CHEMISTRY COMMUNITY

austin-3b

Delta G naught is standard conditions. Delta G is at any point; if equal to delta G naught, it's at equilibrium.

austin-3b

When deltaG is negative it's spontaneous. It wants to get to a lower free energy state.

austin-3b

Specific is the amount per gram and molar is the amount per mole.

austin-3b

Typically when there is no conducting metal on that certain side. It is used as an inert conductor.

austin-3b

It's 144 total (with bonus) but the max is 140.

austin-3b

Does this only apply to find Delta G naught or can you use it to find Delta G, given delta E is not in standard conditions?

austin-3b

You just divide the forward rate over the reverse rate.

austin-3b

I2 is not a metal so it cannot conduct.

austin-3b

Cathode is what is getting reduced. Anode is what is getting oxidized. Usually, the lower E would be oxidized (anode) to try to get a positive E for a galvanic cell.

austin-3b

It will be cumulative, everything from the class.

austin-3b

It represents the rate constant of the reverse reaction.

austin-3b

It is the amount of electrons transferred. When given half-reactions, you would have to scale them to have the same amount of electrons transferred.

Ex.
Ce4+ + 2e --> Ce2+
Fe --> Fe3+ + 3e

Scale to six, you get 6e.

austin-3b

Adding acidic stuff lowers pH and adding basic stuff raises the pH.

austin-3b

Is E at equilibrium zero or equal to E naught?

austin-3b

E naught is E at the standard conditions. E can be at any conditions.

austin-3b

I think it's because second order depends on the concentration. The half life is different for each concentration, you can't just do that to find 1/2, 1/4, 1/8, and so on.

austin-3b

I believe you find the delta E (it's a concentration cell) and know that the amount of electrons transferred is 1. Then, use -nFE.

austin-3b

I believe n is the amount of moles of electrons transferred.

austin-3b

When system does work, it's negative. When work is done on the system, it's positive.

austin-3b

Not sure if we need to plot but it's important to recognize what order it is if given on a graph.

austin-3b

I am confused on how to start these 6L.7 Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions (a) AgBr(s)⇌Ag+(aq)+Br−(aq),a solubility equilibrium (b) H+(aq)+OH−(aq)→H2O(l),the Brønsted neutralizationreaction (c) Cd(s)+2Ni(OH)3(s)→Cd(OH)...

austin-3b

I am confused for part C. For the half-cell diagram, do we include HCl? The solutions manual does not.

austin-3b

How do you balance 6L.3 part d? It seems a little confusing to me.

austin-3b

N is the amount of moles of electrons. If one electron is transferred, n is one and if two are transferred, n is two and so on.

austin-3b

Anode is usually on the left and cathode is usually on the right.

austin-3b

You can use H+ or H2O for acidic solutions but for basic solutions, you use OH- and H2O

austin-3b

Positive E indicates that it is spontaneous and negative E indicates that it is not.

austin-3b

It helps maintain neutrality I believe.

austin-3b

Delta H naught is just the specific standard conditions.

austin-3b

It is usually the case with constant pressure.

austin-3b

It's in joules. P is in atm, volume is in liters. You have L*atms and when you multiply by 101.325 (J/L*atm) correct me if I'm wrong, then you get joules.

austin-3b

When delta G is negative, it's favorable because of spontaneity.

austin-3b

Depends on what you are trying to find. If joules, use 8.314, if not then you use 0.0821.

austin-3b

I believe so.

austin-3b

What type of problems do we need to use Van't Hoff equations for?

austin-3b

At a more negative delta G, it tends to form the substance spontaneously because it's more stable, more favorable. So choose those with a negative delta g, those will be the most stable.

austin-3b

It covers everything up until Friday's lecture I think.

austin-3b

How do you relate the two? On Sapling, the last couple problems deal with this.

austin-3b

You use P1/P2 instead because when you do the proportions with P and V, you get

P1V1=P2V2

V2/V1 = P1/P2

austin-3b

Since you are dealing with an irreversible problem, you use the external pressure (1.43). Then, you multiply that by the delta V (VF-VI). Finally, you multiply that by 101.325 to go from L*atm to joules.

-p(deltaV) * 101.325

austin-3b

This is because when setting q (system) equal to q(surroundings), in this case, the reaction and water, the other side has to be negative.

austin-3b

For irreversible, you always use the external pressure because that is what the system is pushing against.

austin-3b

Spontaneous means a reaction can operate on its own without any extra input. To determine spontaneity, delta H would help, but it doesn't tell the full story. You would have to find delta G. In order to do that you would need to have the delta H (enthalpy) minus delta S (entropy) times T(temp).

austin-3b

It is written as a lowercase w. Uppercase is for degeneracy.

austin-3b

q will always be zero since there's no heat coming in or out.

austin-3b

That is correct. However, when you use the constant to find work (joules), then you use 8.314.

austin-3b

An example of an open system is a glass of water. An example of a closed system is an ice pack. An example of an isolated system is a substance in an insulator.

austin-3b

I'm not really sure how to approach this problem 4.5 In 1750, Joseph Black performed an experiment that eventually led to the discovery of enthalpies of fusion. He placed two samples of water, each of mass 150 g, at 0.00 °C (one ice and one liquid) in a room kept at a constant temperature of 5.00°C....

austin-3b

I believe it's one of the strong acids.

austin-3b

You would do 10 to the negative pka (that dictates the exponent)

10^(-pka)

austin-3b

I would assume no energy change is occurring because nothing can go in or out of the system.

austin-3b

UA Worksheet: The value of ΔH for this reaction is -1107 kJ: 2Ba(s) + O (g) → 2BaO(s) How many kJ of heat are released when 5.75 g of BaO (s) is produced? When asking how many kJ of heat released, do you put the answer as a negative sign or positive sign? I know if it asks for the change in enthalp...

austin-3b

I know if 10^-3 < K < 10^3, you can't determine if reactant or product is favored. Is that case for stability for products vs. reactants? Like if k falls between these numbers, are you unable to determine if the products or reactants are more stable?

austin-3b

If there are zero moles on one side of the reaction (solid only) and you compress the system, does the reaction shift to that side or does it not (because the solid does not contribute to equilibrium)?

Example
A(s) --> B (aq) + C (g)

austin-3b

Is the overall delta H of a reaction in kJs or kj/mol?

austin-3b

Is there a 10^-3 -10^3 cutoff for favorability/stability of products or reactants in a reaction? Like if the k falls in between those values, are neither products nor reactants favored?

austin-3b

How are we supposed to approach this? Hemoglobin (Hb) molecules in blood carry O2 molecules from the lungs, where the concentration of oxygen is high, to the tissues where it is low (see the Interlude following Focus 5). In the tissues the equilibrium H3O+(aq) + HbO2−(aq) ⇌ HHb(aq) + H2O(l) + O2(aq)...

austin-3b

Would the reaction not shift to the left (more A made) if you add the amount of substance B because solids aren't part of equilibrium constant?

A(s) --> B (g) + C(aq)

austin-3b

Beside diatomic molecules and graphite carbon, how do we know if an element is in its standard state?

austin-3b

When given a reaction, how do we deal with an element not in its standard state?

austin-3b

For the calculating the enthalpy of formations, how do we account for phase changes? Are there times where we look at the table of enthalpies of an element and have to account for phase changes (add or subtract)? Or do we just look at the molecule and the phase it is and use it straight up?

austin-3b

We are given two equations to find the enthalpy of formation for dinitrogen pentoxide. How come we multiply the bottom equation (with N2O5 as a product) by half and not the top by 2 order to cancel out the intermediate NO2s? Also, how do we do the second part (and from the standard enthalpy of forma...

austin-3b

How come we don't multiple the last combustion equation by 2 based on the coefficients? Is it because you can use half coefficients for Hess's Law?

H2 + O2--> H2O

austin-3b

The gas constant you use depends on the units given. In this case, you use 0.0821.

austin-3b

Even though steam may have the same temperature as liquid, there is more heat in the system because bonds are being broken. According to the graph, when heat is being transferred to the skin, steam has more energy to release than liquid. This is why it hurts more when you have hot steam.

austin-3b

You are correct. Exothermic reactions where things that go to a lower energy state are generally more stable. Endothermic is vice versa. If elements go to a higher energy state, they are generally less stable.

austin-3b

It has to be 10^-4 or smaller, but you can always check using the 5% rule. If x is less than 5% of initial, the approximation is valid.

austin-3b

Once we calculate the pH of a certain substance, do we need to know how to calculate the pH of its conjugate. Is it like pka and kb where subtracting 14 from the ph gives you the ph of its conjugate?

austin-3b

I think it tells you to refer to them.

austin-3b

6D.15
Calculate the pH of
(a) 0.19 M NH4Cl(aq) (b) 0.055 M AlCl3 (aq).

How are we supposed to know if AlCl3 is an acid or a base?

austin-3b

For increases in pressure (decreases in volume), the reaction goes toward the side with less moles. When finding out which side has the least amount of moles, do you also include aqueous solutions or only gases?

austin-3b

Since it's neutral, you can assume these concentrations are equal. Therefore, you can assign the same variable (x). You do that, square root every side and you get 1 * 10^-7.

austin-3b

Normally, you would exclude it. But if it's a gas, you put it in the ICE Table. Anything that isn't a solid or liquid is included.

austin-3b

Yes, Q is any time. You would measure Q at that moment to determine the direction of the reaction.

If Q<K, the reaction is going right; more products are made
If Q>K, the reaction is going left; more reactants are made

austin-3b

If you remove the product then you lower the Q at that moment. Since Q (the reaction quotient) is lower, it will make more product to get back to the level desired.

austin-3b

I don't think you have to convert unless it asked for Kp. You can use the concentrations for this case.

austin-3b

Since it's k, it's the equilibrium concentration. If it was asking for q, then it would be for the initial concentrations.

austin-3b

P= pressure of the gas (atm)
V= volume of gas (L)
n= number of moles
r= gas constant
T= Temp (Kelvin)

austin-3b

You are correct; even though 61 seems like a big number, you can only conclude that it favors products when it's 10^3 and above and reactants when it's 10^-3 and below.

austin-3b

When Qc is more than Kc, reactants would be formed to get that ratio back to equilibrium.
When Qc is less than Kc, products would be formed to get the ratio up to equilbrium.
When they are equal, it's at equilibrium.

austin-3b

You use Kc when having the molar concentrations of the particles and Kp for when given the pressures (I think only gases have kp values).

austin-3b

Why does the increasing/decreasing of temperature affect the Keq value. Also, why does changing the pressure make the reaction favor the side with more or less moles, depending on how you change it?

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