Question
Given the equilibrium constants for the follo .ing reactions:4Cu(s) Ozlg) < + 2Cu2O(s), 4CuO(s) 2Cu2O(s) Ozlg), Kzwhat is K for the system 2Cu(s) Oz(g) 2CuO(s) equivalent to?(Kz) (Ki):(Ki) ( Kz)(K;) (K2) !
Given the equilibrium constants for the follo .ing reactions: 4Cu(s) Ozlg) < + 2Cu2O(s), 4CuO(s) 2Cu2O(s) Ozlg), Kz what is K for the system 2Cu(s) Oz(g) 2CuO(s) equivalent to? (Kz) (Ki): (Ki) ( Kz) (K;) (K2) !
Answers
Calculate the equilibrium constant $K_{\mathrm{c}}$ and $\Delta_{\mathrm{r}} G^{\circ}$ for the reaction between $\mathrm{I}_{2}(\mathrm{~s})$ and $\mathrm{Br}^{-}(\mathrm{aq})$
Okay if we're looking at iodine and the bromide ion you're gonna want to reverse the half reaction for iodine because it is below and to the right of the bromide. So you're half reactions will look something like this. And then your electrons cancel. You add these two together. You're gonna get 55V and then if you're going to calculate your delta G we're gonna use negative N F. E. And then you've got two electrons that were transferred and then multiply that by Faraday's constant and then multiply that by your 0.55 the volts is the same thing as jewels per Coolum is gonna be negative 106,150 jewels. Now if you want to calculate the standard voltage it's going to be 0.592 over. End times the log of. Kay so you're gonna place this 0.55 here and your two moles of electrons is gonna be your end value. So this will be 20.55 equals 0.592 divided by two Times The log of K. figure out what this .0592 divided by two is and then you're gonna divide by that on both sides like this. So this cancels And then you're gonna have 18.58 is equal to the log of K. And then if you write this um an exponential notation which is the inverse of log notation. You're gonna get K. Is 10 to the 18.58 which is 3.8 times 10 to the 18th
So this question asks, if we have two reactions, let's say a goes to be and has an equilibrium constant K one and C goes to D and has an equilibrium constant K two and now we some these reactions to get a plus c goes to B plus D. What is this equilibrium constant equal to? We'll call it K three. Well, it's simply just equal to the multiplication of the previous two.
In this video we're going to write equilibrium constant expressions for the following reactions. Remember that if we have a solid or a liquid, which applies only the I two, we're going to admit it from the expression um we assume the concentration is constant. So for part a uh we'll do products Overreactions, these concentrations since we do have a quiz species, so we have concentration of cl two and then I minus. And that's going to be squared because of the coefficient. And then on the bottom we have C. L minus and it's squared for part B. The product is CH three, NH three plus And then we have on the bottom ch three and H. Two and then H plus last one for part C. On the top we have gold cyanide and on the bottom A. U. two plus gold ion and C N minus. And that's going to be raised to the fourth power due to its coefficient in the balanced equation. Mhm.
Okay, so we've got some balanced equations here, so we're gonna write some expressions for K. Okay, so Okay. Is always products Overreact mints raised to their coefficients. Okay. And we ignore solids and liquids. So the only thing we're going to pay attention to our gases and things that are Equus if it's a gas will use a partial pressure if it's Aquarius will do its concentration. So this first one looks like we've got a solid and a gas and gas and some acquis. Okay, So it looks like we're gonna go ahead and write the partial pressure of cl two right times the concentration of i minus and we'll square that. Mhm. Who will divide that by the concentration of C L minus squared, Which is supposed to be a quick it's not a gas. Okay, so in part b looks like we've got a couple, we've got three a career solutions, So go ahead and write K. We'll do our products Overreact ints of CH three. NH three plus, and we'll divide this by our reactant, which are what's it all mean, CH three, NH two and H plus. So then the last one, it looks like we've got some A creates things All Aquarius. So again, we're going to go ahead and write product. Okay, is this complex ion? It's an eye on. So we'll do concentration and we'll divide that by concentration of the gold to ion and the cyanide ion. They will raise that to the fourth because it had a coefficient of four.