CHEMISTRY COMMUNITY

MariaCassol1L

He means that when you have a reaction with catalysts the rate of reaction depends on the catalyst and not the reactant concentration as you have more reactants than catalysts, so the catalyst is always busy, and no matter how much you increase the reactant concentration by the rate of the reaction ...

MariaCassol1L

That's because the plots of concentration vs. time, for first and second orders are different. So when you simplify the equations you get different definitions for t1/2, so t1/2 = 0.693/k for first order and t1/2 = 1/k[A]0 for second order.

MariaCassol1L

And adding on, since that is a constant, in the integration, you can take it out of the internal and just multiply the constant in the final answer.

MariaCassol1L

So you are simply graphing different equations. So ln[A]= -kt+[A]0 (ln[A] gives a straight line), you can modify that to get [A]=[A]0e^-kt and if you plot that you get a decreasing exponential.

MariaCassol1L

So to find out order from graphs you must compare three graphs, [A] v time, ln[A] v time, and 1/[A] for a certain reactant. Whichever one, is the closest to a linear graph it is the order for that reactant.

MariaCassol1L

So it is unlikely that we will have to plot the graphs (although we do it in 14BL). But one way this can be applied is that the question gives us three graphs. One that is [A] v time, other 1/[A] v time and another ln[A] v time. So the one that is the closest to a straight line will be the order of ...

MariaCassol1L

Another example is an enzyme catalyst in a biological reaction. So basically if you need the enzyme to make the reaction happen, the rate of the reaction will only depend on how long the enzyme takes to make the reaction happen if you have fewer enzymes compared to reactants (so the enzyme is never ...

MariaCassol1L

So if you think about enzymes as what you need to make the reaction happen. So like an oven when you are trying to make a cake. In this example cake batter is the reactant, but the energy required to make the batter become cake is very high, so you can't just leave the batter out and wait for it to ...

MariaCassol1L

Zero-order reactions happen normally when you have a catalyst (e.g enzymes) that can facilitate the reaction, which means that once your catalyst is saturated no matter how much you increase your reactant concentration the rate will be the same. So yes, mathematically, the reactant concentration doe...

MariaCassol1L

The average rate of the reaction represents the rate over a certain time interval while the instantaneous rate of the reaction represents the rate of the reaction at a specific point. It is also important to notice that the instantaneous rate of the reaction will normally decrease over time until th...

MariaCassol1L

so for the average rate of the reactants, (-1/a)dA/dt, the dA/dt will always be negative since the reactant is decreasing as the reaction progresses, but since we are multiplying it by -1/a the overall rate will always be positive.

MariaCassol1L

Ximeng Guo 2K wrote:Is energy barrier equivalent to activation energy?

Yes, the energy barrier is equivalent to the activation energy. In other words, they both exemplify the energy input required to break certain bonds in the reaction.

MariaCassol1L

Yes, it could happen, but it is extremely kinetically unfavorable, therefore it would take a lot of time.

MariaCassol1L

When volume is constant no expansion/compression work is done, remember w= -PdeltaV or w=- deltanRT or for reversible pathways w=-nRTln(v2/V1). So when deltaV=0, w=0 and deltaU= q because deltaU = q+w. And we calculate using q=nCvdeltaT and so delta=nCvdeltaT

MariaCassol1L

You need to distinguish between moles and particles. They all have the same number of atoms, but diatomic molecules will have half the number of particles because the atoms are bound together.

MariaCassol1L

In the problem that you are referring to, they discuss the change in molar entropy due to phase change from liquid to gas. This means that all liquids will have a relatively large change in entropy since they are entering a significantly more disordered phase and we are considering that the gases yo...

MariaCassol1L

It depends on the information that you are given and what you are trying to solve. So if they give you two K values for two different temperatures, that is a clue that the van't Hoff equation can be used. If you are ever unsure on what equation to use write down everything you know and what do you n...

MariaCassol1L

You should use Kelvin in the van't Hoff equation. If you forget just look at the units in the gas constant which are given in kelvin.

MariaCassol1L

The one thing that will give you "true" spontaneity is the Gibbs Free energy of the reaction because it considers both enthalpy and entropy and which temperature that it is occurring in. Remember that elements want to inhabit the lowest energy state possible but want to inhabit more disord...

MariaCassol1L

The delta H of formation will be equal to the delta H of the reaction only when the reaction you are observing is the formation of a compound from elements in their standard states (think O2 and C forming CO2). For when to multiply the moles, just keep a lookout for the units of the delta H (kJ/mol ...

MariaCassol1L

In #20 when Q=K you are at equilibrium, it is where reactions 'want' to be since it is the lowest energy state and there is no net change in the concentration of products or reactants. Therefore, when Q is not equal to K the reaction will go toward K. If Q<K we know that there is an excess of reacta...

MariaCassol1L

For this question you need to pay attention to the manipulations and the units. A + B --> 2D C --> D so you know you want the Ds to cancel and only have A,B,C in your equation so your first step should be to get the Ds at opposite sides. So you reverse the second reaction and multiply dH and dS by -...

MariaCassol1L

One easy way to know which R to use is by looking at the units, so R=0.08205 L atm mol^-1 K^-1 has both a pressure unit and a volume unit, therefore you need an equation that either requires some value(s) with volume and pressure or you want an answer with volume/pressure. R=8.314 J K^-1 mol^-1 has ...

MariaCassol1L

So in this question, we want to find deltaS and we were given a heat capacity and a temperature change therefore we can use: deltaS= nCvln(t2/t1) all you need to do is plug-in the numbers that you were given and find n using PV=nRT. Remember that your temperatures both should be in Kelvin and that R...

MariaCassol1L

Hi, so you can calculate internal energy change in two ways in this problem. One you can find work by finding the volume change by using Ideal Gas law (PV=nRT), find the volume with the initial temp, and then find the volume with the final temp. (work= - change in volume * pressure, but you have to ...

MariaCassol1L

Steam is in a different phase than boiling water, therefore when water enters in contact with cold skin the heat released is the heat that takes to cool water from boiling temp (100 celsius) to skin temperature. However, when steam enters in contact with the cold skin the difference in temperature m...

MariaCassol1L

Think about this as adding heat or taking away heat from the system. So if the reaction is endothermic, it requires heat, so adding more heat to the system means that the system will want to counter it by using up that heat, so the reaction will favor the product (increase K). If the reaction is exo...

MariaCassol1L

Le Chatelier's principle states that the reaction will try to minimize the effects of changes made upon it. So try to think of this in terms of correcting a change. So if you add reactants to a system, you will make more products to counter the addition of reactants. The same thing happens with pres...

MariaCassol1L

Think of this as changes to the system. So if additional heat is added to an exothermic reaction (that releases heat) the reaction will favor the reactants because the reverse reaction of an exothermic reaction is an endothermic reaction, so by favoring the reactants the system will "use up&quo...

MariaCassol1L

They represent the same concept, the equilibrium constant. Kc is just K using concentrations, while Kp is K using partial pressures. I believe the textbook does not differentiate between K and Kp, and just uses K and Kc, so keep that in mind when doing problems.

MariaCassol1L

The only factor that affects K is temperature. So if you change the temperature of a system the reaction you are observing will have another equilibrium constant entirely. Other factors such as pressure and concentration will affect Q, so that the reaction will probably not be at K anymore and will ...

MariaCassol1L

At first, you should use the Ka/Kb value to see if you can approximate if the equilibrium constant is below 10^-4 then we say that x is so small that the impact it will have on the concentration of the reactant is almost zero, so we don't consider it. However, after you finish your calculation you s...

MariaCassol1L

Here is a quick way to sum it up:
-log(x)=y
x=10^(-y)

MariaCassol1L

This is because of log laws. If we assume that Ka x Kb = Kw is true we take the -log of both sides: -log(Ka x Kb) = -log (Kw) log law: log(ab)=log(a) + log(b): -log(Ka)+ (-log(Kb))= -log(Kw) So since pKa=-logKa, pKb=-logKb, pKw=-logKw you can write: pKa+pKb=pKw I believe Dr.Lavelle has some math rev...

MariaCassol1L

When you reverse a reaction that you know K, you can assume that the equilibrium constant for that reaction will be 1/K if the temperature remains the same.

MariaCassol1L

You use 1/Kc when you are looking at the Kc of a reverse reaction. So for example, if they give you Kc=3 for 2C+O2--->2CO @ a given temperature, then 2CO---> 2C+O2 will have a Kc=1/3 at that same temperature.

MariaCassol1L

Hi, I am taking chem 14BL right now and it mostly overlaps with 14A content, so you might need to freshen up on that, but it shouldn't be a problem for you to take it with 14C.

MariaCassol1L

If a reaction is exothermic it releases heat, while if it is endothermic it absorbs/requires heat. When you are forming bonds the reaction is always exothermic because bonds form because they are in a lower energy form than the single atoms, so the element "want" to form bonds. Breaking bo...

MariaCassol1L

Yes, when the temperature is not given we assume it is at 25 celsius. Just keep in mind that the problem you are solving might require temperature in Kelvin so you might need to convert it.

MariaCassol1L

When pressure is decreased the reaction will favor the side with the least moles of gas as it will want to restore equilibrium by increasing the quantity of that molecule(s).

MariaCassol1L

We established that in this course we are using 1bar=1atm

MariaCassol1L

So, would H20 (g) be included?

MariaCassol1L

pKa is the negative log of Ka. Keep in mind that this notation is universal(in chemistry), so when you have p of something that means you are taking the -log of that thing.
So:
pH= -log[H3O+], pOH= -log[OH-], pKb= -logKb etc...

MariaCassol1L

When setting up ICE tables the changes in concentration will be dependent on the stoichiometric coefficient. This is because we want to keep the values balanced. Using an example: N2 + 3H2 ----> 2NH3 You need 1mol of N2 to interact with 3mol of H2 and that forms 2 mol of NH3 and say you start with 4...

MariaCassol1L

Yes, it is safe to assume that [H3O+] and [OH-] are equal as long as the solution is neutral (pH=7), no matter the temperature (just remember that the Kw will have a different value as 10^-14 applies only to 25C). In the case that you are solving for the pH or pOH and are not sure of those values th...

MariaCassol1L

Rather than trying to memorize it, you should just remember that K is a constant and so Q represents the state of the reaction in relation to that constant. The reaction will always "want" to be in equilibrium and so it will favor either products or reactants in order to achieve the K rati...

MariaCassol1L

It should always be in Kelvin if the problem gives you temperature units in Celsius you should convert it before plugging it in.

MariaCassol1L

P= pressure in atm
V= volume in liters
n= number of moles
R= gas constant (8.31446261815324 J⋅K−1⋅mol−1)
T= temperature in Kelvin

MariaCassol1L

Yes, temperature does matter when we are talking about K. Altough you do not directly include temperature when calculating K, if there is a change in temperature the value of K will change for the given reaction (even if you start with the same concentration of reactants and products as you did befo...

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