CHEMISTRY COMMUNITY

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you would probably need a catalyst to lower the activation energy. I think it wouldn't occur spontaneously due to the high activation energy.

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I would stick to balancing basic solutions with OH- and acidic with H+.

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yes, the rate depends on the slow step since it is the limiting step

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in a concentration cell, the standard potential energy is 0 and the potential energy is very low voltage. the whole reaction is driven entirely by a change in concentration gradient. thus the reaction proceeds because there is a tendency to move from high concentration to low concentration

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No activation energies are not always constant. If you add a catalyst, it tends to lower the activation energy and thus the reaction can progress faster

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Use H2O to balance the Os first and then use H+ to balance the H. if you used H30+ to balance the H then you would once again add Os to the other side.

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Reducing power means the ability to give away electrons. If it has a high tendency to give away electrons then it is a good reducing agent and therefore, likely to be oxidized itself.

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Yes, if something has oxidizing power then that is the same as saying it is a strong oxidizing agent.
Oxidizing agent is the same as what is being reduced.
If it is good at being reduced then it would have a high or more positive reduction potential.

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Once you are done balancing all the elements on both sides. calculate net charge on both sides and then determine which side would need electrons such that the netcharge on both sides is equal. One way to make sure you're writing it correctly is to check if there you added electrons on the LEFT for ...

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MnO4-

Let the charge of Mn = x.
Charge of O= -2
Overall charge = -1

x + (-2*4) = -1
x - 8 = 1
x = -1+8
x = +7

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Adding electrons should be the last step after you have balanced all the elements, H, O. Then look at net charge on both sides and you can add the electrons accordingly.

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Oxidation will be at the anode and reduction at the cathode. The sides could be switched but typically we keep anodes on the left and cathodes on the right. When there is a gain of electrons or the oxidation number decreases then it is reduction. When there is a loss of electrons or an increase in o...

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Use the half reaction from the appendix that has both your species present in it.

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Use platinum since it is more common. And use it when there are no conducting solids or if the species reduced and oxidized are in the same solution.

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Use inert electrodes when both the oxidized and reduced species are in the same solution. Also, use inert electrode when there are gas/ion reactions like in 6L.5 (c).

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Find the half reactions in the appendix and then compare the cell potential values that are given. Whichever equation has a higher E naught value will be reduced and thus the other will be oxidized.

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You need to look for the reduction half reaction in the appendix at the back of the textbook. Look for the 2 equations that have both your species. The one with a more positive reduction potential value will be more likely to be reduced. So the other equation will be oxidized and you can flip it.

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Keep solids towards the ends in general. I'm not sure if you can switch Cl2(g) and Cl2(aq) but to be safe, I wouldn't. It also makes since since the right side is reduction so it is going from Cl2 (g) (oxidation number 0) to Cl2 (aq) (Oxidation number -1).

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Yes, that is correct! both are good methods but the one where you flip the value for oxidation usually makes more sense conceptually.

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You need to find the species from your redox reaction in the appendix 2B and use those potential values.

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If a metal has a more negative reduction potential then it is more likely to be oxidized. therefore, it will be more likely to dissolve compared to the one that has a more positive reduction potential.

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I think you should probably know the derivations of equations that aren't on the formula sheet. Also know the derivations if the objectives include knowing the derivation.

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Standard will have that naught or little circle next to G and E. It means standard conditions (solutions are 1M, gases are 1bar or 1atm, solids and liquids are in pure form, temp is 25 degrees celsius). If it isn't standard then there is no circle.

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That's correct, you don't need the stoichiometric coefficients when writing cell diagrams.

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I think delta G is probably positive since the process is not spontaneous. You have to input energy in the form of photons for the reaction to take place. Therefore, there isn't free energy available to do work. I'm not sure though.

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In an isolated system, the system cannot exchange matter or energy with the surroundings. In a closed system, the system has a fixed amount of matter but it can exchange energy with the surroundings. if the water bottle is insulated really well then it will be an isolated system because it will not ...

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yes, that is correct.
delta H = q when no work is done. So assume they're the same when no work is being done by or on the system.

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delta U is the change in internal energy versus delta H is the change in enthalpy.

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if it just delta T then change in celsius will be the same as change in kelvin. you can try it out.
But if for instance, you're using the ideal gas law PV = nRT then you have to use Kelvin.

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convert torr to atm. 760 torr = 1 atm.
then you will get the answer in L*atm.
multiple this with 101.325 J/ L*atm.
L*atm will cancel out. The final answer will be in joules.

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delta U would be 0 if it is isothermal and no work is done. imagine a gas expanding in a vacuum with constant temperature. since it is not pushing against anything, no work is done and there is no change in temperature so delta U is 0.

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If the system is doing work then it is losing energy so the value will be negative. If work is done on the system, it is gaining energy so it will be positive. similarly, if a system loses heat, it is negative. if it gains heat, it is positive.

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isothermal refers to the system being at constant temperature.

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There is no expansion work done at constant volume. This is because for expansion work to be done the piston needs to move from a lower volume to a higher volume.

There is expansion work done at constant pressure.

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combustion of fossil fuels releases carbon dioxide and other greenhouse gases into the environment. these gases are responsible for trapping heat in the atmosphere which contributes to global warming.

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The constant is important because the final unit of work done must be in Joules not in L*atm.

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A combustion reaction will have the form:

CnHm + (n + m/4) O2 --> nCO2 + m/2 H2O + Heat

so it will be a hydrocarbon + oxygen gas ---> carbon dioxide + water + heat

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in thermodynamics, the external pressure is not constant for a reversible process. Here, internal pressure is equal to the external pressure. If you increase the external pressure by a very small amount, the piston will compress. Reversible process always does more work and the change is gradual and...

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solids < liquids < aqueous < gases

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State functions are important because the path taken to get to the final value does not matter. State functions only depend on the current state. So for enthalpy, you can start with ice and melt it and then cool it and then melt it again and the final enthalpy will still be the same. However, work w...

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When there is compression, the side with fewer moles is favored. In this problem, both sides have the same number of moles so there is no change.

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A strong acid has a smaller pH. A pH of 1 means it is stronger than a pH of 4. That's why a weak acid will have a higher pH than a strong acid.

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The principle simply states that when stress is applied to a system in dynamic equilibrium, the the equilibrium adjusts to minimize the effects of stress. So if a system is compressed, then the equilibrium will favor whichever side has fewer moles. Other examples include adding or removing a reactan...

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Use approximation when value of K is less than 10^-3. Always verify your answer by plugging the concentrations to check if you still get the same K. My TA recommended using the quadratic equation every time just to be safe.

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Q and K both measure the ratio of the concentration of products to concentration of reactants. However, K is a constant and it is the value you get when the mixture is at equilibrium. Q is the same ratio but it is measured at any point during the reaction so it can be compared to K to figure out whe...

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An exothermic reaction means heat is given off (similar to it being on the product side). Think of it as increasing products (or heat in this case) would favor reactants.

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This depends on the information given in the question. I believe even if the molar concentrations of gases is given then you would just calculate Kc. But if the partial pressures are given then calculate Kp.

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Q tells us which direction the reaction will proceed when you measure the ratio of reactants to products at any given time. By comparing Q to K, you can figure out if the reaction will shift towards products or reactants.

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Only if H20 is in the liquid form then do not include it in the ICE table but if it is in gaseous form then include it. Also, don't include any solids or liquids in the ICE table.

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Hi,

In discussion section 2K, we received extra homework problems. The equation was:
N2 (g) + 3H2 (g) --> 2NH3 (g) + heat
If we increase temperature, why does the reaction shift to the left?

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