CHEMISTRY COMMUNITY

kausalya_1k

I believe we look at the ion itself (Within the molecule) to calculate the oxidation number and then determine which species gained an electron and which one lost an electron

kausalya_1k

you would use the integrated rate law in order to find it for values after 1/2 (such as 1/4 and 1/16)

kausalya_1k

I also thought molecularity describes rate law. Could you explain why you think it's the other way around?

kausalya_1k

Just like K, Q can be calculated in terms of partial pressures or concentrations. Therefore there is no need to convert pressure to concentration (also because as someone pointed out earlier, using the PV=nRT will just cancel out the units anyways).

kausalya_1k

the anode is the more negative cell potential and the cathode is the more positive one

kausalya_1k

the order doesn't matter just remember that the anode is on the left and the cathode is on the right

kausalya_1k

Suppose that each of the following pairs of redox couples is combined to form a galvanic cell that generates a current under standard conditions. Identify the oxidizing agent and the reducing agent, write a cell diagram, and calculate the standard cell potential from the standard potentials of the e...

kausalya_1k

I know the standard cell potential (E o) isn't affected by balancing the equation, but what about when you reverse one of the half reactions?

kausalya_1k

The main difference is that in acidic redox solutions you balance the H with H+. However, with basic solutions, you would balance the hydrogens using OH-.

kausalya_1k

ursulavictorino1K wrote:Why does the reaction always go from the anode to the cathode?

Electrons always flow from loss of electrons (oxidation, occurs at the anode) to the side where electrons are gained (reduction, occurs at the cathode).

kausalya_1k

If it's in an acidic solution, you would first balance everything except H and O. Then, balance O using H2O. Then, balance H with H+ If it's in a basic solution, you would first balance everything except H and O. Then, balance O using H2O. Then, balance H with OH- and add anymore H2O's that need to ...

kausalya_1k

We already know that O has an oxidation number of -2
so,
2Cr +7(-2)=-2
Cr=+6
We also already know that H has an oxidation number of +1
so,
2C+6(+1)+1(-2)=0
C=-2

kausalya_1k

Make sure to start with the equation of E cell = E cathode - E anode. Once you put it in the equation, you can solve for the remaining or missing variable. -0.689 V = -1.038 V - E anode E anode = -0.349 V. The negative signs can be confusing, but as long as you plug it in for specific variables and...

kausalya_1k

Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O...

kausalya_1k

W here represents the degeneracy
it can be calculated by (# of states possible)^(# of particles involved)

kausalya_1k

we use Cv for when volume is constant (isochoric) and Cp when pressure is constant (isobaric)

kausalya_1k

standard entropy is calculated when the system is at standard/lab conditions

kausalya_1k

You would use the constant that has all the units that you need to cancel out
so if you are given a volume in L, a pressure in atm, and a temperature given in K, you would use R=.0206 L*atm/(K*mol)

kausalya_1k

since van't hoff is derived from gibbs free energy which assumes constant pressure, we can again assume constant pressure for van't hoff

kausalya_1k

deltaG knot is for when the system is at its standard state

kausalya_1k

A reaction is thermodynamically stable if it's spontaneous (aka when deltaG<0)

kausalya_1k

When deltaG=0, the system is at equilibrium (the forward and reverse reactions occur at the same rate)
When deltaG is negative, the forward reaction is spontaneous
When deltaG is positive, the reverse reaction is spontaneous

kausalya_1k

When deltaG is less than 0 (negative) the forward reaction is spontaneous. When deltaG is positive, the reverse reaction is spontaneous. When deltaG=0, the system is at equilibrium.

kausalya_1k

G knot is for when everything is in its standard state.
An equation that relates the two terms is deltaG=deltaGknot + RTlnQ

kausalya_1k

Hi, I'm confused on question 4C.3: Calculate the final temperature and the change in enthalpy when 765 J of energy is transferred as heat to 0.820 mol Kr(g) at 298 K and 1.00 atm (a) at constant pressure; (b) at constant vol- ume. Treat the gas as ideal. More specifically, I was wondering, how would...

kausalya_1k

the definition of enthalpy is heat at constant pressure, so enthalpy (delta H) is the same as heat (q), which is why in an isobaric problem you can use q=deltaH

kausalya_1k

you would use q=mcdeltaT when the temperature is changing, usually during a time when the phase is changing (so on a graph this area would be slanted)
if the temperature is not changing (isothermal), and the phase isn't changing (so in the plateau of a graph), you would use q=mdeltaH

kausalya_1k

so enthalpy is basically like heat but only during isobaric (deltaP=0) situations
basically, enthalpy is always heat, but heat is not always enthalpy.

kausalya_1k

in isobaric reactions (deltaP=0), enthalpy (deltaH)=heat (q)

kausalya_1k

Would someone be able to briefly explain the difference between isobaric, isochoric, and isothermal? I have these definitions written down and I've read through the textbook, but I feel as though I've memorized the differences and don't fully grasp the concept. Thanks! The way I personally remember...

kausalya_1k

In a closed system, only energy (but not matter) can be exchanged. In an isolated system, neither energy nor matter can be exchanged.

kausalya_1k

in an isolated system, energy and matter cannot be exchanged. So, q=0 and w=0, which means that delta q=0.

kausalya_1k

An open system is where energy and matter can be exchanged. A closed system is where energy can be exchanged, and an isolated system is where nothing can be exchanged between the system and surroundings.

kausalya_1k

You use PV=nRT to solve for n. Then use molar mass of Xe to convert n from moles to grams using stoichiometry.

kausalya_1k

Should we always convert to J (using the 101325 Pa value) on exams? Correct me if I'm wrong, but I don't recall hearing that in lecture?

kausalya_1k

Since heat is on the right side of the equation, adding heat to the system will promote the reverse reaction (produce more reactants).

kausalya_1k

It honestly depends what units P, V, and T are given. Typically we use .08206 L atm/mol K, but it really depends on units for the given values.

kausalya_1k

% ionization=[h3o]/[acid]

kausalya_1k

Yeah
The only time you’d exclude h2o is if it were in liquid or solid form

kausalya_1k

You would use an ICE table when you are trying to find one of the concentrations that is unknown. You will usually be given a K value.

kausalya_1k

i believe that it's possible, but for the scope of this class i don't think we don't need to worry about that

kausalya_1k

Using the ideal gas law,
PV=nRT
we can derive it to calculate concentration:
n/v=P/(RT) since n/V=concentration
then just plug in the partial pressure and temperature (which are given) and the r constant

kausalya_1k

It's because H2O is usually provided in excess during experimental conditions. Also, since it's a liquid, it will have a concentration of 1 always.

kausalya_1k

pOH=-log[OH-] and you would use that for bases
pH=-log[H3O+] and you would use this for acids
pH+pOH=14

kausalya_1k

x was ignored in this case because K is less than 10^-3 so you know that x is going to be too small for it to make a difference in that case. You can test to see if this approximation is accurate as long as your final answer is less than 5% of the initial

kausalya_1k

Liquids are considered pure substances, so their concentration wouldn't change in a reaction. Therefore, their activity is equal to 1 (just like solids) which is why we don't include it in the K calculation.

kausalya_1k

The P is used to denote partial pressures. To solve for K for gases, you would use P to denote partial pressure. However, if you are asked to solve for Kc, you use the [] concentrations (as the c in Kc stands for concentration).

kausalya_1k

Adding a catalyst will speed up both the forward and reverse reaction rates. However, as long as both the forward and reverse reaction rates are still equal to each other, then the reaction is still at equilibrium.

kausalya_1k

I also agree that there needs to be a double arrow in the equation for us to know that it is able to reach equilibrium. The definition of chemical equilibrium is the point at which the forward reaction and the reverse reaction are occurring at the same rate. Therefore, the equation must be able to h...

kausalya_1k

The R represents the ideal gas constant.
Depending on units, it is equal to R=8.314 J·K-1·mol-1= = 8.206 x 10-2 L·atm·K-1·mol-1 = 8.314 x 10-2 L·bar·K-1·mol-1.

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