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Use the data in Appendix L to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given(a) $\operatorname{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)$(b) $\operatorname{CdS}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+\mathrm{S}^{2-}(a q)$(c) $\mathrm{Hg}^{2+}(a q)+4 \mathrm{Br}^{-}(a q) \rightleftharpoons\left[\mathrm{HgBr}_{4}\right]^{2-}(a q)$(d) $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$
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Chemistry 102
Chapter 17
Electrochemistry
Rice University
Drexel University
University of Maryland - University College
University of Kentucky
Lectures
00:44
In chemistry, electrochemi…
03:11
In chemistry, a redox reac…
03:35
For each of the following …
05:18
Using the value of $K_{s p…
02:21
For the following equilibr…
06:30
Use the data in Table 17.1…
06:08
Calculate equilibrium cons…
01:26
03:29
Consider the following equ…
04:32
Use the data in Table 15.3…
02:09
Calculate $\Delta G^{\circ…
01:47
Use the appropriate values…
02:27
Calculate the equilibrium …
05:40
Use data from Appendix $C$…
18:14
Write the equilibrium cons…
02:52
03:12
Among the solubility rules…
07:29
Calculate $\Delta_{\mathrm…
05:47
01:14
Write equilibrium constant…
02:42
Use data from Appendix IIB…
08:36
for a ready Casey Expression I and in a Casey expression, we do not include solids, products of a reactant. So we have a TSH to cubed over h 20 cubed. We could also rate a K p expression, which will be the partial pressure of H two. Cubed over the partial pressure of H 20 cubed relationship between the two k p is equal to K. C R T on Delta End here would be zero. So then that would just simplify. The K P is equal to K C for B. We can write a K C expression. We have a single liquid called a pure liquid, which is not included in a new equilibrium expression. So we've got products each too squared over. 02 k p is equal to the partial pressure of each two squared times. A partial pressure of old too. K p is equal to K C R T. And then end here would be, ah, one minus two. So this would be negative one. We're sorry. Um, it would be three. There's three moles of gas, the product side zero on the react inside. So the relationship would be to the power three Chrissy, we've got ah solid which is not included. R K C expression is going to be each CEO to the fourth. As I see 04 in each two squared KP is going to be equal to partial pressure of HCL to the fourth partial pressure of S T I C 04 times a partial pressure of H two squared K p is equal to K C R T and the Delta End will be four minus three, which is going to be one for d. K. C is equal to one over HD two two plus over CEO minus squared. Since there are no gas is involved, there is no K c expression. Ah, our sorry KP expression. So this would be a K P is equal to K C R T to the zero, which would mean that
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