CHEMISTRY COMMUNITY

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I was wondering by what time and day do we need to submit our take-home final by?

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in acidic solutions, you balance oxygen by adding H2O to the side lacking oxygen and you balance H by adding H+ to the side needing hydrogen. in basic solutions, you add H2O to the side lacking oxygen to balance oxygen. Then to balance hydrogen, add H2O to the side needing hydrogen and add OH- to th...

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zero order means concentration of reactants has no effect on rate so if trial 1 had the same concentration of reactant a but had twice the concentration of reactant b compared to trial 2, trial 1 will still have the same rate as trial 2. essentially rate is unaffected by the change in reactant conce...

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When given experimental data, input the given values into time vs ln[a] (first order), time vs 1/[a] (second order), and time vs [a] (zero order) and look for the graph with the linear graph. this determines the order of that particular reactant.

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From what I remember from Lavelle's lecture, catalysts are purely mechanisms that lower activation energy and are not included in the equation, therefore, they will not be in the rate law

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yes, if both half reactions only show 1 mole of electrons being transferred then n=1 thus the solution manual is wrong

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It will be given which step is the slow step as we can not determine which is the slow step purely based off the equations

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yea I'm curious about this too cuz i didn't turn in week 10 homework yet

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the question will usually indicate which step is the fast step and which is the slow step as we have not learned a process for determining slow steps yet

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use the highest number of moles of electrons transferred according to the half reactions

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the intermediate is a species produced in one step then consumed in the next step so the intermediate species will not appear in the overall reaction equation as it is depleted by the end of the next step.

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the half reaction with the more negative reduction potential will be the anode

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tests are handed back during discussion

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n is the moles of electrons transferred so use the number of electrons found in the half reaction with the most electrons

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lnQ can always be converted to logQ thru the equation: lnQ=2.303logQ

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units of k can be determined by (M/s) / (M^n) where n is the total reaction order

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the anode is the one with the most negative potential

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rate constants are always positive and don't have units

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both should have the same units. it does not really matter what unit it is as long as units match

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so i thought n was the moles of electrons transferred but thats not the case for 6N.7.b. can someone explain

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for b, the solution says In+3 needs 2 electrons to become In +2 and this sets n=2 for the Nernst equation but this number of moles of electrons is clearly wrong

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why do we assume Cu+2/Cu is the anode?

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usually anode will be to the left and the cathode will be to the right. Also oxidation occurs at the anode and reduction happens at the cathode

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you balance reactions in acidic solutions by adding H2O to balance the oxygen and adding H+ to balance hydrogen. In basic solutions, you add H2O to balance the oxygens then to balance the hydrogen, you add H2O to the side needing the hydrogen then add OH- to the other side

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I think Lavelle said test 2 is the 2nd half of outline 4 and all of outline 5 (electrochemistry)

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from what i remember, galvanic and voltaic cells are the exact same thing

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it will always be given as reduction as it allows for easy comparison to determine what will be oxidized and what will be reduced based the voltage value for the reduction reaction

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the molecule that contains the element being oxidized is the reducing agent and the molecule that contains the element being reduced is the oxidizing agent

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you add electrons to the side with the more positive charge to balance the charge on both sides of the equation

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anode is being oxidized so its losing electrons whereas cathode is being reduced so its gaining electrons

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G9not) is under standard conditions whereas G is not and G(n0t) will contribute to the value of G

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a single bar is a porous disk and a double bar is a salt bridge

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galvanic cells from what i understand are basically like batteries that produce a voltage difference and consist of a cation and anion that produces a transfer of electrons across the cell. the potential of this cell is greatest when little to no current is going thru

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porous disk is repped by a single line and double line is salt bridge

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at least from previous classes I took, it is usually safe to assume that O2 will have a -2 charge

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I think when current flows, the voltage decreases as electrons flow thus why max potential is when the switch is open

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usually, a table with the standards will be given to us

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if oxidation number decreases, it means it was reduced by gaining electrons

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G knot is under standard conditions. G is the total so it will include G knot

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test 2 is the week after next week

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temperature is always kelvin

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∆Gº is basically just ∆G but under standard conditions

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K=(RT)^(change in n) x Kc

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when calculating K look for (aq) or (g) and exclude (s) or (l)

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calorimeters maintain the same volume so constant volume

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the energy from steam is higher than the energy for water

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just remember a system doing work isn't gaining work therefore its doing negative work

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From what i remember, most professors usually provide conversions on the test in the form of a formula chart or at least include a needed conversion in the question

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i usually write all the formulas on one piece of paper to remember all the relevant equations for the topic at hand

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You always use Kelvin no matter what equation it is so if you are given celsius convert to kelvin

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POV is from the system so if the system is doing work, its positive and if its having work done on it its negative

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its usually (n)(Cp or Cv)(delta T)

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Work is seen from the system's POV so if a piston is pushing down on a system, the system is the one receiving the work therefore the system isn't the one doing the work hence work by the system will be negative

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use the gas constant that matches the units given in the question

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q at constant volume and q at constant pressure

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You use Kp because K=kp for all cases. Kc=K when certain conditions are met

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To break bonds, energy has to be brought in aka endothermic reaction and to form bonds, energy must leave the system aka exothermic

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i would assume we are starting at 4A rn

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temperature can affect the rate of the reaction leading to different equilibrium concentrations

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yes it was included in the outline questions

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Im guessing its more than one since Ka will usually be [conjeguate base][H30+]/[Strong acid] and [H30+]=[strong acid]

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removing H2 causes an absence of H2 thus causing the reaction to form more H2 since there is an absence of it

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you can still access the textbook thru the 14a course

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its [A-]/[HA] but if [H30+]=[A-] then [H30+]/[HA] will yield the same % value

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reverse reaction is endothermic thus forming reactants when heat is added

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R can have different values depending on what units are needed for the equation

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usually if we're asked to calculate Q then a K value should be provided to help us determine if it has reached equilibrium

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Kb is for weak base reactions and Ka is for weak acid reactions

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K represents the concentrations at equilibrium whereas Q represents concentrations not at equilibrium

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usually questions will include words like "moles" which is used for concentrations. if it doesnt have "moles" i would guess pressures is the way to go

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K is just a number that is used to show the ratio of products to reactants. I dont think K is ever used in equations so it doesnt need units anyways

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according to one of the TA's, a Higher pressure of the reactant just implies that more reactant was added and the presence of reactants/products does not affect the equilibrium constant. Also, for d, a higher amount of reactants will yield a higher amount of products thus initial concentrations will...

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Essentially solids and liquids are in such great excess and it is impossible to come up with a value for their concentration hence why they are omitted

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Aqueous solutions can have their concentrations measured and these concentrations are what allow us to calculate K. However, solids and liquids are in such great excess in solutions where it is too hard to calculate their concentrations hence why they are omitted from the equilibrium constant

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Instead of looking at K as Kc look at it as Kp. Kp= (partial pressure of products) / (partial pressure of reactants) according to this equation, an increase in the partial pressure of reactants would lead to a decrease in the equilibrium constant instead of an increase in the equilibrium constant, w...

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all reactions will eventually reach a point where molar concentrations will become constant so in other words all reactions attain equilibrium after a certain period of time. Temperature, however, will affect the forward and reverse reaction rates thus affecting the equilibrium constant.

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in the student solutions manual, it says that question C is false because the amount of reactants/products doesn't affect the equilibrium constant if the temperature is constant but the question is asking if the equilibrium constant will increase if the reactant starts at a higher pressure so should...

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