CHEMISTRY COMMUNITY

William Satyadi 2A

Yes, if we are given the rate law, we will be able to tell which step is the slow step, because the slow step is the limiting step that controls the rate of the reaction.

William Satyadi 2A

I agree with Alvin, emf stands for electromotive force, which is the same as E, the cell potential difference.

William Satyadi 2A

Furthermore, from the molecularity chart we took notes on from lecture, you should also be able to write the corresponding rate law for the reaction.

William Satyadi 2A

Yes, the two equations you provided can be used interchangeably, as they are the same, just rearranged differently using logarithm rules. From the first equation you provided, we can rearrange it so that ln[A]_f - ln[A]_0 = -kt . Dividing by -1 gives us ln[A]_0 - ln[A]_f = kt . Using log rules, ln[A...

William Satyadi 2A

That is correct; a zero order reaction rate does not depend on the concentration of the reactant present, as long as there is some reactant. When integrating the rate law, we get Rate = $k[A]^{0} = k$ . Since anything raised to the zero power is 1, we can see that the rate just depends on k.

William Satyadi 2A

A zero order reaction graph is linear and has a negative slope when the y-axis is the concentration of the reactant, and a first order reaction graph is linear and has negative slope when the y-axis is ln [Reactant].

William Satyadi 2A

Since Dr. Lavelle covered how to calculate half-life for zero, first, and second order reactions, I would probably study it and practice how to use it, even if the book doesn't require to do so, as I think anything he lectures on is fair game for the final.

William Satyadi 2A

For a first order graph, the ln [A] vs. time graph will be linear and decreasing, since the concentration of the reactant is decreasing. If the graph was changed to represent [A] versus time, then the graph would be a decreasing exponential function.

William Satyadi 2A

After integrating the the rate law for a second order reaction, you can see that 1/[A] = kt+C, in which 1/[A] is on the y-axis and time (t) is the x-axis. Since the reactant is decreasing as the reaction progresses, [A] will get smaller and smaller meaning, \lim_{[A]\rightarrow 0} 1/[A] = \infty , a...

William Satyadi 2A

E is an intensive property, meaning we do not need to manipulate it or multiply it by stoichiometric coefficients.

William Satyadi 2A

Inert electrodes like Pt are needed for some cells which don't have conducting ions. These electrodes allow for electron transfer without affecting the reaction.

William Satyadi 2A

n is the number of electrons transferred in the half reactions, but they need to be balanced first.

William Satyadi 2A

Hey! I don't have the solutions manual so I can't see that it says n=1, but I did find a way to solve it using the same set up as you did. Using the Nernst equation, I got (\frac{0.0592}{2})log(\frac{pH3.0}{pH4.0}) . pH is a measure of H+, and we can use logs to find the H+ concentra...

William Satyadi 2A

Building off of Dylan's point, I'm not sure if this is the case in batteries, but often times, a voltage can be increased using step up transformers. I think Lavelle also pointed out that these batteries were very basic and rudimentary, and I'm sure batteries today are much more complex than the cel...

William Satyadi 2A

The 2.303 comes from the conversion from using logQ instead of lnQ. Either equation works and they're equivalent.

William Satyadi 2A

I believe Lavelle mentioned in class that although it is Joules, it can also be seen as J/mol, so either way is fine.

William Satyadi 2A

I think a conducting metal is required for electron transfer between two half reactions. In Dr. Lavelle's lecture on Wednesday, he stated that some half reactions have no conducting solids, so in one example platinum electrodes were used. Platinum works because it has no reaction with the solution a...

William Satyadi 2A

I believe you are right. In a galvanic cell, the anode side is the oxidized side, as electrons flow out towards the cathode, or reduced, side. Therefore you would be right; the oxidation half reaction would be the anode side and the reduced half reaction would be the cathode side.

William Satyadi 2A

The anode side is the side that is oxidizes, meaning electrons are lost. The electrons move towards the cathode/reduced side.

William Satyadi 2A

An isobaric process is one in which pressure is constant. The work done by or on the system can be calculated by taking the area under the line on the graph. The site below contains examples of what the graph looks like.

https://physics.info/pressure-volume/

Hope this helps

William Satyadi 2A

The microstates are different positions, or states, that molecules can take in a system. An example talked about in class were the differences between solids and gases. Molecules in solids can have only few positions since molecules are packed together so tightly and can't move around. Thus, solids ...

William Satyadi 2A

We also learned a new equations today in lecture, such as the Van't Hoff equation, which can be used to find the G of a reaction. When G is negative, the system is favorable, or spontaneous.

William Satyadi 2A

No work can be done through expansion in a vacuum because work is defined as -P*deltaV.
In a vacuum, there is no opposing force, so therefore the work done is 0. The book defines this as free expansion on page 264.

William Satyadi 2A

According to Table 8.1 on page 262 of the textbook, they note the standard convention of using units of Pascals (Pa) for pressure. However, Example 8.1 on page 264 gives a provides a good practice problem in which atm is used instead of pascals for pressure, and liters instead of meters cubed for vo...

William Satyadi 2A

Building upon what Caroline said, you can also use the equation that Dr. Lavelle derived in class, which was -work= -nRTln(Vfinal/Vinitial).
This will give you the work, and since change in internal energy of the system is 0, q=-w

William Satyadi 2A

When a work is done by a system the work is considered negative because the system is losing energy. For instance, if we considered our bodies as the system, whenever we perform work, move, or exercise we use up some of our energy reserves, subtracting from our total energy. Since we are losing ener...

William Satyadi 2A

I agree with Michelle, since we are supplying heat to a copper kettle, we are adding heat to our system, not to our surroundings. I don't think it counts as a system, but rather the boundary of the outer boundary of the system.

William Satyadi 2A

The way I learned this in the past during physics was by looking at the graphs. If you look at an isobaric system (pressure is constant), the work done by the system can be calculated by finding the area under the graph. If you look at an isochoric system (constant volume), you'll see that the line ...

William Satyadi 2A

The reactants have positive enthalpies because bonds are broken in the reactants, while bonds are formed in products. The breaking of bonds will always require some input of energy, meaning an addition of energy to the system. When bonds are formed, energy is released, meaning the system loses energ...

William Satyadi 2A

Agreed, it might help to look at what molecules in solid, liquid, and gas phases look like. The main difference between solid and liquids is that molecules in liquids are still close but can move around past each other, while molecules in solids are close, compact, and cannot move around much. The d...

William Satyadi 2A

Bonds being broken on the reactants side require energy, which is reported as a positive number. In this case, it was 978 kJ. On the products side, bonds are being formed, which release energy, reported as a negative number (-1086kJ). As a result, the difference between the two products was -59kJ, a...

CHEMISTRY COMMUNITY

Search found 31 matches

Re: Slow/Fast Step

Re: emf

Re: Molecularity

Re: Equation variations

Re: zero order rate?

Re: Zero-order vs. First-order reaction graphs

Re: Half-life of Zero Order [ENDORSED]

Re: First Order Graph

Re: The graph of second order [ENDORSED]

Re: Reaction E [ENDORSED]

Re: Cell Diagrams [ENDORSED]

Re: Finding n

Re: 14.41(b)

Re: Battery voltage

Re: Nernst Equation 2.303RT/F = .059V

Re: Delta G standard

Re: Conducting Metal

Re: Anode vs cathode

Re: Anode and cathode

Re: Isobaric [ENDORSED]

Re: Micro states

Re: How do we find out if a system is favorable? [ENDORSED]

Re: Internal Energy in a Vacuum [ENDORSED]

Re: Units

Re: Formulas Under Different Conditions

Re: 8.17 Work done by a system

Re: system and surroundings [ENDORSED]

Re: Heat Transfer at Constant Volume/Pressure

Re: Bond Enthalpies

Re: Energy and Phase Changes

Re: Exothermic reaction in bond enthalpy example in lecture