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Photon336
 one year ago
Intro to equilibrium tutorial
Photon336
 one year ago
Intro to equilibrium tutorial

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Photon336
 one year ago
Best ResponseYou've already chosen the best response.16What is equilibrium? When we have a state where our forward and reverse reactions are happening at the same RATE, then we can say that our reaction has reached an equilibrium. A common example is a bridge and there are two sides to this bride. One side we label the reactants while the other we name products.say the bridge is two ways and we have cars going both ways. well, lets say we have 8 cars going one way, and 8 going the other and this stays constant. then we say that this is equilibrium has been reached. there could be any number of cars on both sides; the number could be equal; there may be more on one side than another side; we cannot specify this directly (will be explained later). dw:1440634757911:dw TIP: The reversible arrows in your reaction, mean your reaction can go both ways reactants to products and products to reactants. dw:1440634698190:dw What it doesn’t mean it doesn’t necessarily mean that the concentrations of the reactants and products are equal; it just means that our forward and reverse reactions are happening at the same rate. What we can do: We can figure out what the ratio of concentrations reactants and products are at equilibrium, and we introduce a concept called Keq to explain this.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16What is Keq? Keq is our equilibrium constant, like any equilibrium expression Kw, Ksp, Keq, they are dependent on temperature, so the value of our Keq will change as we change the temperature. it shows the concentrations of our products over reactants raised to the stoichiometric coefficients. \[Ac + Bd > Ce + Df \] \[Keq = \frac{ [A]^{c}[B]^{d} }{ [C]^{e}[D]^{f}}\] Example \[CO _{g} + Cl _{2}g \rightarrow COCl _{2} \] this is reversible First make sure our reaction is balanced. (Yes or no) if so let’s move on Now identify the what phase the products and reactants are in; they are all gaseous phase so we can include them all. identify the coefficients, we’re going to raise them to the power of the coefficients we can see that they are all 1. \[\frac{ [COCl _{2}] }{ [CO][Cl _{2]} } = K _{eq}\] .

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16#1 Try one on your own: what’s the equilibrium constant for this reaction below? : \[2NO(g) + O _{2}g > 2NO _{2}(g) reversible \] Observation: do we include solids and pure liquids in our expression? ANSWER: No we do not include them, because think about it: the concentrations of solids and pure liquids don’t change very much. gases are included #2 Try this as well: CaCO3(s) reversible CaO(s) + CO(g) what’s the Keq? \[CaCO _{3}(s) > CaO(s)+CO(s)\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16Keq is a ratio of the concentration of the products over the reactants at equilibrium. so this is useful because it shows if the reaction will favor the products or reactants once at equilibrium. Keq = \[\frac{ Products }{ Reactants }\] generally it’s a fraction and if you’re not a math wiz like me there’s a simpler way to explain it. if Keq >1 this means that in the fraction \frac{ Products }{ Reactants } products must be bigger than the reactants in the fraction. At equilibrium our products will be favored if Keq < 1 \frac{ Products }{ Reactants } reactants must be bigger in the fraction, so that makes our fraction less than 1 at equilibrium there will be more reactants. if Keq = 1 then the products/reactants must = 1 and their concentrations are equal at equilibrium meaning. both forward and reverse reactions are equally favored. #3 In the reaction 2AB(g) +C2(g) reversible →2ABC(g) Keq = 100 what can we say about this reaction, how do you know this? You see where this is going right? equilibrium will lie for that particular reaction. Keq>1 products, Keq < 1 reactants Keq = 1 neither is favored. Try a question. Someone posted this question before on openstudy, I really thought this was an excellent question.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16#4 Which statement is true of a reversible reaction at equilibrium? a) The concentration of reactants is less than the concentration of products. b) The concentration of reactants and the concentration of products are equal c)The concentration of reactants is greater than the concentration of products. d) The concentration of reactants and the concentration of products are constant. e) The concentration of reactants is decreasing and the concentration of products is increasing. I encourage you to do this question.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16Another concept is called: Q which is like K but rather think of it is the ratio of concentrations of products to reactants at any time in the reaction other than equilibrium. for starters let’s say that Q = K., then that’s great because our reaction is at equilibrium. now what if Q < K? what does this mean? there are less reactants than there should be, so our reaction is going to produce more products. Q>K? this means that there is too much product, more than there should be so our reaction is going to produce more reactants. #5 in the following reaction CX(g) reversible → C(g) + X(g) Say Keq = 0.05 What happens when Q = the following and why? Q = 0.05 When Q = 0.005 Q = 1.25

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16You can calculate your equilibrium constant K easily if you know what the equilibrium concentrations are. If the problem gives you equilibrium concentrations, then you set up the expression for Keq and then plug the numbers in. Sometimes they won’t give you the equilibrium concentrations and you may be asked to find what K is given some starting concentrations. this is harder because you have to do ICE tables and stuff like that. #6 For the following reaction (say that we know that these are the equilibrium concentrations): N2 = 0.10 M, O2 = 0.20 M and NO = 0.20M \[N _{2}g + O _{2}g > 2NO\] this reaction is reversible Try another one this will see if you understood another concept mentioned: I totally made the numbers up. #7 CaCO3(s) reversible CaO(s) + CO(g) let’s say if the equilibrium concentrations are 0.40 for CaCO(s) CO(g) 1.20 and CaO(s) 0.80. can you tell me what Keq would be and whether products or reactants would be favored at equilibrium?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16In case you're interested. 1.Keq = [NO2]^2/[NO]^2[O2] 2.Keq = [CO] solids aren’t included 3.Keq > 1 so reaction favors products at equilibrium. 4.d, the concentrations remain constant. 5.a. equilibrium is reached, b. Q<K rxn gies to right, C. Q>K rxn goes to left. 6. [0.40]/[0.10][0.20] = 2 7.solids aren’t included so K = 0.80.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16Thanks for reading this

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16Keq = 1, means both forward + reverse equally favored, (neither could mean that one isn't favored over the other)

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0This was a really good explanation, I always confuse K and Q rules so it was a good reminder!

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Neat Tutorial... Keep up the good work! :) Awesomeness!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Awesome, thanks for sharing!

Photon336
 one year ago
Best ResponseYou've already chosen the best response.16Thanks guys; if there's any suggestions please tell me

Ciarán95
 one year ago
Best ResponseYou've already chosen the best response.1Brilliantly explained @Photon336  really informative and easy to follow! Thanks!

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0@photon Awesome tutorial and very helpful. Keep u the good work!
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