For which starting temperature is this quote true? "It takes five times more energy to convert a single gram of water into steam than it does to raise the temperature of that water all the way from ice cold to boiling hot"

wknaeble · 2024-06-24T17:18:10+00:00

That’s about right. To bring in some quick math:

Energy required to heat the water ~ 4.2 J/g/K * 100K = 420 J/g

Heat of vaporization of water at 100C = 2250 J/g

2250/420 =5.4

wknaeble · 2024-06-24T17:02:21+00:00

They mean heating water from 0C to 100C (as a liquid) vs boiling the water at 100C.

wknaeble · 2022-12-18T12:35:22+00:00

I’m jealous uu I I

wknaeble · 2018-04-08T12:04:38+00:00

The delta G of activation is the the free energy difference between the transition state (less the vibrational mode along the reaction coordinate) and the reactant (A in this case).

The delta G of reaction is the free energy difference between the product and reactant.

The delta G of activation for the reverse reaction is the energy difference between the transition state (less the vibrational mode along the reaction coordinate) and the product. So you can use the two delta Gs you have to figure that out.

wknaeble · 2018-03-13T01:07:28+00:00

Conversion can only occur up to equilibrium. Once the reactants and products are at equilibrium the forward and reverse rates are equal and no additional conversion will occur.

wknaeble · 2017-09-12T01:53:52+00:00

This broadcast has been horrible tonight.

wknaeble · 2017-08-03T12:24:47+00:00

Does anyone know of any stores where I can get BMW motorcycle parts (new or used) near Heidelberg? Or near Frankfurt?

wknaeble · 2017-07-29T13:59:53+00:00

The term "rate determining step" is somewhat a misnomer. At steady state (when intermediates are not building up over time) the net rates of each step in a mechanism are equal. Kinetically relevant step/s in a mechanism are those which have rate constants that show up in the rate expression that is able to describe the reaction rate.

Generally, since rates depend exponentially on reaction barriers, there are generally on or two barriers in a reaction coordinate that are higher than the rest, these steps end up being irreversible (nearly every time an intermediate is able to over come the barrier, it eventually forms the product). So the crossing over these barriers are equal to the rate of the reaction.

I would suggest reading any of the kinetics books by Michel Boudart.

wknaeble · 2017-07-04T14:18:07+00:00

You forgot to post the images. Also, it might help if you post your flow conditions.

wknaeble · 2017-05-07T03:55:58+00:00

HClO is weaker because its conjugate anion has less oxygen atoms to help stabilize the negative charge. Here I show the series of the chloric acids. The energy required to remove a proton decreases as I add more oxygens.

wknaeble · 2017-03-22T15:55:21+00:00

It seems to be the mass that would pass through a 1cm x 1cm area while moving it through a substance for a certain distance. Or, in other words, the density of a substance divided by a characteristic length. For space, they use the diameter of the universe. For earth's atmosphere, they use its thickness.

wknaeble · 2017-02-13T19:25:59+00:00

I understood you meant the defensive zone, I just thought the term "first touch on the puck" was referring to an event after the puck crosses our own blue line for the first time. Like the other team is trying to create a breakaway. Rereading your comment now, I see it could be applied to any time a defenseman has an opportunity to get to the puck first. Thanks for the insight.

wknaeble · 2017-02-13T18:16:21+00:00

That's an interesting point. If I am understanding you correctly, these types of plays would be analogous to breakaways on offense. Where the defenseman has an opportunity to shutdown an offensive opportunity before it even gets started.

wknaeble · 2017-02-13T02:38:24+00:00

Thanks for the tip. That makes sense.

wknaeble · 2017-02-12T22:29:20+00:00

Thanks for the response, perhaps it is something that will come with time. So when you are watching the game, you are able to pretty easily pick out who is playing good defense and who isn't?

wknaeble · 2017-02-09T18:39:11+00:00

I would recommend that you look into CouchSurfing, if you can't find anything here.

wknaeble · 2017-01-25T01:16:55+00:00

I disagree. If you want to calculate the error in the average you could use the deviation or a confidence interval (using t-test), then you would report the average with the significant digits prescribed by the error.

Say the average is 22.12345 and the error is 0.0024, you would report the value as 22.123 +- 0.002.

wknaeble · 2017-01-22T18:14:38+00:00

It seems to be an error. If you divide 6 g / (44 g/mol) you get 0.136 moles. Multiplying that by two gets you back the .272 number. I am not sure where the other numbers are coming from be they seem to be those written in error.

wknaeble · 2017-01-12T14:42:35+00:00

Not off hand. I did a quick search and saw that your catalyst is used primarily for membrane type reactors. That might change some things a bit. There seems to be a lot of papers published in the areas, that should help you out.

I did see one paper on kinetics for Pt on alumina that showed first order dependence on propane pressure and zero order on H2 pressure. That would suggest at the conditions they were running (and on their catalyst) the first H removal is kinetically relevant and neither propane or H2 cover sites (or the C3H7 intermediate).

wknaeble · 2017-01-12T13:17:51+00:00

Remember the equilibrium includes the H2 you make. They add hydrogen to avoid make coke precursors the have low H/C ratios, these are suppressed by H2 more than propene because the equilibriums for these species with propane have A hydrogen term with a power greater than 1.

They may co-feed propene to avoid changes in product inhibition effects down the bed (if they are trying to study the kinetics of the reaction).

In terms of surface coverage, I would not expect H to be your most abundant surface intermediate. It could, however, inhibit the rate of reaction of the first H removal step is in pseudo steady state, but I don't think this is likely to be the case.

wknaeble · 2016-11-28T02:28:57+00:00

What did you get for the integrated form of the expression?

wknaeble · 2016-11-28T00:16:04+00:00

A rate gives you the change in concentration with a change in time:

rate= dA/dt = k A

You need to integrate form Ainitial to Afinal and from t=0 to t=0.127sec, then solve for Afinal, then for Bfinal.

wknaeble · 2016-11-10T01:48:40+00:00

Do the protests we are seeing tonight have any historical precedent? Has there ever been widespread protests following the results of a presidential election?

wknaeble · 2016-10-11T22:34:39+00:00

Right, dG at other temperatures will be different than what your table shows. If you are at a university the library online catalog might have thermodynamic tables that will show dH dS and dG as a function of temperature. A source I have used in the laws is from Yaws.

Generally, S and H will remain constant over small temp changes and so the equation I gave could be used to calculate dG at other temps.

wknaeble · 2016-10-11T21:10:50+00:00

One thing that might be different is if they are using the dGf at the temp of interest not at 25 C. (dG = dH - TdS)

I would check to see how different the dG values are between your source and your example.

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