Gchem Question: #93 from Destroyer

[Vaheb]

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This is from the 2011 edition and it asks:

H2O(l) + H2O(l) --> OH-(aq) + H3O+(aq)

When the temperature is decreased from 25 Celcius, it is found that the Kw has decreased. Which of the following are true?

a) The reaction must be exothermic
b) The reaction must be endothermic
c) The reaction is thermoneutral
d) Not possible. Kw = 1 x 10^-14 at all temperatures
e) None of these.

So I quickly got rid of D because I know that when temperature changes, K will always change.

I just don't understand the solutions:

Recall that increasing the temperature will favor the endothermic reaction, and decreasing the temperature will favor the exothermic reaction. Since the reaction temperature is decreasing, but the Kw has gotten smaller means that the reaction does not favor the exothermic reaction, thus must be endothermic.

I just don't understand how the changing of temperatures (increasing or decreasing) will favor endothermic or exothermic, respectively.

Can someone explain that please? and what Kw has to do with favoring a side of the reaction?

Thank you!

Edit(different topic): Can someone also explain how we can figure out if a given structure like BCl4 is polar or non polar and all that good stuff. Some examples would be nice 🙂 I've forgotten that topic.

 

[jevan]

For the first q-First off a Kw is the same as any K(equilibrium constant) which means that it is (the conc of products raised to its coeffiecient)/(conc reactant to its coefficient)- only diff is Kw refers specifically to water dissociating into its products, H3O+ and OH-.Dont worry about the exact equation with coefficients for the conceptual problems tho, just know that a larger numerator(more products) makes the K larger and vise versa.
Secondly,when ur told a process is exothermic, just imagine that heat is a product (which makes sense bcs the rexn gives off heat)- similarly an endothermic process will use heat, so imagine it as one of the reactants(it gets used up just like any other reactant)
so now that u have that straight when u write out the reaction make sure to put heat on whichever side it belongs...exothermic its a product(right side), endothermic its a reactant(left side)
so using le chateliers principle...a system doesnt like to be changed- so when something disturbs the system it moves in whichever direction will restore its previous state of equilibrium....that said, if i add 10 parts product to a system with 50 parts reactants /50 parts products at equilibrium, the system will take 5 of those xtra products and turn em into reactants so its 55/55, which is just the balance that the system is looking for.
So heat acts exactly the same way as this(u prob already figured out the problem by now)-
so if the dissociation of water were exothermic(heat is product) and heat was removed/we made it colder (products are taken away), we'd shift more reactants to products, making the equilibrium constant larger(bcs K~p/r)
if the process were endothermic (heat is reactant) and we remove heat, some product is now turned into reactant, and the K would be smaller-so it must be endothermic...sorry that was so longwinded

 

[rmm30]

For the first q-First off a Kw is the same as any K(equilibrium constant) which means that it is (the conc of products raised to its coeffiecient)/(conc reactant to its coefficient)- only diff is Kw refers specifically to water dissociating into its products, H3O+ and OH-.Dont worry about the exact equation with coefficients for the conceptual problems tho, just know that a larger numerator(more products) makes the K larger and vise versa.
Secondly,when ur told a process is exothermic, just imagine that heat is a product (which makes sense bcs the rexn gives off heat)- similarly an endothermic process will use heat, so imagine it as one of the reactants(it gets used up just like any other reactant)
so now that u have that straight when u write out the reaction make sure to put heat on whichever side it belongs...exothermic its a product(right side), endothermic its a reactant(left side)
so using le chateliers principle...a system doesnt like to be changed- so when something disturbs the system it moves in whichever direction will restore its previous state of equilibrium....that said, if i add 10 parts product to a system with 50 parts reactants /50 parts products at equilibrium, the system will take 5 of those xtra products and turn em into reactants so its 55/55, which is just the balance that the system is looking for.
So heat acts exactly the same way as this(u prob already figured out the problem by now)-
so if the dissociation of water were exothermic(heat is product) and heat was removed/we made it colder (products are taken away), we'd shift more reactants to products, making the equilibrium constant larger(bcs K~p/r)
if the process were endothermic (heat is reactant) and we remove heat, some product is now turned into reactant, and the K would be smaller-so it must be endothermic...sorry that was so longwinded

Nice👍

 

[SN1]

This is from the 2011 edition and it asks:

H2O(l) + H2O(l) --> OH-(aq) + H3O+(aq)

When the temperature is decreased from 25 Celcius, it is found that the Kw has decreased. Which of the following are true?

a) The reaction must be exothermic
b) The reaction must be endothermic
c) The reaction is thermoneutral
d) Not possible. Kw = 1 x 10^-14 at all temperatures
e) None of these.

So I quickly got rid of D because I know that when temperature changes, K will always change.

I just don't understand the solutions:

Recall that increasing the temperature will favor the endothermic reaction, and decreasing the temperature will favor the exothermic reaction. Since the reaction temperature is decreasing, but the Kw has gotten smaller means that the reaction does not favor the exothermic reaction, thus must be endothermic.

I just don't understand how the changing of temperatures (increasing or decreasing) will favor endothermic or exothermic, respectively.

Can someone explain that please? and what Kw has to do with favoring a side of the reaction?

Thank you!

Edit(different topic): Can someone also explain how we can figure out if a given structure like BCl4 is polar or non polar and all that good stuff. Some examples would be nice 🙂 I've forgotten that topic.

Is the answer endothermic?
The reaction shows bonds being broken so any reaction that involves breaking of a bond is endothermic. Heat is on the left side of the reaction.

You can check if it is right by referring back to the question. If the reaction is indeed endothermic that means heat is on the reactant side and if you decrease the temperature, equilibrium will shift to the left and decrease the concentration of the products. When you decrease concentration of the products, you are also decreasing the K value because K=[product]/[reactant].


And if the reaction was exothermic, that means heat is on the product side. This can NOT be possible since the question tells you that Kw decreased when temperature is decreased. If the reaction was exothermic, then Kw should have increased when you decrease the temperature.

 

[jevan]

Is the answer endothermic?
The reaction shows bonds being broken so any reaction that involves breaking of a bond is endothermic. Heat is on the left side of the reaction.

You can check if it is right by referring back to the question. If the reaction is indeed endothermic that means heat is on the reactant side and if you decrease the temperature, equilibrium will shift to the left and decrease the concentration of the products. When you decrease concentration of the products, you are also decreasing the K value because K=[product]/[reactant].


And if the reaction was exothermic, that means heat is on the product side. This can NOT be possible since the question tells you that Kw decreased when temperature is decreased. If the reaction was exothermic, then Kw should have increased when you decrease the temperature.

umm i think its true that bond breaking itself is endothermic, but in this case and in most processes, new bonds are formed too (H-O bond in H3O+), and other factors like ion-dipole interactions and many other factors too difficult to project will also be factors in net energy change...that said I wouldnt bank on the fact that all dissociations are endothermic...in fact they r usually exothermic...the rest is good stuff tho

 

[SN1]

umm i think its true that bond breaking itself is endothermic, but in this case and in most processes, new bonds are formed too (H-O bond in H3O+), and other factors like ion-dipole interactions and many other factors too difficult to project will also be factors in net energy change...that said I wouldnt bank on the fact that all dissociations are endothermic...in fact they r usually exothermic...the rest is good stuff tho

Yeah I was kind of hesitant about the bond-breaking and bond-making part too
I think this question is a bit tricky and probably won't be on the real DAT. It is a good resource to reinforce on le chatelier, endo, and exo concepts though.

 

[jevan]

Yeah I was kind of hesitant about the bond-breaking and bond-making part too
I think this question is a bit tricky and probably won't be on the real DAT. It is a good resource to reinforce on le chatelier, endo, and exo concepts though.

ya it wont be this heavy for most dat q's, tho i did get a very involved q on this on my dat-but only one and it was one of the hardest ones on the test- but i think knowing the concepts well is very important in general, bcs then u can be confident in tackling any problem they throw at u...specifically i'd try to be conceptually sound with energy transfer/diagram stuff, gas laws, le chatelier,and nuclear chem

 

[jevan]

vaheb- ur second q is little tougher to fully explain, i think its easier to see the polarity of the molecule with a more visual explanation once u determine the structure of the molecule, but it is pretty easy if someone shows u...as for determining the structure tho- see how many of each element u have, and count the total amt of valence electrons in the molecule-
for example;(I'd draw this out if i were u or itll be difficult to follow) given PBr3 we add 5 electrons for the phosphorous plus 7 valence e's each for the 3 bromines(21 for the 3 Br's) so a total of 26 valence e's...
now we have to take some educated guesses as to which atom is going to be at the center, but usually its pretty obvious. Put P at the center and the Br's surrounding it- first things first, make sure to complete the octets on the peripheral atoms- so now we hav each Br surounded by 8 e's and 2 e's of each Bromine are shared with phosphorous in a bond- great we hav 2 e's to use and P needs 2 more to complete its octet...u should count that there are a total of 26 e's now-
geometry around the P is now 4 electron domains (3 bonding, 1 lone pair) which is going to be a tetrahedral-like geometry- meaning the orientation of 4 electron zones around the central atom will be tetrahedral, but the lone pairs dont get included when we name the shape, think of it like theyre kinda invisible....that said,the only way to have a nonpolar tetrahedral derivative molecule is to hav exactly 4 of the same atoms bonded to a central atom like 4 CCl4 or CH4 bcs even if each bond is polar, the net polarity cancels out bcs of symmetry. so in our case this molecule has a net dipole, or is polar.... So, when determing the geometry for PBr3 the molecule is actually not tetrahedral shaped, here it turns out if we just look at the 3 Br's coming off the P it is a trigonal pyramid.... hope this helps

 

[Vaheb]

For the first q-First off a Kw is the same as any K(equilibrium constant) which means that it is (the conc of products raised to its coeffiecient)/(conc reactant to its coefficient)- only diff is Kw refers specifically to water dissociating into its products, H3O+ and OH-.Dont worry about the exact equation with coefficients for the conceptual problems tho, just know that a larger numerator(more products) makes the K larger and vise versa.
Secondly,when ur told a process is exothermic, just imagine that heat is a product (which makes sense bcs the rexn gives off heat)- similarly an endothermic process will use heat, so imagine it as one of the reactants(it gets used up just like any other reactant)
so now that u have that straight when u write out the reaction make sure to put heat on whichever side it belongs...exothermic its a product(right side), endothermic its a reactant(left side)
so using le chateliers principle...a system doesnt like to be changed- so when something disturbs the system it moves in whichever direction will restore its previous state of equilibrium....that said, if i add 10 parts product to a system with 50 parts reactants /50 parts products at equilibrium, the system will take 5 of those xtra products and turn em into reactants so its 55/55, which is just the balance that the system is looking for.
So heat acts exactly the same way as this(u prob already figured out the problem by now)-
so if the dissociation of water were exothermic(heat is product) and heat was removed/we made it colder (products are taken away), we'd shift more reactants to products, making the equilibrium constant larger(bcs K~p/r)
if the process were endothermic (heat is reactant) and we remove heat, some product is now turned into reactant, and the K would be smaller-so it must be endothermic...sorry that was so longwinded

Edit: Nvm, I got it

Thank you so much