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QUESTION: Consider the following equilibrium for the formation of the complex ion, FeSCNFe +3SCN-FeSCN2+Using Le Chatelier's Principle, predict the shift in eq...

Question

QUESTION: Consider the following equilibrium for the formation of the complex ion, FeSCNFe +3SCN-FeSCN2+Using Le Chatelier's Principle, predict the shift in equilibrium when the following changes are incurred on the system: Adding AgCIO4 Adding NaSCN Adding Fe(NO3)z Adding NaHzPOa Adding NaNO: Adding NHASCN Adding Fez(SO4): Removing FeSCN

QUESTION: Consider the following equilibrium for the formation of the complex ion, FeSCN Fe +3 SCN- FeSCN2+ Using Le Chatelier's Principle, predict the shift in equilibrium when the following changes are incurred on the system: Adding AgCIO4 Adding NaSCN Adding Fe(NO3)z Adding NaHzPOa Adding NaNO: Adding NHASCN Adding Fez(SO4): Removing FeSCN



Answers

Consider this equilibrium system:
$$\mathrm{CO}(g)+\mathrm{Fe}_{3} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{CO}_{2}(g)+3 \mathrm{FeO}(s)$$
How does the equilibrium position shift as a result of each of the following disturbances? (a) $\mathrm{CO}$ is added. (b) $\mathrm{CO}_{2}$ is removed by adding solid NaOH. (c) Additional Fe_ $_{3} \mathrm{O}_{4}(s)$ is added to the system. (d) Dry ice is added at constant temperature.

For a Casey expression is going to be, uh, no, we're gonna omit thesis. Ah, lid and pure liquid. So it be sealed too cubed over C o cubed. Because there are gas is involved, we can raid a k p expression, which is going to be the partial pressure seal to cubed, divided by the partial pressure of C O cubed for B A. Okay, see, expression will omit the solids. It would be one over hilarity 02 cubed. And we can write a K P expression that should be won over the partial pressure of both. Too cute. Percy, we'll omit the solids. The Casey is going to be equal to the polarity of S 03 And the KP based on the gas, is gonna be equal to the partial pressure vessel three. And for D ah, well met with solids. The K C here is gonna equal to bury um, two plus times the polarity of s 04 to minus. There is no KP expression for D, as there are no gas is involved in the equilibrium

So here we need to write out the equilibrium, constant expression in terms of the activities and then eliminate the activities of the pure solid and the pure liquids. We just do this by setting their activities to one. So we have the following equation 50 oh h three ad three h plus f the three plus at three. H 20 So we have okay value of 9.1 times 10 to the three Sweet come right out. The equilibrium constant in terms of activities to have K is equal to a of F E three class divided by a H plus cubed so in the above equilibrium, constant activities. We can convert this into concentration. So we have K is equal to f e three plus divided by H plus again to the power three. So to determine these concentration and the K values, we can substitute some numbers in. So what we have is 9.1 times 10 to the three. That's a K value that we know. We don't know f e three plus find that by the concentration of H plus one point, there are times 10 to minus seven. That's cute. So then we just rearranged to solve for F E three plus, that is 9.1 times 10 to minus 18. Mahler, yeah.

Defend the reaction machine iron, um hydroxide. And also we've hydrogen. All right, they're going to write a vehicle leaving expression and also found that vehicle even concentration of iron when PCI seven. Okay, so the equally bring expression you were you close to the concentration i n three plus over the concentration off by jodi. A 1,000,000 to the power off for a reason. Every Joni include water. And also hydro size has been where it's liquid one is a solid. Okay, this case will be easy for us to 9.1 time. Tend to our fleet. All right, so will we be the concentration off iron Feli when the PHS is was a seven point. All right, so we're going to sell for I and, uh, from their equilibrium causing expression. So we have our k multiply by the concentration off. I trolled him. So you re cruise to 9.1 time tentative. Powerfully. Time stands about 10 to 7 to powerfully. Okay, so we're going Teoh self with that and we will find as 9.1 time 10 to power of 1918 motor and this is a concentration off the I and

At the following reaction Here we have every three plus at SC and miners. There's an equilibrium with Effie S C and two plus. So given the following information, we have a concentration of every three plus as one times 10 to minus three. S C and minus is eight times 10 to the minus three and F D SC and two plus as 1.7 times 10 to the minus full. We're using the camera chemical reaction strike yeoman tree in order to calculate equilibrium concentrations or F E three plus and SCN miners where our value is 3.3 times 10 to the power to


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