb 17, 2018 7:12 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidizing & Reducing Agents
- Replies:
**6** - Views:
**151**

### Re: Oxidizing & Reducing Agents

The species that causes oxidation is the oxidizing agent. It accepts electrons released by the species being oxidized. The oxidation agent in a redox reaction gets reduced. The species that brings about reduction is the reducing agent. It supplies electrons to the species being reduced. The reducing...

- Sat Feb 17, 2018 8:54 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Electrolytes
- Replies:
**1** - Views:
**57**

### Electrolytes

In a galvanic cell, what exactly constitutes an electrolyte? Is it the solution the electrodes are in?

- Tue Feb 13, 2018 7:17 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: S=klnW
- Replies:
**7** - Views:
**777**

### Re: S=klnW

You would use this equation if the question asked you to calculate absolute entropy (instead of the change in entropy) of a system.

- Mon Feb 12, 2018 9:41 am
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: General Question [ENDORSED]
- Replies:
**1** - Views:
**105**

### Re: General Question [ENDORSED]

Larger molecules tend to have greater entropy than smaller ones because heavier molecules have more vibrational energy states. In this way, the molecules have a greater number of possible positions they can take up. Using the formula S=k*lnW with W being the number of ways the atoms or molecules in ...

- Fri Feb 09, 2018 11:52 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Homework Question 9.13
- Replies:
**4** - Views:
**174**

### Re: Homework Question 9.13

But how do you know that "for a change in temperature, the energy supplied as heat is related to the increase in temperature by the heat capacity, C". Is that always true? The definition of heat capacity is the ratio of the heat supplied to the rise of temperature produced: C= q/ \Delta T...

- Fri Feb 09, 2018 11:14 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: work
- Replies:
**2** - Views:
**113**

### Re: work

There are two types of work a system can do. There is nonexpansion work and expansion work. Expansion work is the work that concerns a change in the volume of a system. The equation relating work to volume is w=-P ex \Delta V, but this only applies when external pressure is constant. When external p...

- Fri Feb 09, 2018 11:02 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Homework Question 9.13
- Replies:
**4** - Views:
**174**

### Re: Homework Question 9.13

I think you actually mean R is used when volume changes and C v when temperature changes. The equations in which these values are plugged in are both derived from others. On pages 321 and 323, the derivations are explained for the equation for entropy change of a system with temperature change and o...

- Fri Feb 09, 2018 11:51 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Question 9.21
- Replies:
**4** - Views:
**137**

### Re: Question 9.21

This question wants you to use your knowledge about the possible states/number of ways the molecules can be arranged to calculate the overall entropy. This can be done using the equation S=k*lnW, where k is Boltmann's constant: 1.381 x 10 -23 J/K and W is the number of ways that the molecules can be...

- Fri Feb 09, 2018 8:25 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.35 Explained
- Replies:
**2** - Views:
**86**

### Re: 9.35 Explained

This problem is also on the solutions manual error sheet, where there is an explanation. The actual answer is that the change in entropy of B is less than that of C. However, The change in entropy of C is equal to the change in entropy of A. (B<C=A)

- Thu Feb 08, 2018 10:10 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Temperature and Spontaneity
- Replies:
**2** - Views:
**91**

### Temperature and Spontaneity

Is it true that when H and S are both negative that decreasing the temperature increases spontaneity? If so, could someone explain why this is the case? I know it has to do with the equation G=H-TS, but beyond the math, what is the explanation?

- Tue Feb 06, 2018 5:17 pm
- Topic: 9.21
- Replies:
**3** - Views:
**102**

### Re: 9.21

K

_{b}is Boltzmann's constant, so it is 1.381 x 10^{-23}J/K. It doesn't change from problem to problem. Hope this helps!- Mon Feb 05, 2018 5:54 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.17
- Replies:
**10** - Views:
**268**

### Re: 11.17

The problem is: Calculate the reaction Gibbs free energy of N 2 (g) + 3H 2 (g) \rightarrow 2NH 3 (g) when the partial pressures of N 2 ,H 2 , and NH 3 are 4.2 bar, 1.8 bar, and 21 bar, respectively. For this reaction, K=41 at 400. K. I used the equation \Delta G r = \Delta G r (standard) + RTlnQ to ...

- Mon Feb 05, 2018 10:55 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.17
- Replies:
**10** - Views:
**268**

### 11.17

For this problem, my answer was -2736.87 J, which is -2.7 kJ/mol when accounting for significant figures. The solution manual has -27 kJ/mol as the final answer. Is there something wrong with my math that is giving me the incorrect answer? My math looks exactly like what is shown in the manual.

- Thu Feb 01, 2018 9:48 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Total Entropy Change
- Replies:
**1** - Views:
**103**

### Total Entropy Change

The equation for total entropy change is \Delta S total = \Delta S system + \Delta S surr . My question is why doesn't the entropy change of the system always equal the entropy change of the surroundings? I know that only when the total entropy change is positive will the process be spontaneous, but...

- Wed Jan 31, 2018 4:22 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Phase change question demonstrated in Lecture
- Replies:
**4** - Views:
**151**

### Re: Phase change question demonstrated in Lecture

I believe that this just means that when delta H is positive, the reaction is endothermic. This doesn't favor the forward process because you have to add heat to the reactants; it isn't favorable or spontaneous in this fashion. However, it can become favorable when the temperature is high enough.

- Wed Jan 31, 2018 4:14 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.7 vs 9.13
- Replies:
**4** - Views:
**164**

### Re: 9.7 vs 9.13

There is a mistake in the solutions manual for 9.13. So, for both problems you should use C in the equation, not R. If you look at the Solution Manual Errors page from Dr. Lavelle's website, he has it corrected on there. Hope this helps!

- Tue Jan 30, 2018 9:45 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: heat capacity and entropy
- Replies:
**2** - Views:
**64**

### Re: heat capacity and entropy

First, heat capacity is the heat required to raise the temperature of an object by 1 degree Celsius. In general, increasing molecular complexity tends to also lead to higher heat capacities, since larger molecules have more possible ways to vibrate, bend and rotate. The same is true of entropy. More...

- Sat Jan 27, 2018 8:52 pm
- Forum: Calculating Work of Expansion
- Topic: U for Reversible & Irreversible Expansion
- Replies:
**2** - Views:
**93**

### Re: U for Reversible & Irreversible Expansion

Because internal energy is a state function, it is only dependent on the current state of the system. So, the change of the internal energy is the same regardless of if the expansion was reversible or not.

- Sat Jan 27, 2018 4:42 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: equations
- Replies:
**1** - Views:
**101**

### Re: equations

For a change in volume, you would use deltaS=nRln(V2/V1), in which n=moles of gas, R=ideal gas constant, V2=the final volume, and V1=the initial volume. For a change in temperature, you would use deltaS=nCln(T2/T1), in which n=moles of gas, C=molar heat capacity, T2=final temperature, and T1=the ini...

- Tue Jan 23, 2018 8:34 pm
- Topic: 1/22 Lecture (Boltzmann Equation) [ENDORSED]
- Replies:
**1** - Views:
**71**

### 1/22 Lecture (Boltzmann Equation) [ENDORSED]

On the lecture slide explaining the Boltzmann equation, it says "thermodynamic property ('small' error in S)" and "statistical ('large' error in W)." I'm not exactly sure what this is saying. Could someone clarify to me their significance in this context?

- Mon Jan 22, 2018 7:17 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.41
- Replies:
**1** - Views:
**77**

### Re: 8.41

The 7.5 x 10

^{4}comes from the calculation of 400 grams water x (4.184 J/C/g) x (-45). From that I got -7.5 x 10^{4}, but because water is losing heat, you make q_{water}negative, turning the negative value positive (negative x negative= positive). Hope that helps.- Fri Jan 19, 2018 7:51 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase Change Question
- Replies:
**3** - Views:
**100**

### Re: Phase Change Question

For 8.37, you consider the units that enthalpy is reported in (kJ/mol). So, you should divide the heat (measured in kJ) by moles of the substance to find delta H of vaporization.

- Fri Jan 19, 2018 6:13 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.13
- Replies:
**3** - Views:
**105**

### Re: 8.13

The value of q is negative because the problem states that "the cooling system that surrounds the cylinder absorbs 947 kJ as heat." I think this means that the surroundings absorb the heat, and the actual cylinder itself releases it.

- Fri Jan 19, 2018 6:09 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: First law potential energy
- Replies:
**1** - Views:
**97**

### Re: First law potential energy

I believe that the potential energy doesn't change because this is an isothermal expansion of an ideal gas. Even though molecules are moving apart and therefore separation is increasing, the potential energy remains the same because there are no forces between the molecules.

- Fri Jan 19, 2018 5:45 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Confusion About One of the Answers to an Example Done in Lecture
- Replies:
**2** - Views:
**93**

### Re: Confusion About One of the Answers to an Example Done in Lecture

The answer is 49 kJ/degree for both Kelvin and Celsius because both temperature scales are scaled the same. A rise of one degree Celsius is equal to a rise of one degree Kelvin. However, this would not be the case for Fahrenheit. Hope that is helpful!

- Fri Jan 19, 2018 5:34 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.23
- Replies:
**5** - Views:
**298**

### Re: 8.23

The question asks for the heat capacity(C) of the calorimeter and not the specific heat capacity because specific heat capacity is for a certain mass of a sample, which we can't necessarily assign in this problem. The definition of heat capacity is heat supplied/change in temperature. In equation fo...

- Fri Jan 19, 2018 9:28 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: HW Q.8.49 [ENDORSED]
- Replies:
**2** - Views:
**100**

### Re: HW Q.8.49 [ENDORSED]

You use PV=nRT for this problem because of the equation on page 290 of the textbook. delta H = delta U + delta n gas RT where delta n gas =the final number of moles of gas-the initial number of moles of gas. The problem gives you a delta H value you can plug in, and with that, you are able to use PV...

- Wed Jan 17, 2018 4:23 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.31
- Replies:
**3** - Views:
**115**

### Re: 8.31

You would use the equation for heat: q = C x delta T. Plug in the value of (5/2)R for C for part a and (3/2)R for part b. Because these are the molar heat capacities, you also have to convert the grams into moles. Hope this helps.

- Wed Jan 17, 2018 4:16 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bong Enthalpy accuracy [ENDORSED]
- Replies:
**3** - Views:
**110**

### Re: Bond Enthalpy accuracy [ENDORSED]

Bond enthalpies are the least accurate method of calculating reaction enthalpies because they are average values from many different molecules. Bond enthalpies are accurate for diatomic molecules, though. The three methods are Hess's Law, bond enthalpies, and using standard enthalpies of formation, ...

- Wed Jan 17, 2018 4:05 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Bomb Calorimeters
- Replies:
**2** - Views:
**99**

### Re: Bomb Calorimeters

I believe that bomb calorimeters can be considered isolated systems because they do not exchange matter or heat with the surroundings (what is around the calorimeter). Calorimeters are designed to measure the transfer of energy as heat. The heat of the reaction is supposed to be equal to negative he...

- Wed Jan 17, 2018 9:32 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Transfer at Constant Volume/Pressure
- Replies:
**4** - Views:
**134**

### Heat Transfer at Constant Volume/Pressure

Can someone explain to me or clarify the following statement made in the textbook? "At constant volume, the heat transfer is interpreted as delta U; at constant pressure, it is interpreted as delta H." Why is this true? I think I partly understand that when volume doesn't change, no expans...

- Wed Jan 17, 2018 9:28 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: internal energy
- Replies:
**5** - Views:
**140**

### Re: internal energy

Internal energy still is considered a state function despite both heat and work being path-dependent. I believe that because internal energy is not path-dependent (and only state-dependent), it does not matter what combination of heat and work is present to sum to the internal energy.

- Tue Jan 16, 2018 3:38 pm
- Forum: Phase Changes & Related Calculations
- Topic: work, heat, and internal energy
- Replies:
**2** - Views:
**121**

### Re: work, heat, and internal energy

To add on to this, internal energy is the sum of heat and work, as you stated. However, each of these components does not have to be present to calculate internal energy. For example, if no work is being done by the system, then the internal energy is just set equal to heat (U=q). So, no matter what...

- Fri Jan 12, 2018 10:53 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Intensive vs extensive property
- Replies:
**3** - Views:
**124**

### Re: Intensive vs extensive property

An intensive property is one that is independent of the size of the sample. Examples include density or temperature. An extensive property, on the other hand, depends on the size of the sample. Examples of this are volume or enthalpy. Intensive properties might be helpful because they stay constant ...

- Fri Jan 12, 2018 8:50 pm
- Topic: Enthalpy and internal energy.
- Replies:
**1** - Views:
**82**

### Re: Enthalpy and internal energy.

Enthalpy H is related to internal energy of a system. When a system and its surroundings do no work, the heat given off or absorbed by the system is equal to the change in the internal energy of the system. Also, in general, H = U + P*V, where U is internal energy; P, pressure; and V, volume of the ...

- Wed Jan 10, 2018 9:58 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase Changes
- Replies:
**2** - Views:
**131**

### Re: Phase Changes

The temperature part refers to the fact that the enthalpy of sublimation of a substance can only equal the sum of the enthalpies of fusion and vaporization when these two values are measured at the same temperature. Temperature is important to take into account because substances tend to have differ...

- Wed Jan 10, 2018 9:36 pm
- Topic: Most Stable Form for an Element
- Replies:
**4** - Views:
**187**

### Re: Most Stable Form for an Element

I believe the most stable form of an element is, similar to what is stated above, the purest state of the element. The book also states that graphite is the most stable form of carbon at normal temperatures. This relates to the idea that the enthalpy of formation of graphite is zero, while the entha...

- Tue Jan 09, 2018 10:38 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Figure 8.26
- Replies:
**1** - Views:
**103**

### Figure 8.26

In the description that accompanies the heating curve in Figure 8.26, it states that "the heating curve for the liquid is not as steep as for the solid because the liquid has the higher heat capacity." Is this the general trend (that liquids tend to have higher heat capacities than solids ...

- Tue Jan 09, 2018 8:58 pm
- Forum: Phase Changes & Related Calculations
- Topic: Endo and Exothermic Ways to Remember
- Replies:
**28** - Views:
**1668**

### Re: Endo and Exothermic Ways to Remember

This also still sometimes confuses me, and it has since high school. In particular, I had issues with knowing what side of the reaction heat was on. I remember it now as endothermic meaning heat entering (heat is required) and exothermic as heat exiting (heat is a product).