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I. Balance and Express the equilibrium constant (K) for thefollowing reactions:a.) C12H22O11 (s) + O2 (g) <--> CO2 (g) + H2O (l)b.) Al (s) + HCl (aq) <--&g...

Question

I. Balance and Express the equilibrium constant (K) for thefollowing reactions:a.) C12H22O11 (s) + O2 (g) <--> CO2 (g) + H2O (l)b.) Al (s) + HCl (aq) <--> AlCl3 (aq) + H2 (g)c.) Sb (s) + O2 (g) <--> Sb4O6 (s)

I. Balance and Express the equilibrium constant (K) for the following reactions: a.) C12H22O11 (s) + O2 (g) <--> CO2 (g) + H2O (l) b.) Al (s) + HCl (aq) <--> AlCl3 (aq) + H2 (g) c.) Sb (s) + O2 (g) <--> Sb4O6 (s)



Answers

Write equilibrium constant expressions, $K_{p}$, for the reactions (a) $\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})$ (b) $\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$ (c) $2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons$ $\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

In this question, we need to write the expression for the equilibrium constant for different for four different equilibrium equations. Now, important here, when we write this expression, K. C. There's a few things to take note of when we do so. First of all we can write the concentrations of the products at the top of this expression. So as the numerator, I'm going to write the concentrations of the reactant as the denominator. Now, as soon as we have more than for instance, two of more than one product, then we need to multiply these concentrations. Um Then the other aspect is that as soon as we have strike geometrical efficient in front of one of the products or reactions that will be included in the expression. We need to write this these as exponents. And then the other aspect to take into account is that we emit the concentration of concentrations of solids, pure solids and liquids because they have constant concentration at a given temperature. Right? So let's take number eight. in this case the reactant is a solid pure solid. So, we're going to emit this concentration so this is going to be actually over one. And we're just going to multiply the concentrations of the products. So it's NH three and H C. L. So, this is just going to be the concentration of NH three times the concentration of hydrochloric acids. Right? Number B number be K C will be equal to at the top as numerator. We will have C. 02 carbon dioxide concentration times the concentration of into. And then very important here we have a strict geometrical vision of two. So, this must be the exponent in the case of thai atomic nitrogen gas. Right then, as denominated, we will have the concentration of, you know, two. And once again we have two as a psychometric coefficient for N 02. And therefore it should be the exponent. Yeah. All right. So this brings us to number C. Yeah. The K C expression. Full number C. He's equal to. All right. At the top these only one product and that is a solid. So we will not include that. Therefore one will be the numerator of this expression. Then as denominator, these system components solid and water. And the water is a pure liquid and then carbon dioxide gas. So we will not use the first two. We will emit the sodium carbonate, solid and the water liquid because these are pure substances of which the concentration is a constant at a given temperature. And we will therefore only include the concentration of the carbon dioxide here. Then the number 30 number D. K C. Is equal to. Mhm. In this case, once again we have um a solid which is the product only one product and that's a solid iron hydroxide. So therefore the numerator will be one and then the denominator will consist of Iron three Plus concentration times the hydroxide ions concentration. But remember we have a strict geometrical vision of three years, so this must be the exponent for the concentration of the hydroxide iron. So let's quickly just wreak. Up here we had to write the um equilibrium constant expression for equilibrium equations for this purpose. The concentrations of the products um act as numerator is in our expression, concentrations of the reactant as denominators. Um Story geometric coefficients must be used as exponents and then pure liquids and solids must be or the concentrations of pure liquids and solids must be emitted.

So here we are asked to write equilibrium constants. In our first case we have the reaction of silver, solid, nickel with carbon oxide and caches form leading to the formation of a complex in gaseous form. So we never include solids and are equal expression and we can express since they don't have a significant change in concentration and we express gases in terms of pressure. In our second case we have the reaction of Hydrofluoric acid with liquid water, which we're not going to include since its concentration doesn't change, leading to the formation of Hydro knee um and Aquarius fluoride and nine. And our equilibrium constant, which would be the acid dissociation constant in this case would be defined. And in our last case we have the reaction of chlorine gas with bromide. And uh I on and I am leading to the formation of grooming and liquid, which we are not going to include an equilibrium expression And Chloride and nine. So we're going to use equilibrium constant. Uh our general equilibrium constant and we adjust for stoke geometric coefficients. And since chlorine gas where we are going to use pressure and this gives our final answers

So here we are asked to write equilibrium constants. So we write products. Overreact and suggesting for strike geometric coefficients, not including for solids or liquids, which concentrations don't change significantly over the course of reaction. And for gas is typically we're going to use pressures so we can have the pressure of chlorine. In our second case we have a complex science dissociating and we have to use concentration for a quiz species. And our third case essentially we have the reaction of carbon oxide gas with water and gases form leading the formation of carbon dioxide and hydrogen gas and are equal and constant is going to be in terms of pressures. And in our last case we have the reaction of for hydro knee um with chloride and magnesium dioxide in solid form, which we aren't going to include in our equilibrium expression, leading to the formation of magnus two plus and water in liquid form. So we are not going to include these two species in our equilibrium expression. And you can find their equilibrium expression is the falling. And this gives our final answers

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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