How do I find the equilibrium constant using two given equations?
Easy, I’ll leave the actual calculation to you:
How can I find the chemical equilibrium constant for [math]2NO_ {2} \rightleftharpoons 2NO +O_{2}[/math]?
After a lot of digging around I finally found the answerWell start by recording our concentrations for the reaction before equilibrium and after equilibrium in an ICE Table[math] 2NO_ {2} \rightleftharpoons 2NO +O_{2}[/math]Initial 6.8 0 0Change -2x 2x xEquilibrium 6.8–2x 2x x(Everything is in mole)Since we have x which equals 1.2 we practically can find all the Moles we need for the equilibrium equationThen using the volume given get the concentration and calculate the equilibrium
Cant find equilibrium constant for reaction. Can anyone help?
answering your emailed request: reduction:Mn2+(aq) + 2 e– ---> Mn(s) E = –1.18 reduction: Fe2+(aq) + 2 e– ---> Fe(s) E = –0.440 voltaic cell: oxidation: Mn(s) --> Mn2+(aq) + 2 e– - E = +1.18 reduction: Fe2+(aq) + 2 e– ---> Fe(s) E = –0.440 your standard cell voltage would be E = + 0.74 volts for the reaction: Fe+2(aq) & Mn(s) --> Fe(s) & Mn+2 (aq) Calculate [Fe2+] if [Mn2+] = 0.075 M and õcell = 0.800 V the nernst: E = Eo - (0.0592 / n) log Q 0.800 V = 0.74 V - (0.0592 / 2 mol electrons ) log [products] / [reactants] 0.80 V - 0.74 V = - (0.0296 ) log [Mn+2] / [Fe+2] 0.06 V = - (0.0296 ) log [0.075 M] / [Fe+2] 0.06 V / (- 0.0296 ) = log [0.075 M] / [Fe+2] - 2.027 = log [0.075 M] / [Fe+2] inverse log of both sides (aka 10^x) 0.00940 = [0.075 M] / [Fe+2] [Fe+2] = 0.00940 / [0.075 M] [Fe+2] = 0.1253 Molar your answer rounded to 2 sig figs is [Fe+2] = 0.13 Molar =======================================... Finally, what is the equilibrium constant for this oxidation/reduction reaction? @ equilib, the cell is dead, voltage = zero Q = K the nernst: 0 = Eo - (0.0592 / n) log K Eo = (0.0592 / 2 mol e-) log K 0.74 volts = (0.0296) log K log K = 25 inv log: K = 1 X 10^25 ========================= p.s. you can not get K from [Mn2+] = 0.075 M & [Fe+2] = 0.125 Molar because it was not at equilibrium under these conditions since the reaction was still producing voltage
What is the equilibrium constant of SO2Cl2(g) <=> SO2(g) + Cl2(g)?
Given this reaction: SO2Cl2(g) <=> SO2(g) + Cl2(g) A solution is made containing initial [SO2Cl2] = .025M At equilibrium, [Cl2] = 1.3*10^-2M Calculate the value of the equilibrium constant. I just don't even know where to begin! Please help me understand this!
For chemical equilibrium, can the constant K ever be zero?
Hi: The equilibrium constant is not 0. You must have added NCS- to the solution (or added Fe3+ to a soln of Na+? NCS-). So what was the original conc of NCS- ([NCS-]o)? At equilibrium the [CNS-] is = to [NCS-]o - 0.002M (needed to form the Fe cmplx). This can be approximated to [NCS-]o 0.002M small compared to [NCS-]o but check. K = [FeNCS2+]/[Fe3+][CNS-] K = [0.002]/[0.198][CNS-]o (only one sig fig for [FeNCS2+]?) This will give you a non zero number hopefully close to the lit value for K. Good luck, dr p
What are the new equilibrium equations when the following stresses are applied?
Original Equilibruim: Fe^(3+) + SCN^(1-) <----> Fe(SCN)^(2+) 1. Stress: Remove heat (ice bath) and color red is observed Equilibrium equation: ??? 2. Stress: Add heat (hot water bath) and color orange is observed Equilibrium equation: ??? 3. Stress: Add Fe(NO3)3 and color brown is observed Equilibrium equation: ??? 4. Stress: Add KSCN and color dark red is observed Equilibrium Equation: ??? 5. Stress: Add AgNO3 and color white w/ precipitate is observed Equilibrium Equation: ??? 6. Stress: Add NaOH and cloudy yellow color is observed Equilibruim Equation: ??? Please Help Me Out. Thank You very much.
What is phase equilibrium? What are some examples?
I would expand upon the other postings with this:A substance is in “phase equilbrium” when more than one phase of a substance (solid, liquid, gas) is present *and* the amount of material in each phase does not change.Water is a good example because it happens in real life and is easy to understand. Imagine the co-existence of solid and vapor forms of water in a freezer (without auto-defrost) compartment that hasn’t been opened in a while. In a closed contain of water, the vapor and liquid phases quickly come to equibrium as long as the temperature and air pressure are constant.There is a “triple” point for water in which ice, liquid, and vapor can all be present: Triple point - Wikipedia. This is a specific temperature and pressure. The Wikipedia article describes this point for water, mercury, carbon dioxide, and many more.Note also that the cool phase diagram there has lines between the phases. These are points of phase equilibrium (temperature and pressure). And, even more exciting, it shows that the various phases of ice (yes, there are multiple forms) are stable under different conditions. Pressure matters a lot!Equilibrium does not mean, however, that there is no change. Equilbrium is dynamic. In the freezer, some of the ice will sublime and some of the vapor will condense to the solid form, giving the “frost” on the sides of the freezer.In the closed container, water is continually evaporating and condensing, but we can only see it in the beads of water or condensation that may appear one along the insides of the container.One more cool fact about the diagram: the “critical point” at the right shows the temperature and pressure where the liquid and vapor phases are “merged”. At that place, one can think of a very think gas or a very vaporous liquid. There’s no difference in this hot, high pressure condition.
Calculate [Fe^2]+ when the cell reaction reaches equilibrium?
Mikey! to the rescue!! =) AKA bbk lol sorry anyways back to the question so by definition of this question 1. Ag/Ag+ half reaction E = 0.8 + 0.0591log [Ag+] 2. Fe2+/Fe3+ half reaction E = 0.771 - 0.0591log([Fe2+]/[Fe3+]) don't need to use n since n=1! The overall reaction is Fe2+ + Ag+ ===> Fe3+ + Ag and K = [Fe3+]/([Fe2+] [Ag+]) Therefore: 0.80 - 0.771 = 0.0591log[Fe3+]/([Fe2+][Ag+]) or log K = 0.491. so K = 3.10 So your trying to find the equilibrium concentration so use the I.C.E table then you will get at equilibrium [Fe2+] = 0.0055 - X [Ag+] = 2.5 - X [Fe3+] = 0.0055 + X * adding cause its producing [Ag(s)] = nothing its a solid it doesn't apply to equilibrium therefore use the quadratic equation 3.10 = [0.0055-x] / [2.5-X][0.0055-X] now use the formula X = + or - the thing you get, but likely it will be the - since the plus will give you a big number and when you subtract the [Fe2+] it will be a negative TADA! you should get [Fe2+] = 0.00126 *remember you should try this problem yourself. I might have miscalculated*
The standard solution of FeSCN2+ (prepared by combining 9.00 mL of 0.200 M Fe(NO3)3 w/1.00 mL of 0.0020 M KSCN
To the previous answerer: good job copying and pasting, YOU DIDNT ANSWER THE QUESTION. look to the previous answerer's post for the theory. here's how to answer it: equations needed: [SCN–]eq = [SCN–]i – [FeNCS2+]eq [FeNCS2+]eq = (Aeq/Astd) * [FeNCS2+]std [SCN–]i is given= .00070 M Aeq is given= .150 Astd is given= .530 [FeNCS2+]std = .0002 plug it in! [FeNCS2+]eq = (.150/.530) * .0002 = .000056604 M [SCN–]eq = .00070 M - .000056604 M = .0006434 M <-- answer
Why does the temperature remain constant during a phase change?
The temperature of any substance is related to its average kinetic energy where energy is the ability to do work. When a solid is heated, the average kinetic energy of the substance increases because of which the temperature increases. But during phase change the energy given or taken in is called latent heat which is used only to form bonds between molecules or break the bonds between molecules. This heat won’t affect the average kinetic energy of the substance. Since average kinetic energy of the substance will remain same during phase change, the temperature will remain same too.