Consider the following equilibrium: COBr2(g) ⇌CO(g)+Br2(g) Kc=0.190 at 73^∘ C (a) A 0.50 mol s

Consider the following equilibrium: COBr2(g) ⇌CO(g)+Br2(g) ...

Consider the following equilibrium: $\operatorname{COBr}_{2}(g) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=0.190$ at $73^{\circ} \mathrm{C}$ (a) $A$ 0.50 mol sample of $\operatorname{COBr}_{2}$ is transferred to a 9.50-L. flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species. (b) The volume of the container is decreased to 4.5 L and the system allowed to return to equilibrium. Calculate the new equilibrium concentrations. (Hint: The calculation will be easier if you view this as a new problem with 0.5 mol of $\mathrm{COBr}_{2}$ transferred to a 4.5 -L flask. (c) What is the effect of decreasing the container volume from 9.50 L. to 4.50 L?

Consider the following equilibrium:
$\operatorname{COBr}_{2}(g) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=0.190$ at $73^{\circ} \mathrm{C}$
(a) $A$ 0.50 mol sample of $\operatorname{COBr}_{2}$ is transferred to a 9.50-L. flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.
(b) The volume of the container is decreased to 4.5 L and the system allowed to return to equilibrium. Calculate the new equilibrium concentrations. (Hint: The calculation will be easier if you view this as a new problem with
0.5 mol of $\mathrm{COBr}_{2}$ transferred to a 4.5 -L flask.
(c) What is the effect of decreasing the container volume from 9.50 L. to 4.50 L?

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state of a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, meaning the concentrations of reactants and products remain constant over time. It is a dynamic state where, despite ongoing molecular activity, no net change in the composition of the system is observed.

Equilibrium Constant (Kc) and its Expression

The equilibrium constant, Kc, is a numerical value that expresses the ratio of the concentrations of products to reactants, each raised to the power of their respective stoichiometric coefficients, at equilibrium. It provides a measure of the position of equilibrium and indicates whether the formation of products or reactants is favored in a given reaction at a specific temperature.

ICE Table Method

The ICE table is a systematic tool used to organize the initial concentrations, changes, and equilibrium concentrations of reactants and products in a chemical reaction. It helps simplify the process of setting up equations based on the stoichiometry of the reaction, which is particularly useful when solving for unknown concentrations at equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle states that if an external change is imposed on a system at equilibrium, such as a change in concentration, pressure, or temperature, the system will adjust in a way that counteracts the imposed change. For reactions involving gases, changes in volume (and thus pressure) can significantly shift the equilibrium position, favoring the side with fewer moles of gas when volume decreases.

Transcript

00:01 Okay, so in this question we have the following equilibrium expression and they tell us the k value is 0 .19 at 73 degrees celsius.

00:10 And they want us to calculate the equilibrium concentration of all of these.

00:15 Right, so the first thing that we should do is we should set up an icebox.

00:19 So, i is the initial concentration.

00:21 C is change in concentration and e is the equilibrium concentration.

00:27 Right? so the first thing that we should do is find out what the concentration is.

00:31 That we have for cobr2, right? because it tells we have 0 .5 modes of c obr2 and we have a volume of 9 .5 liters.

00:42 This means that all we got to do is take the modes and divide it by the leaders and we get the concentration, right? so the concentration of c obr2 is equal to 0 .5 modes divided by 9 .5 liters, 9 .5 liters, which is 0 .053 molarity.

01:01 0 .053.

01:03 This is also can be written as moles per liter.

01:08 So we know that at the beginning we have no co and no br2, right? because at t equals 0, we only have cobr2, but after some time we'll have co and br2 and we'll have a decrease in cobr2, right? so after some time we'll have a decrease in cobr2, it's minus x.

01:27 And now it's plus x for co and br2.

01:31 The reason why it's plus x is because we see that in this equation, every one mole of co br2 that disappears, we get one more of co, right? so that's why it's plus x.

01:43 So the amount of cobr2 that disappears, we get the same amount of co popping up, and that is the same for br2.

01:51 Now let's write the equilibrium expression.

01:54 Well, we see that 0 plus x is just x, 0 plus x is x, and we see that 0 .053 minus x is the equilibrium expression for c .o .53 .5 .0 .5 .000 is the equilibrium expression for c.

02:02 O br2.

02:04 Now we have the k value, right? so we can solve for x.

02:08 Right.

02:08 So k is equal to the concentration of the products over the concentration of the reactant.

02:20 So k is equal to 0 .19.

02:24 We know what co and br2 is.

02:26 It's x, right? so i'm just going to write it as x square, because x times x is x square.

02:31 And then we have the concentration of cobr2, which is 0 .05, oops.

02:37 0 .05.

02:40 Now what we can do is we can cross multiply.

02:44 So 0 .19 is the same thing as 0 .19 over 1.

02:48 So we cross multiply and we get x.

02:54 Let me a second.

02:56 X square is equal to 0 .19 times it by 0 .053, which is 0 .11007 minus .19x, right? so what we do now is we just plug this into the quadratic formula.

03:20 Actually, the first we need to set it equal to 0 .0...

Please give Ace some feedback