CHEMISTRY COMMUNITY

Laurence Tacderas 1K

There may be problems where you need the final concentration of A to solve for the rate constant k or the time, but you are only given the final concentration of B. So then, given a balanced chemical equation, you would convert that final concentration in terms of A using stoichiometry.

Laurence Tacderas 1K

If you're given 2 experiments between two reactants A & B and there is one experiment that has both concentrations changing, solve for the order of reactant A first and then use that to help you find the order of reactant B (in the experiment which has both changing).

Laurence Tacderas 1K

They're definitely really helpful because they can tell you what order the reaction is simply based on what is being measured vs time. Then you can go from there and calculate the rate constant from the slope of the line.

Laurence Tacderas 1K

Positive means non-spontaneous, Negative means spontaneous, and 0 means at equilibrium

Laurence Tacderas 1K

There was an announcement that review sessions start this Saturday and continue through week 10

Laurence Tacderas 1K

N will help to determine the rate of the reaction, while K will determine the rate at varying concentrations

Laurence Tacderas 1K

Just remember that what ever reduces/oxidizes (or is the "reducing/oxidizing" agent) is the one getting oxidized/reduced. Then you can tell if it is the cathode (reduction) or anode (oxidation) half reaction

Laurence Tacderas 1K

The deltaG will be equal to zero, and you'll know because the reaction won't be spontaneous in either direction

Laurence Tacderas 1K

I believe it's up to either section 14.7 or 14.8 (for the sixth edition)

Laurence Tacderas 1K

I found certain review sessions to be rather helpful, generally they were the ones that went over specific topics. I would also recommend going to sessions, both TA and UA led, that provide practice questions that mirror the difficulty of the tests.

Laurence Tacderas 1K

Probably the Gibbs Free Energy portion of chapter 9

Laurence Tacderas 1K

I believe as long as you did your work in pen you can have your TA re-check the scoring that gave you

Laurence Tacderas 1K

Since strong acids fully disassociate, just know that they won't even have a Ka value because there's no reactants at the end of the reaction. If Ka=[P]/[R] and there's no R, then you won't have to worry about Ka for strong acids, only weak acids.

Laurence Tacderas 1K

Since deltaU=3/2nRT, if the reaction is at constant temperature, the internal energy will be equal to zero

Laurence Tacderas 1K

Yes just make sure your're using the correct heat capacities and units

Laurence Tacderas 1K

The tea can receive and transfer matter to its surroundings, as well as energy in the form of heat

Laurence Tacderas 1K

Moles per liters, or moles/L (or often written as M for molarity)

Laurence Tacderas 1K

I checked and saw that the midterm will cover problems up to #47 of chapter 9 (6th edition). Everything else seems to involve Gibbs Free Energy which won't be on the midterm.

Laurence Tacderas 1K

They'll be returned in section this week

Laurence Tacderas 1K

As people have said above, the fact that molar heat capacity is intensive, we can apply it to any similar reaction regardless of the number of moles reacted.

Laurence Tacderas 1K

I saw that too. I think if you start with enthalpies and then go back to the first law it might be more coherent.

Laurence Tacderas 1K

Since Kw is a constant value, the strengths of acids and their conjugate bases (as well as bases and their conjugate acids) are complimentary. That would make sense if you think about it: the stronger an acid/base, the more likely it will stay disassociated and thus its conjugate base will be weaker...

Laurence Tacderas 1K

Yes, buffer questions are mentioned in the outline so there's a chance they'll show up on the test

Laurence Tacderas 1K

If K is less than 10^-3, then the reaction heavily favors product. Thus, x, or the change in molar concentration, is so low that it can be considered negligible. Essentially, the equilibrium concentration of reactant is the same as its initial concentration.

Laurence Tacderas 1K

When decreasing the volume of the container, the reaction will favor the side which produces less moles. This is due to the fact that reducing volume increases concentration (moles/volume=conc), so, following Le Chatlier's Principle, the reaction will try to minimize changes in concentration.

Laurence Tacderas 1K

Q, the reaction quotient, uses the molar concentrations of the reactants and products at any point during the reaction, while K is the value when the reaction is at equilibrium. If the two values are equal, then the reaction is at equilibrium. If not, then more products or reactant will form so Q ca...

Laurence Tacderas 1K

Partial pressures of the reactant and products affect Kp, not Kc. Changing the pressure by reducing the volume can affect both Kp and Kc.

Laurence Tacderas 1K

The flask shows the reaction at equilibrium - while the number of X 2 molecules and X atoms are different compared to their initial amounts (no X initially), the total number of moles is the same before and after the reaction. By taking the mole fractions (the number of moles X 2 or X / the total nu...

Laurence Tacderas 1K

You would have to convert either the partial pressure of the gas to equilibrium molar concentration to determine Kc or convert the equilibrium molar concentration of the aqueous solution to determine Kp. However, molarity and partial pressures cannot be used in the same equilibrium constant expressi...

Laurence Tacderas 1K

To create a balanced chemical equation, X must have a stoichiometric coefficient of 2 (since the flask shows the diatomic molecule X₂ decomposing into X atoms).

CHEMISTRY COMMUNITY

Search found 30 matches

Re: stoichiometry

Re: Solving for the rxn order

Re: Graphs

Re: Delta G rules

Re: Final

Re: n and k

Re: Reducing agent/oxidizing agent

Re: Gibbs At Equilibrium

Re: Test 2

Re: Best Midterm Review

Re: HW for W7

Re: Midterms

Re: Ka value

Re: When does delta U equal zero?

Re: Heat Capacity

Re: Cup of Tea

Re: Quiz 1 Number 3c

Re: Midterm content

Re: Test #1

Re: Heat Capacity

Re: Homework

Re: The Conjugate Seesaw

Re: buffer questions on test

Re: Shortcut method (finding concentration change)

Re: Change in Pressure

Re: K and Q [ENDORSED]

Re: Equilibrium constant

Re: 6th Edition 11.7 part c

Re: Heterogeneous equilibrium

Re: 11.7C