CHEMISTRY COMMUNITY

jisulee1C

lavelle said the final will be on CCLE so we would submit the final online on CCLE

jisulee1C

Usually it is written if it is a slow step but if not assuming it's this problem we are looking at the activation energies. The step that takes the most activation energy will be the slowest step because it requires a lot of energy to overcome the barrier.

jisulee1C

There was an email sent out the final will be on CCLE and is 3 hours and 30 minutes (starts Sunday 11:15am and ends 2:45pm, extra 30 minutes for logging on). I'm pretty sure it's just filling out and answering online on CCLE so we won't have to print it out.

jisulee1C

since the slow step is the rate determining step, changing the activation energy of an already fast step will not change the rate because it is not changing the rate of the slow step

jisulee1C

adsorption refers to the physical contact of reactants with the surface of a catalyst in a different state and it is when reactants sit on the surface of a catalyst versus absorption which is when the reactant breaches the surface of the catalyst therefore I would interpret it as being a liquid on t...

jisulee1C

since there are three peaks there are three overall steps in the reaction
for there to be intermediates they have to form in one reaction and be used up in another
therefore there are two intermediates involved in the three step reaction

jisulee1C

a homogenous catalyst is in the same phase or state as the reactants and if it were a homogenous catalyst would absorb being in the same state of matter therefore we wouldn't consider adsorption since there is no contact between substances of different physical states while a heterogeneous catalyst ...

jisulee1C

At the pH of 6 you are assuming that the acetic acid dissolves into a higher concentration of it's conjugate base CHCH3COO- because the pKa (of acetic acid) =4.75 and the conjugate acid has a change of -1.

jisulee1C

An intermediate and catalyst are not the same thing as in an intermediate is produced and used up during the reaction while a catalyst is not used up in the reaction. You can tell if a catalyst is used if the activation energy is lower than before, you are assuming the intermediate steps involve cat...

jisulee1C

It should be given. If not given you can substitute values in to this step's rate law. When the rate law, which now includes these substituted values, matches with the overall rate law, then the rate-determining step has been found. This is assuming that every step is the slow one and finding the ac...

jisulee1C

It is written in the problem whether or not the step is slow or not. If a step is slow you can assume there is a high activation energy, or the forward reaction is not spontaneous and therefore the reverse reaction does not occur for a slow step.

jisulee1C

you can look at the graphs, look at the units for the rate constant, and also determine using a table of concentrations

jisulee1C

electromotive force (emf) is mentioned as the same as the cell potential of a cell (voltage).

jisulee1C

even if both are negative the more negative would still be the anode
if the total Ecell is negative then you can assume that it is not spontaneous in the forward reaction and however may proceed spontaneously in the opposite direction

jisulee1C

termolecular reactions are rare because all three molecules must collide simultaneously for it to occur. the more molecules needed, the lower the probability of the reaction occurring because they would all need to collide with a certain energy, speed, and in a certain direction to cause the reaction.

jisulee1C

Isochoric process = constant volume and isometric means the same thing as isochoric.

jisulee1C

Isothermal means a constant temperature, so we can use the formula ΔS = nR ln(V2/V1).Since entropy is a state function, only the initial and final states matter, not the pathway taken to get there. In this case for irreversible free expansion, the expansion is very fast therefore the surroundings do...

jisulee1C

In the nernst equation: E = Eo -(RT/nF)ln Q since the anode foes on top and cathode foes on the bottom for Q, as the reaction progresses, the concentration of the left (oxidation) anode as the solid electrode dissolves while the concentration in the right (reduction) cathode decreases as the electro...

jisulee1C

Because the two solutions HCl (a strong acid) and NaOH (a strong base) are reacting, the most enthalpy is produced because the answer choice B has both in their highest concentrations (both at 2M). Both strong will create the largest delta T in their strongest concentration because a neutralisation ...

jisulee1C

Once you plot the second order graph using 1/concentration versus time you can find slope which = k.

jisulee1C

at constant temperature and pressure, the two are related through this equation: ∆S=∆H/T (from ∆G = ∆H - T∆S, when ∆G = 0)

jisulee1C

He said for the class pre-equilibrium would be used.

jisulee1C

k (the rate constant) units depend on the order of a reaction (which can be found online). For a first order reaction the units are 1/s, for a second order reaction the units are 1/ms. as for the K (equilibrium constant) there are no units.

jisulee1C

Yes, you need to have a conducting solid metal for both the anode and cathode.

jisulee1C

Increasing the pH would mean that the solution is becoming more basic with more OH- ions. And because the OH- ions are products of the first step of the mechanism (the fast equilibrium step), increasing the pH, increases OH-, which increases the amount of products favoring the reverse reaction there...

jisulee1C

The third step when looking at the illustration is doesn't take a lot of potential energy and therefore is one of the fast steps. However, since the overall reaction rate depends on the slow step as the rate-determining step, since adding a catalyst to accelerate third step doesn't accelerate the sl...

jisulee1C

the order can also be determined by the given rate constant and the units or from a table with given concentrations and rates

jisulee1C

I don't think there will be the same three hour time limit as was scheduled before because he wrote "The three-hour closed-book exam that was scheduled for Sunday is also cancelled."

jisulee1C

In its place will be a short, straightforward, open-book, take-home final posted on my class website. Again, I will email you the link when the take home final is posted on my class website along with instructions. He sent ^^ this in the most recent email so for now it's going to be a take home fina...

jisulee1C

Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst and are purely mechanisms that lower activation energy and are not included in the equation, therefore, they will not be in the rate law

jisulee1C

The slow step is the rate determining step and therefore we use the slow step when determining the rate of the reaction mechanism.

jisulee1C

Changing the mass of the cathode doesn't change the concentration of the cathode solution therefore changing the mass of the electrode would have no change / affect on the cell potential.

jisulee1C

Usually for the reaction mechanism it states whether or not the reaction is fast or slow. If the reaction step is slow than that is the rate determining step for the reaction. In addition you would know that the reverse reaction for the slow step is almost nonexistent while for the fast reaction the...

jisulee1C

I am pretty sure your TA will send an email about picking up test 2.

jisulee1C

Catalysts provide an alternative pathway with a lower activation energy therefore affecting the kinetics of the reaction by speeding up the rate of both the forward and reverse reactions.

jisulee1C

Since most tables give the reduction potential for the half reaction, the higher (more positive) reduction potential determines the cathode which is the species that will undergo reduction.

jisulee1C

the rate limiting step is the slowest step

jisulee1C

You are assuming for this problem that the heat (q) lost by the water is equivalent to the heat (q) gained by the calorimeter (in this case the copper pot?) therefore you can use mc(delta)T=mc(delta)T to calculate the heat lost by the water and find the final temperature of the water.

jisulee1C

A catalyst cannot be an intermediate because intermediates are produced by the reaction and used up in another following step of the reaction mechanism while a catalyst is not used up during the reaction. Because a catalyst is not used up, a small amount can be used to speed up the reaction and cont...

jisulee1C

Lavelle just sent out an email with a link to his week 10 review powerpoint slides.

jisulee1C

I would memorize the strong acids. There are 7 strong acids: chloric acid, hydrobromic acid, hydrochloric acid, hydroiodic acid, nitric acid, perchloric acid, and sulfuric acid. Knowing which acids are strong are important to solve problems for dissociation, Ka, and pH of the solution.

jisulee1C

I think this equation is derived using the Arrhenius equation for two different temperatures but it is not on the objectives sheet as something we need to do. Instead, I would memorize the equation because it is useful and not on the equations sheet.

jisulee1C

Because a catalyst only affects the reaction rate because it provides a separate pathway with a lower activation energy the reaction rates will increase (be faster) while changing the rates (due to kinetics) has no effect on thermodynamics and therefore there is no change for delta H of the reaction.

jisulee1C

I believe all review sessions will be canceled.

jisulee1C

The integrated rate laws are on the constants and equations sheet but you have to know which rate law corresponds to which order reaction. In addition, it also has the half life equations. I think for outline 6 it states that one of the objectives is to ... Know how to derive the differential and in...

jisulee1C

I think Lavelle finished his course outline and the topics so if anything maybe practice problems and explanations will be posted online or on Bruincast. If posted on Bruincast you can usually access the media files through UCLA CCLE.

jisulee1C

The curve utilizes the total number of points that you can earn and factors in the highest student's score but I think it varies each quarter depending on the student's scores on tests, the final, and midterm.

jisulee1C

What calculation do we also have to incorporate when going from a gas to a liquid or vice versa? When calculating q for the reaction you can use the phase change graph and calculate the different q for both the heating and the phase change to calculate the q for the phase change from liquid to gas ...

jisulee1C

it depends on whether the reaction is the elementary rate law. If the reaction is the elementary reaction one can assume that the exponent is the order in relation to the reactant. Therefore if b is squared the it is in second order while if a is to the power of 1 it is in first order. If you cannot...

jisulee1C

Usually the proposed reaction mechanism has steps that are labeled with either slow or fast which can be used to determine which step is the rate determining step and therefore calculate the rate of the reaction.

jisulee1C

It depends on how the problem is given to you. If there is the chart, with different concentrations and rates you can determine the order with respect to a reactant. Or if the rate constant is given in units you can determine the overall reaction order.

jisulee1C

None of the homework questions in this section required you to use the collision frequency equation so I am assuming that being familiar with the concepts is more important.

jisulee1C

After finding the equilibrium constant for the reaction which can be calculated by dividing the forward reaction constant by the reverse rate constant = 0.676. Because the equilibrium constant is less than 1, reactants (the reverse reaction) is favored there this is an endothermic reaction because i...

jisulee1C

To determine which half reaction is the cathode and anode you have to look at the standard reduction potential. For Co2+ reaction the standard reduction potential is -0.28V and for the Ce4+ reaction the standard reduction potential is +1.61V. Since Ce4+ reaction has a more positive higher voltage it...

jisulee1C

Since the problem is asking to calculate the reaction Gibbs Free energy at 1200K and standard conditions are at 298K the given K is at standard condition. Since the reaction is not at standard condition to calculate the total Gibbs Free energy of the reaction you have to calculate both the standard ...

jisulee1C

Yep. The K would be 1/41 because that's the inverse.

jisulee1C

Since an adiabatic process has no transfer of heat q = 0 therefore the only energy would be as work. For a reversible expansion the temperature stays constant so I think the temperature would be greater for an irreversible expansion compared to a reversible expansion.

jisulee1C

Pb lead which has 82 electrons has a higher molar entropy than C carbon which has 6 electrons because overall lead has more particles (82 protons / 82 electrons and some number of neutrons) than carbon which has only 6 protons / 6 electrons and some lower number of neutrons. Because lead has more pa...

jisulee1C

Gibbs Free Energy. If delta G is negative you know that the forward reaction is spontaneous.

jisulee1C

Use the Cv of the ideal gas for both because they are ideal gases. Ideal gas, Cv = (3/2) R. and plug into the equation to calculate the change of entropy of change of volume.

jisulee1C

For the problem given it's not a multistep reaction. In fact, you are asked to manipulate the K of the given chemical reaction to find the other K values. For example A since the reaction is reversed you would find the inverse of the given K to find the new K value for the new reaction.

jisulee1C

This would depend on the problem. But for the given enthalpies such as the enthalpy of formation they usually give it in KJ or J/ mol and you can manipulate that value depending on how much reactant they give. If no values are given, 1 mol can be used.

jisulee1C

There are common strong acids and bases. For acids there is HClO4 , HCl, HBr, HI, HNO3, and H2SO4 although there is the pattern of halogens forming strong acid because they easily dissociate it's better to memorize the more common strong acids and bases.

jisulee1C

A state function means that only the initial state and final state are needed to determine the value. I think Lavelle's example of how only the initial and final destinations matter when you hike a mountain and how the pathway you took is irrelevant explains conceptually what a state function is.

jisulee1C

q is equal to delta H if there is constant pressure

jisulee1C

Yes, you use the five percent rule to check if the product concentration formed is less than five percent of the initial concentration which allows you to use the approximation. Use the reactant with the smallest concentration given. If it passes the five percent rule for the smaller concentration t...

jisulee1C

In a reversible reaction there is a change of entropy and because it is isothermal delta U will be zero. However, in an irreversible reaction there is no work done because only free expansion occurs. Therefore since w = 0 and delta U = 0 the q = 0 so the q of surroundings =0 and delta S of the surro...

jisulee1C

According to the second law of thermodynamics the entropy of a system will increase over time because the universe is always tending toward disorder. Thus standard molar entropies are positive.

jisulee1C

A greater pKa results in a smaller Ka because pKa = -log Ka. If you plug in values you can see that a stronger acid will have a larger Ka and thus a smaller pKa.

jisulee1C

I am assuming that since the problem in the review session included the structure of glucose, you won't need to remember the structure and instead will be tested on whether or not you understand the concept of calculating delta H of the reaction using bond enthalpies.

jisulee1C

In an acid base titration the acid is reacting with the base to form water H2O. Therefore when writing out chemical reactions, the acid and base can react to form water. Depending on whether the acid or base is strong or weak will influence the reaction however in the example HBr + KOH = H2O + KBr f...

jisulee1C

In a reaction there are variables that change. If the pressure is constant, the temperature or volume or number of moles or other things could be changing. In fact if the pressure is constant it would be an isobaric process where heat could be used to both do work and change the internal energy of t...

jisulee1C

Delta H would be zero if the total heat energy of the reaction is zero but remember that only in constant pressure is delta equivalent to q. Delta U is always zero in an isothermal process because the internal energy is proportional to temperature and since the temperature remains constant the chang...

jisulee1C

If the bond enthalpies are not available you can calculate the standard reaction enthalpy using the standard enthalpy of formation of all products and subtracting that from the standard enthalpy of formation of all reactants.

jisulee1C

In lecture, Lavelle highlighted that a state property (such as enthalpy) is not dependent on the path taken to obtain that state. Some state property examples include E, P, V, T, density, and heat capacity. On the other hand some that are not state peroperties include work (w) and heat (q).

jisulee1C

When solving the quadratic formula, two answers are usually calculated. You should use the positive x value calculated for concentration because concentration itself cannot be a negative value.

jisulee1C

Bond enthalpy are given to us in the problems however most of the time the bond enthalpy given is the breaking of bonds in gases. Therefore usually an enthalpy for phase change must be added for liquids and solids.

jisulee1C

I am assuming that if we were asked to calculate enthalpy of a reaction, either the enthalpy of formation or bond enthalpies would be given to us in the problem.

jisulee1C

Kc means that concentrations were used for the values when finding the equilibrium constant. K would vary depending on what form of values were used, for example Kp for partial pressures and Kc for concentrations.

jisulee1C

Expansion of volume would be the opposite of decreasing volume therefore you would be decreasing the pressure which would shift the reaction to favor the side with more moles of gas because the reaction will shift to minimize the effect of the stress.

jisulee1C

Strong acids are protonated fully and dissociate fully into hydronium ions while weak acids are not fully deprotonated therefore do not fully dissociate and in equilibrium there are leftover acid molecules that show the acid is too weak to fully dissociate.

jisulee1C

For example if product is added the reaction will shift to reactants to undo the stress and vice versa. In addition, lowering or raising the temperature would depend on the reaction and whether it is endothermic or exothermic. For an exothermic reaction raising temperature would shift the reaction t...

jisulee1C

For gases, the same rule applies except that you would be using partial pressures and the Kp instead of concentrations and Kc when comparing to Q.

jisulee1C

ICE charts can be used to find equilibrium molar concentration using molar ratios and the change in molar concentration if the initial molar concentration for either reactants or products are given and the equilibrium constant is also given.

jisulee1C

Solids and liquids (pure substances) are not included in the K expression because the molar concentration does not change in a reaction.

jisulee1C

If Q is less than K at some point during the reaction, then the forward reaction is favored while if Q is greater than K at some point during the reaction, the reverse reaction is favored.

jisulee1C

Le Chatelier's principle states that the chemical reaction will adjust to minimize the effects of changes such as the change of reactant added, change of product added, or change of pressure. Specifically for pressure, if the pressure of the reaction increases or the volume decreases then the reacti...

jisulee1C

Kc stands for the equilibrium constant when using concentration values to calculate. Often K is used to represent either Kc or Kp.

jisulee1C

Bond angles will depend on electrons and the molecular shape. Lone pair electrons have greater repulsion than bonded electrons and will therefore create a smaller bond angle.

jisulee1C

Equilibrium calculations are not on the final.

jisulee1C

Any element in the beyond 3s2 will be able to form more than four bonds and is therefore an exception because the d orbital begins in principle energy level 3 and is therefore able to accomodate more than the standard 8 electrons.

jisulee1C

The VSEPR formula depends on the valence electron bonding to the central atom. Therefore depending one whether it is lone pair electron or bonded electron pair it will change the molecular shape of the molecule.

jisulee1C

Abbreviations are fine for the final, including ox.

jisulee1C

Nope for a positive charge you don't need to add anything.

jisulee1C

The hybridization depends on the number of electron density regions around the central atom so having a single versus double versus triple bond won't really affect the hybridization for the central atom.

jisulee1C

HOCl would act as the stronger acid.

jisulee1C

Polyprotic acids can donate more than one proton. Examples include H2CO3.

jisulee1C

If there are no lone pairs then the molecular shape is the same as the molecular geometry. In the geometry the arrangement considers a bonded electron pair the same as a lone pair of electrons. The geometry will be either linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral.

jisulee1C

Coordination number is the number of parts at which ligands are attached to the central metal atom. mono dentate has one binding site (an example would be chloride ions), bidentate has two donor atoms, and polydentate ranges in the bonding sites. For 9C part c en is bidentate therefore assume there ...

jisulee1C

Because the coordinate complex has an overall negative charge it is an anion and therefore when naming the overall complex with negative charge -ate is added to the metal. If the symbol of the metal originates from the Latin name, the latin stem is used, therefore for iron Fe, the latin ferrum is us...

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