O valor de K, na decomposiçăo do hidrogenossulfeto de amônio...

O valor de $K$, na decomposiçăo do hidrogenossulfeto de amônio é $1,8 \times 10^{-4}$, a $25^{\circ} \mathrm{C}$. (a) Quando o sal puro se decompōe num balio, quais as concentraçöes de equilibrio do $\mathrm{NH}_3$ e do $\mathrm{H}_2 \mathrm{~S}$ ? (b) Se o $\mathrm{NH}_4 \mathrm{HS}$ for colocado num balão que contém $0.020 \mathrm{~mol}$ $\mathrm{L}$ de $\mathrm{NH}_3$ e depois o sistemat entra em equilforio, quais as concentraçöes do $\mathrm{NH}_1$ e do $\mathrm{H}_2 \mathrm{~S}$ neste equilibrio?

O valor de $K$, na decomposiçăo do hidrogenossulfeto de amônio é $1,8 \times 10^{-4}$, a $25^{\circ} \mathrm{C}$.
(a) Quando o sal puro se decompōe num balio, quais as concentraçöes de equilibrio do $\mathrm{NH}_3$ e do $\mathrm{H}_2 \mathrm{~S}$ ?
(b) Se o $\mathrm{NH}_4 \mathrm{HS}$ for colocado num balão que contém $0.020 \mathrm{~mol}$ $\mathrm{L}$ de $\mathrm{NH}_3$ e depois o sistemat entra em equilforio, quais as concentraçöes do $\mathrm{NH}_1$ e do $\mathrm{H}_2 \mathrm{~S}$ neste equilibrio?

Key Concepts

Chemical Equilibrium

Chemical equilibrium is the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant concentrations of the reactants and products. This concept is fundamental in understanding how reactions behave over time without any net change in the macroscopic properties of the system.

Equilibrium Constant (K)

The equilibrium constant is a numerical value that expresses the ratio of the concentrations (or pressures) of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients. It provides insight into the extent of reaction and how far a reaction proceeds under specific conditions.

ICE Table Method

An ICE (Initial, Change, Equilibrium) table is a systematic tool used to organize and calculate the concentrations of reactants and products at equilibrium. By tabulating the initial amounts, the changes that occur as the system moves toward equilibrium, and the final equilibrium concentrations, it simplifies solving equilibrium problems with unknown quantities.

Stoichiometry in Equilibrium

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. In the context of equilibrium, it is used to relate the changes in concentrations of different species based on the balanced chemical equation, ensuring that the conservation of mass is maintained while calculating equilibrium concentrations.

Transcript

00:01 Okay, so in this question, they want us to calculate the equilibrium concentration of nh3 and the h2s.

00:08 Right? so they tell us decay is 1 .8 times 10 to a negative 4 at 25 degrees celsius.

00:16 In order to solve this question, we need to set up an icebox.

00:20 They didn't give us an initial concentration of anything.

00:23 So we can just infer, let's say we have one molar of nh4hs, and at t equals 0, we should have 0.

00:32 Of both of these right so we should have zero product at time equals zero and as time progresses we would have a decrease in the amount of nh4 hs because it's being converted into nh3 and h2s right so we have plus x plus x the reason why it's all 1x is because all of these are 1 to 1 ratios they're all 1 moles right so let's write the equilibrium expression now so we have 1 minus x 0 plus x is x and 0 plus x is x now let's write the k expression we have k is equal to the concentration of the product over the concentration of the reactant.

01:13 But we've got to be careful here because if we look at this, you see that this is a solid and any solid or liquid are not included in the k expression instead they just thought of as one.

01:23 Right.

01:23 So if this is a one, this means that the expression, the k expression is only the concentration of nh3 times h2s.

01:33 Right, we know k is equal to 1 .8 times 10 to the negative 4 and we know both h3 and h2s is equal to x right so if we multiply them together you'll be x square so x would be 0 .013m now we can use x to solve for the equilibrium concentration so we know that nh3 is equal to concentration of h2s because they're both equal to x right and x is equal to 0 .013 so this is the concentration of nh3 and h2s now, the question they only asked us to solve for the concentration of these two, but i'm going to solve for the concentration of this because it will help us in part b.

02:19 So, concentration of nh4hs is equal to 1 minus x, which is equal to 1 minus 0 .013.

02:30 And we get 0 .987 as our concentration.

02:39 Now for part b, they're asking us what the equilibrium concentration would be if we started off with, a certain amount of nh3 right they say 0 .02 moles per liter so let me rewrite the equations we have nh4hs and we have nh3 plus h2s right so i see e we know the initial concentration of nh4 hs is 0 .987 right it's from here that's why i calculated before and we know the initial concentration of h2s is 0 .03, right? and the initial concentration of nh3 is not 0 .013.

03:30 The reason why is because they said that in the question, what if it already had 0 .02 molars, right? so it'll be 0 .013 plus 0 .02, which is equal to 0 .03...

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