The Periodic Table

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12:56 AM

Q: Boltzmann factor vs. graph of Maxwell–Boltzmann distribution

In discussions of the Arrhenius equation describing how rates change with temperature, there is often a reference to collision theory. In the Arrhenius equation, $$ k = A e^{-\frac{E_\mathrm{A}}{R T}}$$ there is a Boltzmann factor interpreted as the fraction of collisions of sufficient energy ...

9 hours later…

Bounty Hunter

10:12 AM

Q: Quadrupole moment of a molecule

What is a quadrupole moment of a molecule and how does it arise? How is it measured for a particular molecule? I've read the Wikipedia article on quadrupoles and understand that it has to do with the distribution of charge, but I'd like a more chemistry-oriented explanation.

physical-chemistry polarity

5 hours later…

user10535

3:22 PM

Howdy, how do you calculate the moles of a product if you have the formula and you know the moles of the limiting reagent?

A.K.

3:34 PM

@user10535 (moles of product produced per mole of limiting agent) x moles of limiting agent

user10535

@A.K. Thank you very much

Thermochemistry:
Given the chemical equation, the standard enthalpy of the reaction, and number of moles of the products on the left hand side of the equation, how do you find the amount of heat evolved by the reaction?

1 hour later…

user10535

4:43 PM

Martin - マーチン

5:34 PM

@A.K. @user10535 there is no quantity 'moles'.... just saying :P

A.K.

5:51 PM

@Martin-マーチン ^

user10535

6:05 PM

Okay, so what is the right way to solve?

M.A.R. ಠ_ಠ

6:26 PM

@user10535 balance the chemical equation and see how many moles of each reactant is needed, then there's the enthalpy of formation formula.

user10535

I have the balanced equation and I know the mole ratio

What I don't get is how q=mcdelta T relates

I know that in this case (because the instructions say as much) to use the specific heat of water

And I have the moles

But I do not have delta T

M.A.R. ಠ_ಠ

@user10535 That usually happens when the heat released/absorbed is supposed to change the temperature of this glass of water I'm drinking or something

user10535

And I am solving for q

M.A.R. ಠ_ಠ

Oh, but you should calculate q based on the reaction data alone

user10535

What does that mean?

M.A.R. ಠ_ಠ

6:29 PM

You get q, then see how T changed

user10535

I don't have q

How else can I get q?

M.A.R. ಠ_ಠ

@user10535 delta H_f(products) — delta H_f(reactants)

^ That's an emdash, not a minus sign. @Loong is going to kill me

user10535

Oh, good

So I have that number

And here is what the question is asking for:

It wants the heat evolved for the given amounts of the substances that react

M.A.R. ಠ_ಠ

Note that if they say, for example, that 29 moles of products were produced from nitroglycerin explosion, you conclude that it's four moles of nitroglycerin, not one

user10535

And I have to use the enthalpy of reaction that I calculated earlier

And I have the enthalpy

So how do I do that?

M.A.R. ಠ_ಠ

6:32 PM

Does my "note that" help?

user10535

So I multiply by the ratio of the limiting reagent to the product?

Multiply the enthalpy number?

M.A.R. ಠ_ಠ

Yep, the normal calculations of stoichiometry

user10535

By the way

What units are q usually in?

Joules?

M.A.R. ಠ_ಠ

KJ per mol if it's calculated for the 'standard' reaction, and kilojoules if you just threw a random bunch of substances and they're not the 'standard' quantity

1 J is a pretty small amount of energy. It'd be more like the reactants sneezing than a legit reaction

Remember, it's not allergy if your face is not three times larger

user10535

Thank you very much. Those are all my questions for today

M.A.R. ಠ_ಠ

6:37 PM

Salut

M.A.R. ಠ_ಠ

7:03 PM

Happy Nowruz y'all

orthocresol

7:38 PM

@user10535 The heat transferred q is always in units of energy. Most commonly kJ.

It's an extensive quantity (think q = mcΔT, you need to multiply by the mass, so that makes it extensive).

Hot nitric acid (HNQ)

7:51 PM

Q: Why is the principal energy of an electron lower for excited electrons in a higher energy state?

Several places state the 'principal energy of an electron' can be calculated as such: $$E = \frac{2π^2mZ^2e^4}{n^2h^2}$$ Another equation I found was: $$E = -\frac{E_0}{n^2},$$ where $$E_0 = \pu{13.6 eV}~(\pu{1 eV} = \pu{1.602e-19 J})$$ As seen in these equations, the greater the principal n...