CHEMISTRY COMMUNITY

Sally Nason - 1K

Find the redox equations on the sheet for each of the elements given and the most negative value for E has the most reducing power.

Sally Nason - 1K

You have to set the rate equal to k times the concentration of H2O (the moles of H2O (1.8/18.016) times 1/.750L) times the concentration of O2 squared (moles of O2 (1.8/32) times 1/.750L and square that). The equation should look like: 2.7 x 10^2 = k*[(1.8/18.016)*(1/.750)]*[(1.8/32)*(1/.750)] and t...

Sally Nason - 1K

Since you use the equation w = -P*deltaV for irreversible expansion, you would likely be given the pressure or the change in moles to use for substitution with PV = nRT. For reversible expansion, we use the equation w = -nRT*ln(V2/V1) so you would likely be told it is isothermal and the moles would ...

Sally Nason - 1K

For work you should recognize expansion versus compression but that is as specific as you need to know. The other equations, such as those in the beginning of chapter 8 in the textbook, are not on the equation sheet.

Sally Nason - 1K

If you are able to draw the Lewis structure of molecules and know what to look for within the structure to determine the functional group, you should be set.

Sally Nason - 1K

There can only be one "slow" step because it is considered the slowest of all the steps. Slow is a relative term and describes the speed of one step of reaction in relation to the speed of the other steps.

Sally Nason - 1K

You can also think about molecularity in regards to the order of a reaction, as first order is unimolecular, second order is bimolecular, and third order is termolecular.

Sally Nason - 1K

If the question tells you what order the reaction is in, always go with that because it is directly stated in the question.

Sally Nason - 1K

I think this refers to questions like 15.1 on the homework where you are given the equation: N2 (g) + 3H2 (g) -> 2NH3 (g) and are asked what is the relation between the rate of consumption of H2 and the rate of consumption of N2. In this case, the rate of consumption of N2 is 1/3 times the rate of c...

Sally Nason - 1K

The derived equations are on the equation sheet. Derivations are helpful for sure so be comfortable with them but you will not need to know how to derive them.

Sally Nason - 1K

Could someone please explain how we know that HBr, NO2, and HOBr are first order? All we are given are two reaction equations with no numbers.

Sally Nason - 1K

Also in a lot of the homework problems, you use the change in the concentration of the product to find the change in the concentration of the reactant.

Sally Nason - 1K

I don't think you need to ever write it in terms of products because even if you are given the increase or decrease in concentration of products, you can convert it to reactants.

Sally Nason - 1K

/Is it always necessary to change units to mol from mmol?

Sally Nason - 1K

It will become more obvious that oxygen is being reduced once you balance the hydrogens and oxygens in the equation. The half reaction isn't balanced so it is difficult to tell what is happening. The balanced reaction should look like H2O (l) + O3 (g) + 2e- -> O2 (g) + 2OH- (aq). The oxygen is going...

Sally Nason - 1K

I also used the appendix. I think we will receive a sheet with half reactions that we will need for the test.

Sally Nason - 1K

In this equation, n stands for the moles of electrons.

Sally Nason - 1K

These are given on the equation sheet.

Sally Nason - 1K

I had trouble with this one as well. You would think that since the outdoors is so much larger that any temperature change wouldn't affect the entropy as much but the question really just wants you to look at the equation.

Sally Nason - 1K

Every time there is a temperature change or a phase change you need a separate equation. Usually when an ice cube has a phase change it first has to gain heat or lose heat to reach the right temperature for the phase change. That would be 2 equations. Then, if the ice cube changes temperature after ...

Sally Nason - 1K

Usually when the question says the process is isothermal, delta U will be equal to zero. An isothermal system will also have a delta S(universe) of zero.

Sally Nason - 1K

For cis compounds there are 12 different orientations but for trans compounds there are only 3 different orientations. Therefore, the cis form will have higher residual energy (use Boltzmann entropy calculations).

Sally Nason - 1K

During a phase change, the temperature does not change but heat must still go into/leave the system in order to cause the phase change. Since heat is entering or leaving the system, the entropy will change and therefore needs to be accounted for.

Sally Nason - 1K

I also got -2.7 kJ/mol! I think you are right and the answer key has an error in it!

Sally Nason - 1K

If the value for delta G is negative then energy needs to be added in order to form the compound so therefore it would not be stable, as it doesn't occur spontaneously. However, a negative delta G for a compound means the compound can form without any stimulus or added energy and is therefore a stab...

Sally Nason - 1K

I agree. I've put it in several different calculators and have gotten -2.7 kJ/mol every time.

Sally Nason - 1K

When you need to go up the hill (low to high), an input of energy is required to do so, so therefore this motion cannot happen on its own and it is not spontaneous. However, if you go down the hill (high to low), the reaction can happen on its own because no energy needs to be put into the system an...

Sally Nason - 1K

You can also think of it like delta G goes from negative to positive with 0 as the "threshold" so to find at what temperature this threshold is at, you set deltaG equal to 0.

Sally Nason - 1K

I was also confused by this in class but in the equation G = G^o + RTln(P/P^o), I believe that the P is the atmospheric pressure and the P^o is always 1.00 atm because it's standard. Therefore, when P is more than 1.00 atm, (P/P^o) is positive and when P is less than 1.00 atm, (P/P^o) is negative.

Sally Nason - 1K

When I plug (1.38x10^-23)*ln(6^6.02x10^23) into my calculator, I get a result that says "overflow" but the answer is 14.9 J/K. How is this possible?

Sally Nason - 1K

If there is a heat reservoir in the surroundings, the temperature will be constant and therefore heat will be reversible. If a system is doing work, it is losing energy but heat from the surroundings can be added to the system to replace the energy lost. For instance, during a phase change, work is ...

Sally Nason - 1K

Why is the net production 2/3 mole of gas? There are 2 moles of reactants and 4 moles of products so how is the change 2/3?

Sally Nason - 1K

How do you figure out the enthalpies of formation? The enthalpy for ZnCL2 isn't in the appendix of the book.

Sally Nason - 1K

From what I understand, there are three forms of movement that affect the internal energy of a molecule, and therefore the heat capacity. These are: translational, linear rotational, and nonlinear rotational. For more complex molecules, all three motions are occurring and therefore the molecule has ...

Sally Nason - 1K

In this question, the part we are trying to figure out is the work that can be done by the system, or on the system. In general, if the system does work, such as expansion, the work value will be positive and if the system has work done on it, such as compression, the work value will be negative.

Sally Nason - 1K

Since enthalpy is a state function, it doesn't matter what path is taken to the final point, the only values that matter are final and initial. That is why you can "skip" intermediate steps by adding equations.

Sally Nason - 1K

When you subtract the enthalpies of formation of the reactants from the enthalpies of formation of the products, the result is the enthalpy of the reaction.

Sally Nason - 1K

For monatomic gases you can use the formula C[p][/M] = (5/2)*R as well as C[V][/M] = (3/2)*R

Sally Nason - 1K

Is the work done during compression always positive and the work done during expansion always negative?

Sally Nason - 1K

It would be a closed system because while the mercury cannot leave the thermometer, heat can be transferred to or from the thermometer, therefore not making it isolated.

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