CHEMISTRY COMMUNITY

Devin Patel 2D

K is the equilibrium constant while k is the rate constant in kinetics. Furthermore, K = k/k'. k' being the rate constant for the reverse reaction and k being the rate constant for the forward reaction.

Devin Patel 2D

Yes! A catalyst lowers the activation energy for both the forward and reverse reactions. This is why the equilibrium constant doesn't change with a catalyst because the rates of both reactions are equally affected.

Devin Patel 2D

Since the intermediates are consumed during the reaction, we never include them in the rate law. However, let's say an intermediate is a reactant in the slow step. We then have to evaluate where this intermediate came from and use that in the rate law.

Devin Patel 2D

The enthalpy of formation is more accurate compared to using bond enthalpies because the bond enthalpy values given are the average bond enthalpies. These averages are just estimates so they aren't the most accurate. I believe Hess's law is as accurate using enthalpy of formation.

Devin Patel 2D

Technically, the correct unit is bar, however, atm is very close to bar, so we can use both atm and bar when working with partial pressure.

Devin Patel 2D

I've always seen concentrated denoted by brackets, especially in equations and formulas. In problems and questions, concentration is often presented in molarity, or moles/liter. However, yes, when it comes to denoting it in formulas it's usually always in brackets as far as I know.

Devin Patel 2D

For first-order and zero-order reactions, the slope will be -k and for a second-order reaction, the slope will be +k.The reason the slope is -k for first and zero-order reactions is that the concentration of the reactants decreases over time, but the slope is positive for second-order reactions beca...

Devin Patel 2D

Essentially, at t = t1/2, [A] = (1/2)[A]0. Then when deriving the equation for the half-life, ln(1/2*[A]/[A_0]) = -kt_1/2, we would have to find ln(1/2), which equals -0.693. After canceling out negatives on both sides we have the value of 0.693.

Devin Patel 2D

In terms of mathematics, the zeroeth rate law is essentially Rate = k. This is because the exponent correlated to the concentration of the reactant in the rate law, Rate = k[A]^n, would be zero and [A]^0 = 1. On a graph of [A] vs time, there would also be a straight line with a negative slope if it ...

Devin Patel 2D

Yes, the rate constant can absolutely be altered. One of the ways you can change the rate constant is by altering the temperature. The rate constant can also be changed by altering the activation energy. This can be done by adding a catalyst enzyme (which would lower the activation energy) or by rem...

Devin Patel 2D

So to find delta G, you would need E∘ = (RT/nF)ln(K) and ΔG∘ = -nFE∘. When taking the natural log of a number less than 1, the result is a negative number, and when taking the natural log of a number greater than 1, the result is positive. A K value of less than 1 would result in a negative value th...

Devin Patel 2D

For me, the lectures are helpful, but they may not be to everyone, and so I advise doing what I'd do when the lectures are a bit confusing. The textbooks are always super helpful and many youtube videos are also very helpful, especially for conceptual understanding. A good thing is that most of thes...

Devin Patel 2D

Once you balance the redox reaction or even balance the half equations, you should be able to see the coefficient in front of the electrons or see how many electrons are transferred. This number would be the n in the NFE equation.

Devin Patel 2D

If the E naught value is positive then yes, the redox reaction is spontaneous and favorable. This is true because when E naught is positive, G naught would be negative, thus showing that it is spontaneous and favorable.

Devin Patel 2D

Yes, I agree with the previous poster. It'd be a great idea to look at the standard reaction potentials and the ones with the higher reduction potential will have a greater ability to act as an oxidation agent.

Devin Patel 2D

I also agree with previous posters that internal energy for an open system would be deltaU = q + w. However, I think this situation is pretty confusing because of all the things to take into account for an open system such as adding in mass or taking mass out. I am not knowledgeable on how this chan...

Devin Patel 2D

This is not on our equation sheet but it can be deduced. This formula works in constant volume conditions because normally deltaU = q + w. In constant volume, w=0, so deltaU = q. q=nCvdeltaT, therefore deltaU = nCvdeltaT.

Devin Patel 2D

Delta H = q at constant pressure or in an isobaric condition. In this condition then Delta U = Delta H + w.

Devin Patel 2D

I'm fairly positive that you only consider the moles of Gas. What makes me believe this is that we would be using the Ideal Gas Constant and this doesn't apply to liquids or solids, or aqueous solutions. I believe in the textbook problems they also only consider moles of gas.

Devin Patel 2D

Yes, so when you do P x deltaV the units are in (atm)(L). SO to get it into Joules you need to multiply that answer by 101.325 because 1L.atm = 101.325J.

Devin Patel 2D

delta H and q are the same when pressure is constant. Essentially, qp = delta H, when the change in pressure is zero.

Devin Patel 2D

I believe that temperature is always in Kelvin for calculations involving Gibbs Free Energy.

Devin Patel 2D

This would not be possible. When looking at the equation: ∆G = ∆H-T∆S, we can see that by having a negative ∆S, the negatives cancel out thus making the positive change in enthalpy add the temperature times the change in entropy. Thus ∆G could only be positive and non-spontaneous.

Devin Patel 2D

We can relate Gibbs free energy and the Equilibrium constant through some equations. One of which is delta G naught = -RTlnK. From here we can make inferences about their relation. We just recently learned this today in a lecture and I think it answers your question.

Devin Patel 2D

Delta G is the change in Gibbs Free energy and this is the change for a system during any point in a reaction except at 298K and 1 atm, which is the standard conditions. In these standard conditions, the change in Gibbs Free energy is delta G naught. It is important to note this difference because w...

Devin Patel 2D

I agree with the previous post, I believe that the heat comes from the surroundings, especially in the diagram he provided during the lecture. However, I'm not entirely sure where the heat would come from in a different scenario, as I don't want to assume that it always comes from the surroundings s...

Devin Patel 2D

This integral equation is used when there is a reversible pathway, however, it is not necessary to use this exact equation or formula. We learned in the lecture that we could use a much easier equation for the reversible pathway which is: w = -nRTln(V2/V1). This equation is definitely one that is ea...

Devin Patel 2D

I think you are right on that. Due to O2 having only one type of arrangement, the entropy is 0. Thus, for molecules with only one type of arrangement, the entropy could be 0, because ln1 = 0. Therefore, for diatomic molecules that have two of the same atom, for example, O2, Cl2, or H2, there is only...

Devin Patel 2D

If a reaction is spontaneous, that means it can occur without an input of energy, and thus it is favorable. An equation is usually spontaneous when a reaction is exothermic and has high entropy. This is in relation to Gibb's free energy which I'm sure we will go over soon.

Devin Patel 2D

The change in internal energy is able to remain the same because the energy lost due to work is replaced as energy in the form of heat. This creates no change and works when you think of the universe as an isolated system. The first law of thermodynamics states that energy cannot be created or destr...

Devin Patel 2D

Percent protonation is usually calculated when dealing with bases and the Kb constant, and percent deprotonation would be when dealing with acids and the Ka constant. This is because acids can give H+, thus deprotonated, and bases will accept the H+ and get protonated.

Devin Patel 2D

I was confused about this problem as well for the longest time but then I realized they were asking for the answer in Joules and we have to convert it so it's in joules. You'd just have to multiply your answer by 101.325 J/atm. I didn't really know we had to do this based on the content of the lectu...

Devin Patel 2D

So I was looking at the solution for this question, it advises me to use this formula to find work which is: w = -nRTln(Vfinal/Vinitial). I don't remember learning this in the lecture and I used the formula with the integral instead, but it wasn't correct. I was wondering why the integral formula we...

Devin Patel 2D

I'm pretty positive we learned q=mC(Delta T) on Monday's Week 4 lecture. There were quite a bit of sapling questions on this topic so going over Monday's lecture would really help because it is almost entirely about this topic.

Devin Patel 2D

Since the K value is extremely small and much smaller than 10^-4, we can assume that the x is insignificant. However, only remove the x for the denominators in this case, because we still need to find the actual x value. The x value will just be extremely small.

Devin Patel 2D

Catalysts would not affect the change in enthalpy, whether the reaction is exothermic or endothermic. Because enthalpy is a state function, the path taken isn't relevant to delta H. Catalysts affect only the path taken by lowering the activation energy. So, although the reaction would proceed quicke...

Devin Patel 2D

q is the denotation for heat (in joules), which is not a state property, and H is the denotation for enthalpy, which is a state property.

Devin Patel 2D

An endothermic reaction will always have a positive delta H because, for a reaction to endothermic, it absorbs heat. An exothermic reaction will always be a negative delta H because, for a reaction to be exothermic, it gives off heat.

Devin Patel 2D

Hi,
So on professor's website, under the solutions manual errors document, it says that this question has an error in the book. H2 is supposed to be a product, which makes the moles on each side equal. Hope this helps!

Devin Patel 2D

I believe that if the diatomic molecule is in their stable state in standard conditions, then the standard enthalpy of formation will be zero.

Devin Patel 2D

If you know for certain that you're working with strong acids and strong bases, you don't necessarily need the ICE box because they fully disassociate. So I would use an ICE box for weak acids and bases because they only partially disassociate and it would also help visualize those calculations bett...

Devin Patel 2D

Hi, instead of multiplying by 2, they divided by 0.500. This is essentially the same thing and looked confusing to me at first but then I realized. Hope this helps.

Devin Patel 2D

This is a great question. I came from Chem 20B and we never really went over acids and bases, and we definitely didn't go over which were strong or weak and how to find out. Thanks for bringing this to my attention so that I can memorize the strong ones.

Devin Patel 2D

In the lecture, I believe we were informed that if the K value is less than 10^-3, we can assume that x is negligible. However, as many others have stated, just to be safe you can use the 5% rule.

Devin Patel 2D

Hi! Without giving the answer away, what I did to start was to set up the equilibrium constant ratios for every reaction. For example, the first one would be [HI]^2/[H][I]. I did this for all of the reactions and then noticed which ones would have similar components to the original reaction given to...

Devin Patel 2D

Yes, Q is the reaction quotient and is essentially the ratio of products/reactants whenever the reaction is not at equilibrium. K is the ratio when the reaction is at equilibrium. The reaction quotient can help us figure out if the reaction is not at equilibrium and can inform us whether the reactio...

Devin Patel 2D

Hi, so you would use usually use partial pressure for gases and when you are solving for Kp and concentrations for solutions and when you are solving for Kc. Or, you can convert partial pressures to concentration and vice versa.

Devin Patel 2D

A forward reaction is when the reaction is forming product or in nontechnical terms, going towards the right, towards the product. A reverse reaction is vice versa, as it is when the reaction is forming reactant. So, when a reaction favors products, it means that the concentration of products is gre...

Devin Patel 2D

Hi, like many others have stated, doing practice problems for me are very helpful. It gives me a chance to apply the knowledge we learn and it is very helpful for exams. I also plan to space out my learning. I've been guilty of cramming my studying in the final days before an exam and it can be pret...

Devin Patel 2D

Hey everyone, I know this has been asked and answered many times but I'd still like to get clarification on this as I've never experienced this before. Last quarter I took Chem 20A and we had no extra resources outside of class. This class has many and so I'm a bit confused and overwhelmed by the di...

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