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Photon336
 one year ago
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Photon336
 one year ago
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Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@rushwr @sweetburger

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0When stating an answer please provide a justification as to why you got the answer and why you feel the others are wrong.

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1for 381. I will go with C. Because we know when it says a reaction is non spontaneous \[DeltaG > 0\]

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1since they have not mentioned anything about a heat change or anything we can't predict anything about enthalpy and entropy soo those answers are cut off

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1the last one happens when the reaction is in equilibrium !

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0381. C \[\Delta G = \Delta H  T \Delta S \] The spontaneity of a reaction does not dictate how fast the reaction will go but rather the tendency of that reaction to occur. if something is spontaneous/ a reaction it will happen. because there is free energy. delta G shows us the free energy available. These three answers below are wrong because, you need two know both the sign of delta H and Delta T S to determine whether a reaction is spontaneous. delta S = negative wrong Delta H is positive wrong Delta G = 0 Delta G = 0 means that it's neither spontaneous in the forward or reverse direction. Delta G must be greater than zero, that would mean that the reaction is spontaneous in the opposite direction not the forward direction.

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0When ΔG is negative, a process or chemical reaction proceeds spontaneously in the forward direction. When ΔG is positive, the process proceeds spontaneously in reverse. When ΔG is zero, the process is already in equilibrium, with no net change taking place over time.

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1382. I'll go with D. STP is know as the standard conditions . But here it says it's not the same so wrong !

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1hey what is delta Hrxn? Is it the enthalpy for the reverse reaction ?

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0Hrxn is the enthalpy of reaction

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1Standard enthalpy of formation is the change that occurs when one mole of a compound is formed from its constituents in their most abundant allotropes and naturally existing physical states under standard conditions for a compound. Since there is 2 moles of HF formed the enthalpy value should be multiplyed by 2 making it the answer A

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0382. B dw:1439097489909:dw can clearly see why A is correct. B cooling to absolute zero would imply that the molecules just stop moving, at absolute zero so that's true. C STP = Standard temperature and pressure 1 atm 25C correct. D is correct, elements yeah the standard Heat of enthalpy is zero because they just aren't formed from anything, well at least fro enthalpy.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@rushwr 383. \[H _{2}\] \[F2\] those are gases diatomic elements

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0they form naturally so they wouldn't have a \[\Delta H rxn\] i believe because they don't form from anything

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1@Photon336 for the 382 one they are asking for the incorrect statement so isnt it D cuz they have said STP is not the same as standard conditions. But they are the standard conditions so it's wrong right?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0Delta Hrxn is for one mole of a substance formed we have two moles so it's \[270 \frac{ kj }{ mol }*2 mol = 540 kj \]

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1but what i thought for 383 was the first answer.

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1yeah my answer have mentioned my explanation above !

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0The whole diatomic gas thing only works for O2 N2 and H2

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0I'm pretty sure^^

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0381. C 382. B 383. A

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0made a mistake it should be just kj/mol not just kj. tara well i guess since it was A, the values for delta Hf i think were 0 for the reactants. I guess it's like those gases exist as they are naturally i think so Delta H rxn = 0

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0I said B. for 382. but wasn't sure really @taramgrant0543664 shouldn't cooling a substance to absolute 0 reduce the entropy to 0 in all cases?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0I also posted 384/385 in one of the attachments.

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1385. I think it is D In adiabatic system \[\Delta Q = 0\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0yep that's right 385 is D.

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0HF is a heteronuclear diatomic gas so it acts a little different I think about that absolute 0 one I think it's that one because you can't actually cool a substance to absolute 0 because it hasn't been done they are close but even being off by a couple of points there is still disorder but it does get closer to 0

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0384. C \[\Delta G = RTlnK\] \[K _{p} > 1 \] then LnK > 1 that makes Delta G < 0 Kp > 1 would mean that the products are favored. if /[K_{p} < 1\] then Lnk < 1 that makes Delta G >0 because RTlnK would be positive. so what this means is that it would be non spontaneous, Kp < 1 means that the products aren't favored but rather the reactants. so this all would make sense.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@taramgrant0543664 @rushwr for 382. they said that you can only cool something to absolute 0 for crystalline substances not the others I guess like gases and liquids.

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0As temperature approaches absolute 0 it would become a "perfect" structure with no chaos or entropy

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0i guess that's not possible for gases/liquids then because like the molecules would have the be moving more slowly, then Intermolecular forces would be greater, i think that if you cooled something it would get to a solid.
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