CHEMISTRY COMMUNITY

Neil DSilva 1L

There was a typo in my original post. It read: - reduction of Fe2+ to Fe3+ (0.77 V) It should have said (and it's updated): - reduction of Fe3+ to Fe2+ (0.77 V) To answer your other questions: The zinc will be oxidized in the anode half-cell and the iron (iii) will be reduced in the cathode half-cel...

Neil DSilva 1L

When you're looking at a rate law, a first order reaction will have an overall order of one (the sum of all the exponents will be 1) and a second order reaction will have an overall order of 2 (the sum of all the exponents will be 2). The exponents can be determined experimentally (for reactions wit...

Neil DSilva 1L

Anytime there is a possibility of the molecule having an isomer, you have to indicate if it's cis- or trans-. Notice that there are two ways you can draw the molecule if cis- or trans- isn't specified (you could draw the fourth carbon going up from the third carbon and it would have the same molecul...

Neil DSilva 1L

You are essentially correct. You basically just look at all the half reactions you can form with the given chemicals and compare them to find the highest combination. You would get the highest combination by taking the half reaction with the largest negative value and the half reaction with the larg...

Neil DSilva 1L

I don't think you labeled the problem correctly. I found this in my course reader in the 2010 Midterm, Q2B (you may want to edit your title). To answer your question: you're right that you have to flip the original equation that you have, but notice the coefficients are different. The original equat...

Neil DSilva 1L

Here's a picture showing you both the E/Z forms of the molecule. I just took the molecule from the answer in the practice final and rotated it on its side to focus on the double bond. The stars indicate the priority molecule on either side of the double bond.

Neil DSilva 1L

I'm pretty sure Professor Lavelle said that both the common names and IUPAC naming are acceptable. So knowing either one should work.

Neil DSilva 1L

Integration is just reversing the process of finding the derivative, so you would add one to the exponent and divide by the new exponent (whereas in deriving something, you multiple by the exponent and then subtract one from the exponent).

Neil DSilva 1L

To go from [A]^-2 to -[A]^-1, you have to integrate it, not derive it.

But you're right, if we were to derive [A]^-2, we would get -2[A]^-3.

Neil DSilva 1L

I'm not sure about your first question. For your second question, you can write a molecular formula from a line structure by going from left to right and writing down what you see. For example, for 2-methylpentane, the molecule starts with a CH3 on the leftmost side, followed by two CH2's (which the...

Neil DSilva 1L

The overall redox reaction given in the problem is 2Fe(s) + 2H_{2}O(l) + O_{2}(aq) \rightarrow 2Fe(OH)_{2}(s) . The way that I was able to determine which half reaction to use was by looking at the products and reactants given and the overall standard cell pot...

Neil DSilva 1L

Yes, we would use the same Greek prefixes used to label multiple substituents. So three double bonds would be indicated by "tri-en-".

Neil DSilva 1L

I'm not sure if it means that the reaction depends only on the activation energy or the catalyst, but the way I picture zero order reactions is like this: Picture a hemoglobin molecule (this is going to be a very simplified picture of the body). Each hemoglobin molecule can only carry four oxygen mo...

Neil DSilva 1L

By definition, energy and matter cannot be transferred into/out of an isolated system, so whatever energy is in the system remains constant. Therefore, if an isolated system has +15 kJ of energy at some point in time, it will have +15kJ at any other time. Yes, the amount of energy would remain const...

Neil DSilva 1L

They have difference numbers of hydrogen atoms attached to the carbons and a different number of double bonds and that's probably the biggest difference, benzene is a completely different class. The number of carbons is just one part the nomenclature, you also have to consider what else makes up the...

Neil DSilva 1L

Martha, I don't think rates can be negative either. If k is always positive and concentrations are always positive, the rate is always positive as well. Correct me if I'm wrong, but even if the rate is for decomposition / consumption and has a negative sign, the rate is still actually positive, beca...

Neil DSilva 1L

No, I believe the stoichiometric coefficients wouldn't change anything in regards to the half-life. For first order reactions, the half-life is only dependent on k (t = ln2 / k). So if we're given the time, that's all we need to solve the problem. To go from 1 whole quantity to half that quantity ta...

Neil DSilva 1L

I think we exclude the Ar because it occurs as a reactant and a product, so it would just cancel out. Molecularity is determined by the number of reactants that take part in an elementary reaction, so even though the Ar is canceled out, it is still used up in the elementary step, so it contributes t...

Neil DSilva 1L

The negative sign is added so that the slope can be negative while k remains positive, so if the slope was -3:
$slope = -3 = -k$
and k is just 3 (still positive).

Just remember, the proportionality constant, k, is always positive.

Neil DSilva 1L

Also, in first/zero order reactions, as Tatyana mentioned, the straight line plot has slope -k, but we have to add the negative sign because k has to remain positive, but the line has a downward slope.

Neil DSilva 1L

Another way to think of it is that reactions rates are always positive and since k is a proportionality constant that relates some given concentration(s) (which are always positive) with the rate (also always positive), it's not possible for k to be negative.

Neil DSilva 1L

The Arrhenius Equation doesn't depend on the order of the reaction. It just shows the temperature dependence of k, so yes, the Arrhenius Equation does apply to reactions of any order.

Neil DSilva 1L

On Prof. Lavelle's Constants and Equations, the last line gives the two relevant equations for this problem. For second order reactions, use \frac{1}{[A]} = kt + \frac{1}{[A]}_{0} and t_{1/2} = \frac{1}{k[A]_{0}} (the sheet uses R instead of A, but I used A because that's what the solutions manual u...

Neil DSilva 1L

Since the logic for your method is the same as the logic used in parts a and b, I believe you could use your method to solve a problem like that. But I believe you mixed up the equation in your description. (\frac{1}{2})^{n} = \frac{[A]}{[A]_{0}} (so the ratio of final concentration to initi...

Neil DSilva 1L

That's just the initial concentration of A subtracted by the change in A.

The change in A is stoichiometrically determined using the change in B (A decreases twice as fast as B increased).

Neil DSilva 1L

Yeah, I didn't think the "unique rate law" was a thing, so I was just checking.

Neil DSilva 1L

The units for the rate of milk spoiling are \frac{[milk]}{hr} . I believe we assume a first order reaction, so rate = k[milk] . Then, the units of k and k' are both \frac{1}{hr} . This is why you can rewrite \frac{k}{k'} as \frac{t'}{t} since k and t (time in hours) are inversely related. We...

Neil DSilva 1L

You're referring to the unique average rate (not unique rate law), right? Other than that, everything else you said seems correct.

Neil DSilva 1L

I'm assuming you're talking about the Cl at the end of the first step, and yes, it should have a negative sign since it's the intermediate chloride ion that the solutions manual references.

Neil DSilva 1L

The value plugged into the rate law is the concentration of $N_{2}O_{5}$ and we get that by converting the grams to moles and dividing by the total volume. The grams given already tells us the quantity of $N_{2}O_{5}$ in the reaction, so we don't need to multiply it by two.

Neil DSilva 1L

The derivation for that equation can be found on page 589 of the textbook. Essentially, you're just subtracting two Arrhenius equations for different T and k values (in the same way we were able to subtract one Van't Hoff equation from another). T and k correspond to one pair of values and T' and k'...

Neil DSilva 1L

Problem 14.47 refers you to Table 14.3 (on page 581) and that might help you understand how to write the rate law for an elementary step. For elementary reactions, we use the coefficients to write the rate law for the step. So in the case of 14.47b, the solutions manual is correct. Since each of the...

Neil DSilva 1L

Another instance where we would use the coefficients to come up with a rate law is if we're given the elementary steps of the reaction mechanism and we're told what the slow step is. But since that's not the case with the example given, again, we don't deal with coefficients.

Neil DSilva 1L

In response to your first question, page 569 states that the rate law can only be determined from experimental data and cannot be determined from a chemical equation. I believe that means that regardless of what the coefficients are, you need to have experimental data to form a rate law and you woul...

Neil DSilva 1L

The overall order of a reaction is determined by adding the powers in a rate law expression. The coefficients are not used in determining rate laws such as those on page 569 ("The rate law for a reaction is experimentally determined and cannot in general be inferred from the chemical equation f...

Neil DSilva 1L

The ln10 comes from a modified version of the first-order half-life equation. Instead of using ln2 (which equals 0.693, the number we see in the half-life equation) which comes from the following analysis: ln\frac{[SO_{2}Cl_{2}]_{0}}{0.5\times [SO_{2}Cl_{2}]_{0}} = ln\frac{1}{0.5} = ln\frac{10}{5} =...

Neil DSilva 1L

If you look at page 56 in the course reader at the example problem with $NH_{4}^{+}$ and $NO_{2}^{-}$ , we divide because we're finding the ratio of Rate 2 to Rate 1 (and this gives allows us to analyze the rates).

Neil DSilva 1L

This is how I visualized finding W for BH2F. There's three possible states and only one molecule, so W = 3.

Neil DSilva 1L

For those who see this post but haven't found the answer, Chem_Mod answered it here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=8&t=5148&sid=fdc826ba084c85c0219e9914a9020715 The phase of water is determined by the temperature at which the reaction occurs. Most reactions occur at roo...

Neil DSilva 1L

It will always be greater than zero and at least 1.

Neil DSilva 1L

If the water and the metal weren't at the same temperature, there would still be heat flow from the metal (higher temperature) to the water (lower temperature), since heat is the energy transferred as a result of a temperature difference.

Neil DSilva 1L

They are basically the same units with different magnitude, so I would assume that either is fine, as long as you have the correct number of significant figures.

Neil DSilva 1L

Skatwal posted a good solution to this problem in response to another student's question here: viewtopic.php?f=130&t=5018 We can determine if work is done by or on a system by looking at the sign of w . If w is negative, the system is losing energy and expansion work is being done by the system....

Neil DSilva 1L

A few of our classmates have already asked this question. One of the original posts is linked here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=76&t=5078&sid=38898720bd519479c315c42e4fa70642 . For the quizzes in the workbook, you may use the appendices in the textbooks. If a similar ...

Neil DSilva 1L

If the number isn't used in the calculations, it isn't used in determining sig figs. We round off using sig figs because the measurements obtained are limited to the accuracy of the tools used to obtain them, so our answers can only be accurate for a certain amount of significant digits. If a number...

Neil DSilva 1L

In the first step, the system is expanding, so energy is lost to the surroundings in the form of work (the energy of the system is decreasing).
In the second step, we can assume the system is being compressed, so energy is being forced into the system (the energy of the system is increasing).

Neil DSilva 1L

The question asks for the enthalpy for the combustion of one mole of butane. The balanced reaction (with all whole number coefficients) is the combustion of two moles of butane, so you divide the enthalpy you get using the balanced reaction by two to get the answer for one mole . Yes, when you balan...

Neil DSilva 1L

Another one of our classmates asked a similar question here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=76&t=5078&sid=4c6a5f2ce3077b9f24614bdee4d51b6d . For the quizzes in the workbook, you may use the appendices in the textbooks. If a similar question is asked on the actual quiz, w...

Neil DSilva 1L

I have no idea what the right answer to this would be, but I think the degeneracy is never going to be zero. If you consider a two-state system, the degeneracy is W=2^{n} . No matter what you plug in for n, you're never going to get to zero. If you plugged in zero, you would get a degeneracy of 1. Y...

Neil DSilva 1L

Bond enthalpies are calculated when all the substances are in the gas phase, but the problem has carbon in the solid phase (as graphite). The atomization of carbon is just the energy required to turn the graphite into gaseous carbon (it's essentially another way of saying sublimation), so all the su...

Neil DSilva 1L

You can have a negative $\Delta$ values. It follows the same sorta thinking as $\Delta H$ or $\Delta V$ . It just means the change was a negative change and the final value is less than the initial value. So, for example, if there are 4 moles of gas in the reactants (instead of 2), $\Delta n = -1$ .

Neil DSilva 1L

The units for work is joules. Using the 8.3145 J/K just allows some of the units to cancel and give us a final answer in joules. When deciding which value of R to use, just look to see what the question is asking and use the value that will allow you to use the least amount of steps for simplicity.

Neil DSilva 1L

The equation for entropy is always written the same way (without a negative sign -- to see why there is no negative sign in the equation, look back in your course reader or review the lecture for how the equation is derived): \Delta S = nRln\frac{V_{2}}{V_{1}} However, entropy can have a negative va...

Neil DSilva 1L

Remember that there are three ways to solve any enthalpy problem. Depending on the situation, you can use Hess' law, enthalpies of formation, or bond enthalpies. In this case, using the enthalpies of formation makes the most sense (since we aren't given multiple equations and their enthalpies to do ...

Neil DSilva 1L

2 C_{8}H_{18} + 25O_{2} \rightarrow 18H_{2}O + 16 CO_{2} is the combustion reaction for octane, but since enthalpies are per mole , the reaction for which enthalpy is calculated is: C_{8}H_{18} + \frac{25}{2}O_{2} \rightarrow 9H_{2}O + 8 CO_{2} . So there's 8 CO 2 produced and you would divide the ...

Neil DSilva 1L

Even though you don't need to find the exact initial volume for this problem, it's helpful to remember that since you're dealing with an ideal gas you can just use PV=nRT and the given initial conditions to find any unknown conditions. In this case, you're given P, T, and n (and you always have R). ...

Neil DSilva 1L

Yes, the specific heat capacity for air should be 1.01 J/C.g (then grams cancels out). This is mentioned in the Solution Manual Errors file linked here: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Solution_Manual_Errors_5Ed.pdf . There are also a few other errors in the solutio...

Neil DSilva 1L

I also found this (http://www.eoht.info/page/Joule%E2%80%99s+second+law) online about Joule's second law. It states that $U = f(T)$ (internal energy is a function of temperature). It follows the same logic as some of the other posts, if you feel like reading more into it.

Neil DSilva 1L

All of this assumes two states. So we have S=k_{b}lnW as our equation. For 1 mole (N A ), W = 2^{N_{A}} \therefore S = k_{b}ln(2^{N_{A}}) We use the log rule to get N_{A}k_{b}ln(2) N_{A} \times k_{b} = R , so we get Rln(2) This is for one mole, so if we have n moles, we just ...

Neil DSilva 1L

It's not necessarily saying that the heat gained by the water is negative. It just means that the heat lost and heat gained are equal and opposite.

Neil DSilva 1L

I'm going to assume you're looking at the solutions manual. -20,000J just comes from the previous step, where the enthalpy per mole of zinc consumed is determined and then multiplied by the number of moles of zinc. 8.5g Zinc releases 20 kJ and this is equivalent to 20,000 when converted to joules.

Neil DSilva 1L

The gases in the four cylinders of an automobile engine expand from 0.22 L to 2.2 L during one ignition cycle. Assuming that the gear train maintains a steady pressure of 9.60 atm on the gases, how much work can the engine do in one cycle? You can follow the same steps as Sample 7.1 on the previous ...

Neil DSilva 1L

\Delta H values are determined by taking the difference in enthalpies of the products and reactants, which are both measured at 298 K (which is a stadard condition, 25^{\circ }C ). You can just assume that is the temperature when given a \Delta H value. (There is a note about this on pg. 261 of the...

Neil DSilva 1L

Justin,
Wouldn't open beakers (in the diagrams) indicate open systems that are exposed to the atmosphere and an implied 1 atm of external pressure?

Even then, I think Skatwal's explanation is the best way to approach this problem. Using $w=-P\Delta V$ just seems too complicated.

Neil DSilva 1L

Hey Martha, Could you clarify how you got your \Delta V values? In part a, you used T_{f} < T_{i} making \Delta V negative (but temperature is constant so wouldn't T_{f}=T_{i} ?). Then in part b, you said \Delta V is positive, since there are more liquid particles than gas particles (how did you com...

Neil DSilva 1L

Problem 7.63 deals with the \Delta H^{\circ }_{c}(H_{2}, g) which is the enthalpy of combustion for one mole of H_{2} . (This is the same equation and enthalpy as the reaction for the formation of one mole of water, but since the problem specifically deals with the enthalpy of combustion, I ...

Neil DSilva 1L

The problem gives you the final reaction (the one we are trying to find \Delta H for). Then it gives you the \Delta H_{c}^{\circ} of two hydrocarbons and hydrogen gas. This value tells us the \Delta H for the combustion reactions for each of the hydrocarbons and/or hydrogen gas. So, for example, whe...

Neil DSilva 1L

R is the gas constant. It can be found online, on the back of the periodic table that came with the course reader, and in your textbook.

Neil DSilva 1L

Recall Dr. Lavelle's example of an endothermic reaction: melting ice. Ice melts because it takes the energy from its surroundings; that's why water gets cooler when it has ice melting in it. The same kind of thing doesn't happen when you combust something; the surroundings aren't getting colder. Com...

Neil DSilva 1L

Additional Note: - Be sure to use the 5% rule to validate your assumption each time you ignore the +X or –X when dealing with the concentrations. There were two other times in this problem where we should have used the 5% rule, but we omitted them for the sake of time. Corrections to the Video: - @ ...

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