CHEMISTRY COMMUNITY

Lisa Tang 1C

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....

Lisa Tang 1C

You can find the absolute entropy by using the Boltzmann equation S=k_B*lnW. k_B is Boltzmann's constant (1.38x10^-23J/K).

Lisa Tang 1C

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...

Lisa Tang 1C

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.

Lisa Tang 1C

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

Lisa Tang 1C

For the problem, you would just plug in the values given into the equation ln(k₂/k₁)=E_a/R((1/T₁)-(1-T₂)). The activation energy isn't negative, so it would just be ln(k₂/(1.5x10¹⁰)=38/.008314((1/298)-(1/310)).

Lisa Tang 1C

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...

Lisa Tang 1C

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...

Lisa Tang 1C

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 ...

Lisa Tang 1C

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.

Lisa Tang 1C

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!

Lisa Tang 1C

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?

Lisa Tang 1C

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 ...

Lisa Tang 1C

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...

Lisa Tang 1C

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.

Lisa Tang 1C

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...

Lisa Tang 1C

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...

Lisa Tang 1C

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

Lisa Tang 1C

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...

Lisa Tang 1C

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+.

Lisa Tang 1C

This problem asks for the half-reactions and the cell diagram of H⁺(aq) + OH^-(aq) --> H₂O(l), the Bronsted neutralization reaction. Where did the O₂ in both the half reactions come from (in the solutions manual)?

Lisa Tang 1C

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

Lisa Tang 1C

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.

Lisa Tang 1C

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...

Lisa Tang 1C

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

Lisa Tang 1C

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

Lisa Tang 1C

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 ...

Lisa Tang 1C

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...

Lisa Tang 1C

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...

Lisa Tang 1C

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...

Lisa Tang 1C

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...

Lisa Tang 1C

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)

Lisa Tang 1C

Is it true that when $\Delta$ H and $\Delta$ 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 $\Delta$ G= $\Delta$ H-T $\Delta$ S, but beyond the math, what is the explanation?

Lisa Tang 1C

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!

Lisa Tang 1C

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 ...

Lisa Tang 1C

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.

Lisa Tang 1C

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...

Lisa Tang 1C

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.

Lisa Tang 1C

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!

Lisa Tang 1C

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...

Lisa Tang 1C

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.

Lisa Tang 1C

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...

Lisa Tang 1C

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?

Lisa Tang 1C

The 7.5 x 10⁴ comes from the calculation of 400 grams water x (4.184 J/C/g) x (-45). From that I got -7.5 x 10⁴, but because water is losing heat, you make q_water negative, turning the negative value positive (negative x negative= positive). Hope that helps.

Lisa Tang 1C

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.

Lisa Tang 1C

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.

Lisa Tang 1C

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.

Lisa Tang 1C

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!

Lisa Tang 1C

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...

Lisa Tang 1C

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...

Lisa Tang 1C

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.

Lisa Tang 1C

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, ...

Lisa Tang 1C

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...

Lisa Tang 1C

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...

Lisa Tang 1C

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.

Lisa Tang 1C

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...

Lisa Tang 1C

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 ...

Lisa Tang 1C

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 ...

Lisa Tang 1C

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...

Lisa Tang 1C

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...

Lisa Tang 1C

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 ...

Lisa Tang 1C

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).

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