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The dissociation of molecular iodine into iodine atoms is represented as$$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$At $1000 \mathrm{K},$ the equilibrium constant $K_{\mathrm{c}}$ for the reaction is $3.80 \times 10^{-5} .$ Suppose you start with 0.0456 mole of $\mathrm{I}_{2}$ in a 2.30 -L flask at $1000 \mathrm{K}$. What are the concentrations of the gases at equilibrium?

$$\begin{array}{l}{\left[\mathrm{I}_{2}\right]=0.0194 \mathrm{M}} \\{[\mathrm{I}]=8.58 \times 10^{-5} \mathrm{M}}\end{array}$$

Chemistry 102

Chapter 14

Chemical Equilibrium

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10:03

In thermodynamics, a state of thermodynamic equilibrium is a state in which a system is in thermal equilibrium with its surroundings. A system in thermodynamic equilibrium is in thermal equilibrium, mechanical equilibrium, electrical equilibrium, and chemical equilibrium. A system is in equilibrium when it is in thermal equilibrium with its surroundings.

00:54

In chemistry, chemical equilibrium (also known as dynamic equilibrium) is a state of chemical stability in which the concentrations of the chemical substances do not change in the course of time due to their reaction with each other in a closed system. Chemical equilibrium is an example of dynamic equilibrium, a thermodynamic concept.

03:10

The equilibrium constant f…

02:08

$K_{c}=5.6 \times 10^{-12}…

01:45

The Equilibrium Constant a…

01:28

Hydrogen gas and iodine ga…

01:27

So for the falling balanced chemical equation, we want to know what the equilibrium concentration values are from both our product and a reactive. So since we're looking at values at equilibrium, you want to make in ice table. But before we can fill anything in, we have to convert to concentrations that we know what are starting concentration for our reactant is so we were told that four the starting concentration initial concentration for I a dying. We have 0.456 moles. And we were also told the volume 2.3 leaders. So this is equivalent to 0.198 So we know that this is our starting initial concentration for are I don gas. So you put this in here. You This is zero. This is our product and this is initials. We have a no product initially, our change. We aren't sure. So we're gonna put an X because we do not know this value exactly. Whatever the changes, we could have a positive two times that change. Yes, we have a two right here. That means at equilibrium with C, the initial minus the change. And here we see just two times that change value. Thanks. So since we were given the equilibrium, constant equals 3.8 times 10 to the native five. We can plug in our products over our reactant and we raises to a power of two on our numerator because our products have in our story geometry coefficient of to. So if we were to calculus, calculate this out and sulfur X, you find that X is equivalent to 4.2 89 rounded and we see that is times 10 to the negative four. So now that we have this X value, we can go ahead and find the concentration of are reacting and the concentration of our product at equilibrium. So we know that to find equilibrium, we saw that we had 0.19 eight minus x you know, X now to be four point to 89 times 10 to the native four, which is equivalent to zero point 019 four. And here we have two times eggs. You see here we know x to be 4.289 times 10 to the native for and this so much smaller value 8.58 times 10 to the negative five

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