CHEMISTRY COMMUNITY

RohanGupta1G

use the half life to solve for k and then use the integrated rate law.

RohanGupta1G

Usually i look at what information is given and then am able to determine which equation works.

RohanGupta1G

The key is you need to remember to balance the equation. Then you can use the integrated rate law to solve for time.

RohanGupta1G

As the others said, C is zero order because looking at experiments 1 and 4, even though the concentration of C changes, the initial rate remains the same. This means C has no impact and therefore its exponent is 0.

RohanGupta1G

For when the units aren't given, the SI unit of time is seconds.

RohanGupta1G

Im pretty sure the unique rate refers to only 1 mole of reactant in the reaction. Since there are three moles of oxygen we multiple the unique rate by 3.

RohanGupta1G

It shouldn't really matter but sometimes like in problem 15.3 you need the answer to be in moles so you can run further calculations with your answer.[/quote]

RohanGupta1G

i shouldn't really matter but sometimes like in problem 15.3 you need the answer to be in moles so you can run further calculations with your answer.

RohanGupta1G

As someone already said, the units are already Moles of N02 L^-1S^-1 so you don't need to divide by moles of N02

RohanGupta1G

When the reactant is first order, you do not need to include the exponent, when the reactant is 0 order, you do not need to write the reactant at all.

RohanGupta1G

Figure out if the E nott value is positive

RohanGupta1G

As others have said, standard conditions refers to the temp and pressure etc, that are accepted so that accepted values for E nott and many other values from tables are valid.

RohanGupta1G

Yeah i think when they are both negative you gotta just choose the one that is less negative for the cathode

RohanGupta1G

I believe the difference in concentration creates a potential difference because on one end reduction occurs, and on the other oxidation is occurring which allows for the travel of electrons and in turn allows for current to flow.

RohanGupta1G

I don't think Erxn is zero at equilibrium. Erxn=Estandard at equilibrium. It only equals 0 for hydrogen

RohanGupta1G

Erxn is Estandard at equilibrium. Q=1 at equilibrium. This checks out mathematically using the equation Estandard=Erxn+rt/nf lnq. if q=1, then lnq=0 and Erxn=Estandard

RohanGupta1G

heat capacity is q=c(delta t)
specific heat capacity is q=mc(delta t)
basically specific heat capacity is specific to your sample because it incorporates mass.

RohanGupta1G

I think the question will specify if the gas is ideal. If it doesn't specify whether the gas is ideal or not, it will give explicit condition of the gas.

RohanGupta1G

That is correct because the work done is a vacuum is 0 because there is no pressure or opposing force.

RohanGupta1G

im not positive but maybe because it says to assume ideal gas behavior? could just be a poorly written question as well.

RohanGupta1G

i think they are using the q=mc(dt) equation to solve for dt. this works because we are given the m, the c, and the q leaving the dt. I think your way probably works too, but is a lot more work.

RohanGupta1G

As Suchita said, it depends on how the question is worded, and how you complete the calculations, however kj/mole should be good.

RohanGupta1G

Direction of heat flow determines the sign as well as whether or not the reaction is spontaneous.

RohanGupta1G

As jannie had said, gasses do have more entropy and therefore the reactants in this case have more entropy than the products because there are more moles of gas on the reactants side.

RohanGupta1G

As the others said you would set the ice side equal to the water side and use the respective heat capacities in each side.

RohanGupta1G

Delta u= q+w
where u is energy and q is heat and w is work. Thus the ways to change energy is by adding or subtracting heat or work.

RohanGupta1G

Yeah like nicole said, it depends on how you word the answer. If you add the word released then your number is positive, but if you are referring to the change in energy for the reaction then it should be negative

RohanGupta1G

As others have said, the reaction releases energy causing the surrounding area absorb the energy and feel warmer .

RohanGupta1G

Not sure if this is right or not but the way I was thinking about it was the lower he pressure, the easier it is to add heat and therefore the lower the heat capacity... similarly the greater the volume, the more energy it can absorb and therefore the specific heat capacity is lower... again I’m not...

RohanGupta1G

I believe the delta symbol always means change in the given variable which in this case is U(final)-U(initial)=(delta) U. U is the energy of the system.

RohanGupta1G

I’m pretty sure if the piston allows the volume to increase then pressure will remain constant.

RohanGupta1G

Exothermic phase changes are ones that release heat such as a gas turning into a liquid or solid, or a liquid turning into a solid. As the temperature lowers, energy is released and the phase changes.

RohanGupta1G

If you aren't apart of the solution, you're apart of the precipitate

RohanGupta1G

I don’t think we need to know what you drew however we need to know how to draw and interpret a phase change graph plotting energy and heat and you definetley should be able to determine whether this change is endothermic or exothermic.

CHEMISTRY COMMUNITY

Search found 34 matches

Re: Test 3 #8

Re: Calculating Gibbs Free Energy

Re: Test #3 Q3

Re: 15.17

Re: Units for Time

Re: 15.5

Re: Proper units

Re: Proper units

Re: 15.3

Re: Writing rate laws

Re: Dissolving metal in a solution

Re: standard conditions [ENDORSED]

Re: Cell potential

Re: Concentration Cells

Re: E rxn vs E standard at equilibrium [ENDORSED]

Re: K and Q at equilibrium

Re: State Properties

Re: ∆U for isothermal reversible

Re: Work

Re: 9.13

Re: 8.99 Concept

Re: Standard Gibbs Free Energy of Formation (Units)

Re: sign of entropy

Re: 9.33 (a)

Re: heat capacities

Re: Energy of systems

Re: Heat Released

Re: Exothermic

Re: Molar heat capacity

Re: Delta U？

Re: Work

Re: Phase Changes

Re: Post All Chemistry Jokes Here

Re: Paths and Changes of State