This is possibly the best question I have been asked in chemistry - a smart kid from Indian once explained it to me and I have forgotten. He then explained it to the other teacher who then explained it to a text book author. Short answer just accept it!
I’ve been watching all your videos and they are all so helpful! THANK YOU!!! I have a question, about changing temperature. Is this true? Equilibrium shifts to the left when: temperature is increased in an exothermic reaction (kc decreases) and temperature is decreased in an endothermic reaction (kc decreases) Equilibrium shifts to the right when: temperature is increased in an endothermic reaction (kc increases) and temperature is decreased in an exothermic reaction (kc increases)
8 countries now (UK, USA, Russia, Turkey, Lebanon, Spain ,Czechia, Thailand). There are few independent schools in Oz - and almost all of them are religious... Alas, I am not.
Love all the videos, i just wanted to add that i like to remember the ways in which temperature effects the equilibrium point as... If delta H = exothermic which is a (-) negative sign, then it is favoured with a (-) negative or lower temperature. Makes more products And vice versa so... If delta H = endothermic which is a (+) positive sign, then it is favoured with a (+) positive or higher temperature. Makes more products
for the "evil question", I didnt work out the delta H being postiive or negative because of the bonds formed and made, instead my thinking was. An increase in pressure should cause equilibirum to shift to products as there are less gas particles, but overall, despite that the mixture ends up being dark brown, implying tempreture has a dominating effect on the equilibrium mixture. Would that explanation give me the marks or do I have to say something along the lines of what you said Mr. Thornley?
Hi Mr. Thornley, I was just wondering in the video at 3:32 you say that the equilibrium will shift to the left if you remove some oxygen from the reactants side. However for the first example with the colors you said that when the gas is removed from the products side, the equilibrium shifts over to the right side. So why is it for the example at 3:32, when you remove O2 from the reactants side the equilibrium still shifts to the left, should'nt it shift towards the right towards the porducts side to achieve equilibrium again?
Position of equilibrium means shifts to the left or right. Stresses often change the position of the equilibrium, making more or less product. Kc = [products]/[reactants] raised to their coefficient power. This does NOT change when the reaction is stresses UNLESS the stress is temperature change.
consider the hypothetical reaction quotient ([A])/([B]^2[C]) Let's say that as a result of increasing pressure, concentration was doubled. i.e. (2[A])/((2[B])^2(2[C])) 2([A])/8([B]^2[C]) In case that^ wasn't clear, this doubling of concentration is seen to make the denominator of the equation bigger than the numerator, and thus K is a smaller number now. Therefore, the reaction must shift to the right and more products must be made in order to keep K the same, constant number.
That questions at 6:28 --> wouldn't doubling N2 mean that the reaction opposes the stress and favours the exothermic forward reaction, increasing the temperature, which would affect the Kc? Great videos though
Um... with your Contact process example you said the opposite to what you said in the other two examples. In other words you said that when temperature increases it shifts to the exothermic but doesn't an increase in temperature cause the reaction to shift to the endothermic (and in this case left) side? P.S. Your videos are awesome :)
how does shift of equilibrium differ from changing the equilibrium constant? I know that only temperature affects Kc but I don't really understand why. Kc=[products]/[reactants] so if the equilibrium position shifted to the right, wouldn't more products be formed so wouldn't Kc be increased. So, if pressure shifted the equilibrium position, shouldn't Kc change?
Rich I think you have got this all wrong. Endothermic is to the right while Exothermic is to the left. E.g When the temperature increases, Le Chatelier's principle says the reaction will proceed in such a way as to counteract this change, ie lower the temperature. Resulting in endothermic reactions to move forward, and exothermic reactions to move backwards. Therefore forward reactions move to the right while backward reactions move to the left.
if i start off with a reaction at 70% products, 30% reactants when at equilibrium, and then change something e.g concentration of reactants, i get that the equilibrium shifts to the right so e.g 80-20% but if the rates balance out after that, i can't explain why it isn't 70-30% again? My head just goes back to thinking 'it'd just be 70% and 30% of a different number, same relationship though?' hope this makes sense, can someone explain this to me? :p
Um... with your Contact process example you said the opposite to what you said in the other two examples. In other words you said that when temperature increases it shifts to the exothermic but doesn't an increase in temperature cause the reaction to shift to the endothermic (and in this case left) side? P.S. Your videos are awesome :)
![Le Chatelier's Principle [IB Chemistry SL/HL]](http://i.ytimg.com/vi/Eq_ASIJYjPA/mqdefault.jpg)
![Le Chatelier's Principle [IB Chemistry SL/HL]](/img/tr.png)
![7.1/R2.3.3 The Magnitude of Kc [SL IB Chemistry]](http://i.ytimg.com/vi/fZe03sYAyAM/mqdefault.jpg)
![5.1 Energy change when the temperature of a substance is changed [SL IB Chemistry]](/img/n.gif)
Amazing videos! I can't imagine doing IB without your help for chemistry and EconPlusDal for economics! Thanks a lot! We appreciate it!
Ikr! :P
This is possibly the best question I have been asked in chemistry - a smart kid from Indian once explained it to me and I have forgotten. He then explained it to the other teacher who then explained it to a text book author. Short answer just accept it!
Videos are fantastic. I am reviewing chemistry with this and it has helped me put it into a new perspective.
U explain this so much better than my chem. teacher!:)
I’ve been watching all your videos and they are all so helpful! THANK YOU!!!
I have a question, about changing temperature. Is this true?
Equilibrium shifts to the left when: temperature is increased in an exothermic reaction (kc decreases) and temperature is decreased in an endothermic reaction (kc decreases)
Equilibrium shifts to the right when: temperature is increased in an endothermic reaction (kc increases) and temperature is decreased in an exothermic reaction (kc increases)
I will now forever remember Le Chatelier's principal and it's connection to equilibrium thanks to the equilibrium moustache of Mr Le Chatelier!
6 COUNTRIES! But do you ever plan to come to Australia to bless our land with your amazing teaching skills...
8 countries now (UK, USA, Russia, Turkey, Lebanon, Spain ,Czechia, Thailand). There are few independent schools in Oz - and almost all of them are religious... Alas, I am not.
Love all the videos, i just wanted to add that i like to remember the ways in which temperature effects the equilibrium point as...
If delta H = exothermic which is a (-) negative sign, then it is favoured with a (-) negative or lower temperature. Makes more products
And vice versa so...
If delta H = endothermic which is a (+) positive sign, then it is favoured with a (+) positive or higher temperature. Makes more products
for the "evil question", I didnt work out the delta H being postiive or negative because of the bonds formed and made, instead my thinking was. An increase in pressure should cause equilibirum to shift to products as there are less gas particles, but overall, despite that the mixture ends up being dark brown, implying tempreture has a dominating effect on the equilibrium mixture. Would that explanation give me the marks or do I have to say something along the lines of what you said Mr. Thornley?
I think so - I would award marks for this.
Hi Mr. Thornley, I was just wondering in the video at 3:32 you say that the equilibrium will shift to the left if you remove some oxygen from the reactants side. However for the first example with the colors you said that when the gas is removed from the products side, the equilibrium shifts over to the right side. So why is it for the example at 3:32, when you remove O2 from the reactants side the equilibrium still shifts to the left, should'nt it shift towards the right towards the porducts side to achieve equilibrium again?
Hi Mr Thornley, can you please explain the difference between the position of equilibrium and Kc?
Position of equilibrium means shifts to the left or right. Stresses often change the position of the equilibrium, making more or less product. Kc = [products]/[reactants] raised to their coefficient power. This does NOT change when the reaction is stresses UNLESS the stress is temperature change.
at 3:32 why is it that if you remove o2 the reaction shifts towards the reactants side to relieve the stress?
consider the hypothetical reaction quotient ([A])/([B]^2[C])
Let's say that as a result of increasing pressure, concentration was doubled.
i.e. (2[A])/((2[B])^2(2[C]))
2([A])/8([B]^2[C])
In case that^ wasn't clear, this doubling of concentration is seen to make the denominator of the equation bigger than the numerator, and thus K is a smaller number now. Therefore, the reaction must shift to the right and more products must be made in order to keep K the same, constant number.
No greater satisfaction than when you find the "evil" question trivial :)
This is beautiful
That questions at 6:28 --> wouldn't doubling N2 mean that the reaction opposes the stress and favours the exothermic forward reaction, increasing the temperature, which would affect the Kc?
Great videos though
Sick stuff bro. I enjoy chemistry myself.
Um... with your Contact process example you said the opposite to what you said in the other two examples. In other words you said that when temperature increases it shifts to the exothermic but doesn't an increase in temperature cause the reaction to shift to the endothermic (and in this case left) side?
P.S. Your videos are awesome :)
Does pressure changes ONLY apply to gases regarding equilibrium?
in the 2NO2 N2O4 reaction, since the particles are already equal, would increasing the pressure have no affect at all?
if 2 things turn into 1 thing then bonds must have been made.
Do we not care about limiting reagents?
For the IB test, do we have to know the Kp=Kc(RT)^delta(n) ?
how does shift of equilibrium differ from changing the equilibrium constant? I know that only temperature affects Kc but I don't really understand why.
Kc=[products]/[reactants] so if the equilibrium position shifted to the right, wouldn't more products be formed so wouldn't Kc be increased.
So, if pressure shifted the equilibrium position, shouldn't Kc change?
Does Kc only increase when the products are favored?
For the same reaction, with the same coefficients - yes
@@ibchemvids aha~ got it tysm!!!!
So for the temperature part, the reaction became redder because it favors the reverse reaction because it is endothermic correct?
Rich I think you have got this all wrong. Endothermic is to the right while Exothermic is to the left. E.g When the temperature increases, Le Chatelier's principle says the reaction will proceed in such a way as to counteract this change, ie lower the temperature. Resulting in endothermic reactions to move forward, and exothermic reactions to move backwards. Therefore forward reactions move to the right while backward reactions move to the left.
A Question.
Why is Temperature the only factor that changes Kc?
Because of Le Chatalier (according to Wikipedia). This makes sense to me, the fact that nothing else changes it makes less sense~
yup
if i start off with a reaction at 70% products, 30% reactants when at equilibrium, and then change something e.g concentration of reactants, i get that the equilibrium shifts to the right so e.g 80-20% but if the rates balance out after that, i can't explain why it isn't 70-30% again? My head just goes back to thinking 'it'd just be 70% and 30% of a different number, same relationship though?' hope this makes sense, can someone explain this to me? :p
Flying spaghetti monsters - I have never seen that equation before
You should come to my school, plz
NO NO NAWW NO NOOOO
Nice mustache equilibrium symbol joke!
North! Hope all is well.
thank you for this video it helped a lot thanks
IB exams comming up? haha
yup
Um... with your Contact process example you said the opposite to what you said in the other two examples. In other words you said that when temperature increases it shifts to the exothermic but doesn't an increase in temperature cause the reaction to shift to the endothermic (and in this case left) side?
P.S. Your videos are awesome :)