(a) Con base en las constantes de disociación del apéndice D, determine el valor de la constante

step 1 This isn't too hard, but it takes a little time. Using dissociation constants from Appendix B, e

(a) Con base en las constantes de disociación del apéndice...

(a) Con base en las constantes de disociación del apéndice $\mathrm{D}$, determine el valor de la constante de equilibrio de las reacciones siguientes. (Recuerde que, cuando se su-man las reacciones, las constantes de equilibrio correspondientes se multiplican.) (i) $\mathrm{HCO}_3^{-}(a c)+\mathrm{OH}^{-}(a c) \rightleftharpoons \mathrm{CO}_3{ }^{2-}(a c)+\mathrm{H}_2 \mathrm{O}(l)$ (ii) $\mathrm{NH}_4{ }^{+}(a c)+\mathrm{CO}_3{ }^{2-}(a c) \rightleftharpoons \mathrm{NH}_3(a c)+\mathrm{HCO}_3{ }^{-}(a c)$ (b) Se acostumbra emplear flechas únicas en las reacciones cuando la reacción directa es apreciable ( $K$ es mucho mayor que 1) o cuando los productos escapan del sistema, de modo que nunca se establece el equilibrio. Siguiendo esta convención, ¿cuál de estos equilibrios se podría escribir con una sola flecha?

(a) Con base en las constantes de disociación del apéndice $\mathrm{D}$, determine el valor de la constante de equilibrio de las reacciones siguientes. (Recuerde que, cuando se su-man las reacciones, las constantes de equilibrio correspondientes se multiplican.)
(i) $\mathrm{HCO}_3^{-}(a c)+\mathrm{OH}^{-}(a c) \rightleftharpoons \mathrm{CO}_3{ }^{2-}(a c)+\mathrm{H}_2 \mathrm{O}(l)$
(ii) $\mathrm{NH}_4{ }^{+}(a c)+\mathrm{CO}_3{ }^{2-}(a c) \rightleftharpoons \mathrm{NH}_3(a c)+\mathrm{HCO}_3{ }^{-}(a c)$
(b) Se acostumbra emplear flechas únicas en las reacciones cuando la reacción directa es apreciable ( $K$ es mucho mayor que 1) o cuando los productos escapan del sistema, de modo que nunca se establece el equilibrio. Siguiendo esta convención, ¿cuál de estos equilibrios se podría escribir con una sola flecha?

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Key Concepts

Equilibrium Constant

The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations or activities of products to reactants at equilibrium, each raised to the power corresponding to its stoichiometric coefficient. It quantifies the position of equilibrium and indicates the extent to which a reaction favors products or reactants.

Multiplicative Combination of Equilibrium Constants

When multiple reactions are added together to form an overall reaction, their individual equilibrium constants multiply to yield the equilibrium constant of the net reaction. This principle allows complex reactions to be understood and computed via the combination of simpler equilibrium processes.

Arrow Notation in Chemical Equilibria

Arrow notation in chemical equations conveys the direction and reversibility of a reaction. A single arrow is generally used when the reaction proceeds virtually to completion, often due to an overwhelming equilibrium constant or because products are continuously removed from the system, whereas a double arrow indicates that the reaction is reversible and has reached a dynamic equilibrium.

Transcript

00:01 This isn't too hard, but it takes a little time.

00:03 Using dissociation constants from appendix b, excuse me, appendix d, determine the value for the equilibrium constant for each of two reactions that are given.

00:16 Remember that when reactions are added, the corresponding equilibrium constants are multiplied.

00:25 Okay, reaction one is carbonate.

01:02 Okay, that looks good.

01:07 Checking in the back of the book, our ka is 5 .6 times 10 to the minus 11.

01:21 We know that this reaction, and these are both a q, we know that our kw is 1 times 10 to minus 11th, what i say minus 11, minus 14th.

01:53 So keq is our unknown.

02:06 Keq will be equal to our ka times our kw which will be equal to 5 .6 times 10 to the minus 11th.

02:33 Hang on just a second here and make sure i got the right answer.

02:58 Okay times one times 10 to the 14th minus 11.

03:08 This equals 5 .6 times 10 to the third and that is for our first reaction.

03:26 Hang on again here.

03:35 Our second reaction is, as follows.

04:15 And for this, we know that our ka is 5 .6 times 10 to the minus tenth.

04:25 Then let's look at the following reaction, which will be 1 over our ka2, 1 .785 times 10 to the minus tenth.

05:08 Then i'm going to use the, excuse me, the same equation.

05:14 Well, i guess i better write it down...

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