help with questions

[bturville]

For the first one, its just half...simple arithmetic from the given activation energy. 149kj
For ii, it should be 320kj. Work out the EA for the reverse by subtracting the delta H from the activation energy, to get 160kj/mol, then multiply by two because theres 2 moles. Any questions with delta H or EA you should always draw an energy diagram if you get stuck.

I did that in my head, hopefully its correct

[Rosie]

Another question:
Q. considering the reaction: Hg_(l) + 0.5I_2(g) ----> HgI_(g) 

i.) what is the Ea for the reaction 0.5Hg_(l) + 0.25I_2(g) ----> 0.5HgI_(g)?
ii.) what is the Ea value for the reaction 2HgI_(g)  ------> 2Hg_(l) + I_2(g)?

solution (this is how i worked it out):
i.) Ea of forward reaction is 298kJ/mol
delta H for reaction is +138kJ/mol, because reaction has been halved, the delta H is squared,
therefore, 298 + 11.75 = 309kJ
why am i wrong. i thought that when working out these calculations, you just alter the delta H without changing the activation energy.   

[dusty_girl1144]

Another question:
Q. considering the reaction: Hg_(l) + 0.5I_2(g) ----> HgI_(g) 

i.) what is the Ea for the reaction 0.5Hg_(l) + 0.25I_2(g) ----> 0.5HgI_(g)?
ii.) what is the Ea value for the reaction 2HgI_(g)  ------> 2Hg_(l) + I_2(g)?

thanks

yeah the delta H for this reaction is +138kJ/mol and the activation energy for the reaction is 298 kJ/mol. There were lots of parts to this question that's why i forgot to add things in.

for some reason im thinking i) is just half of 298
and
ii) 298 X 2

 

Another question:
Q. considering the reaction: Hg_(l) + 0.5I_2(g) ----> HgI_(g) 

i.) what is the Ea for the reaction 0.5Hg_(l) + 0.25I_2(g) ----> 0.5HgI_(g)?
ii.) what is the Ea value for the reaction 2HgI_(g)  ------> 2Hg_(l) + I_2(g)?

solution (this is how i worked it out):
i.) Ea of forward reaction is 298kJ/mol
delta H for reaction is +138kJ/mol, because reaction has been halved, the delta H is squared,
therefore, 298 + 11.75 = 309kJ
why am i wrong. i thought that when working out these calculations, you just alter the delta H without changing the activation energy.   

and for some reason im guessin ur answer is something to do with not Ea but rather with the equlibrium constant....? as i remember that is square with "K" the reaction equation is doubled. square rooted with halved and put to the reciprical when equation is reversed...

[bturville]

and for some reason im guessin ur answer is something to do with not Ea but rather with the equlibrium constant....? as i remember that is square with "K" the reaction equation is doubled. square rooted with halved and put to the reciprical when equation is reversed...

yeah rosie, you're getting confused with calculations with K and calculations with ∆H.

What were the answers given, anyway? What i had above should be right.

[onlyfknhuman]

and for some reason im guessin ur answer is something to do with not Ea but rather with the equlibrium constant....? as i remember that is square with "K" the reaction equation is doubled. square rooted with halved and put to the reciprical when equation is reversed...

yeah rosie, you're getting confused with calculations with K and calculations with ∆H.

What were the answers given, anyway? What i had above should be right.

i 2nd your answer

[deledio]

From the TSFX Industrial Chemistry Test:
(Reaction is exothermic)
Would you expect the bonds in the ractants to be stronger or weaker than the bonds within the products?

Their answer: The bonds in the products are stronger as it takes more energy to convert the
products to the activated complex than it does to convert the reactants to the
activated complex.

Is that right?

[Mao]

sounds right

[bec]

From the TSFX Industrial Chemistry Test:
(Reaction is exothermic)
Would you expect the bonds in the ractants to be stronger or weaker than the bonds within the products?

Their answer: The bonds in the products are stronger as it takes more energy to convert the
products to the activated complex than it does to convert the reactants to the
activated complex.

Is that right?

If you draw the energy change graph, you can visualise this pretty well. The activation energy is greater in the reverse reaction - which means that the energy required to break the bonds in the products is GREATER than the energy required to break the bonds in the reactants.

(Don't get "strength of bonds" confused with energy content - in exothermic reactions, the products have less energy but stronger bonds.)

And....if I'm completely wrong, someone correct me please...

[dusty_girl1144]

From the TSFX Industrial Chemistry Test:
(Reaction is exothermic)
Would you expect the bonds in the ractants to be stronger or weaker than the bonds within the products?

Their answer: The bonds in the products are stronger as it takes more energy to convert the
products to the activated complex than it does to convert the reactants to the
activated complex.

Is that right?

i woulda thought the bonds in the reactants are stronger. if u look at an exothermic graph... exo releases energy yeah? .... so i woulda gathered that it needed greater energy to break the bonds of the reactants then products....



im soooo lost now....

[Glockmeister]

but there talking about the energy within the bonds. The activation energy in a exothermic reaction is less than the activation energy required for an endothermic relation (remembering that the backward reaction of an exothermic reaction is an endothermic one). So therefore it takes more energy to be able to break such bonds in a endothermic reaction, thus the bonds in an endothermic reaction are stronger than one in an exothermic reaction.

[Rosie]

Ok, more questions:

1. The Earth's ocean's contain significant amounts of dissolved carbon dioxide. The dissolving process can be described by the following equilibria:

CO_2(g) <----> CO_2(aq)
CO_2(aq) + H₂O_(l) <-----> H⁺_(aq) + HCO₃^-_(aq)

What is the likely effect of increasing the concentration of atmospheric CO₂ on the pH of seawater at the ocean surface and the final concentration of dissolved CO₂?

The answer is: pH decreases and the final dissolved [CO_2(aq)] will be lower than the previous equilibrium concentration.
Can someone explain to me why this is so.


2. pH calculations are most accurate for strong monoprotic acids with concentrations greater than or equal to 10^-7 M at 25 degrees. Can someone explain this to me as well.

thanks

[Pandemonium]

increasing [CO2(g)] would cause the forward reaction of CO2(g) <----> CO2(aq) to be favoured because you're adding more reactant.

therefore, [CO2(g)] decreases and [CO2(aq)] increases.

and because we're increasing [CO2(aq)], then the forward reaction of CO2(aq) + H2O(l) <-----> H+(aq) + HCO3-(aq) would be favoured.

therefore there is a net gain of [H+] and [HCO3-] with loss of some [CO2(aq)].

if [H+] increases, pH decreases.

therefore the answer is as stated.

[Pandemonium]

pH calculations are most accurate for strong monoprotic acids with concentrations greater than or equal to 10-7 M at 25 degrees. Can someone explain this to me as well.

at 25 degrees Celsius, we know that Kw is pretty much spot on 10^-7 M.
however, when we shift the temperature up, Kw increases slightly and also, if we cool down the water/solution, Kw decreases slightly. this causes the pH to decrease (when heated) and increase (when cooled).
but we don't actually know the extent that it decreases/increases whereas with 25 degrees C we're pretty much certain it's 10^-7 M

[Mao]

Pandemonium: these are correct, though I'd like to point out a minor thing,

all of these changes are due to LCP, which only partially oppose the change.

hence, the initial increase in [CO2(g)] would cause a subsequent decrease in [CO2(g)], but overall, it is still greater than originally.

the subsequent increase in [CO2(aq)] would cause another subsequent decrease in [CO2(aq)] as per equation 2, but overall, it is still greater than the original concentration

hence, by adding [CO2(g)], the concentration of [CO2(g)], [CO2(aq)], [H+(aq)] and [HCO3-(aq)] have all increased.

the answer supplied is incorrect, reason as followed:

in stating that "H decreases and the final dissolved [CO2(aq)] will be lower than the previous equilibrium concentration", we imply that

hence,

so an increase in CO2, which doesn't noticeably increase the pressure [or we'll all be epically fucked], and has made K smaller, which is absurd.

[Mao]

pH calculations are most accurate for strong monoprotic acids with concentrations greater than or equal to 10-7 M at 25 degrees. Can someone explain this to me as well.

at 25 degrees Celsius, we know that Kw is pretty much spot on 10^-7 M.
however, when we shift the temperature up, Kw increases slightly and also, if we cool down the water/solution, Kw decreases slightly. this causes the pH to decrease (when heated) and increase (when cooled).
but we don't actually know the extent that it decreases/increases whereas with 25 degrees C we're pretty much certain it's 10^-7 M

firstly, at 25 degrees. in pure water at 25 degrees.
self ionisation of water is endothermic, hence as T increase, the concentration of hydronium and hydroxide increase hence Kw also increase, causing the pH to drop even though it is still neutral.

this explains the temperature part of the original question

Why strong acid: weak acid only partially dissociate/ionise, and that assumption is made when we do the calculations, that the equilibrium concentration of the weak acid is equal to the original concentration. Though it is still accurate for acids weaker than vinegar, it still has a degree of uncertainty in it. For strong acids, however, such as HCl, it pretty much ionises completely in water (a very small degree of uncertainty), and is more "accurate"

why monoprotic: the second/third ionisation of polyprotic acids often has very low Ka [weak acid], and ionisation only occur to a small extent. This impacts the calculations [and make it indeed very hard]. Hence calculations involving monoprotic acids are usually considered more "accurate" than calculations involving polyprotic acids, since usually the second/third ionisation are ignored.

Why 10^-7M: at 25 degrees, the concentration of hydronium ion in water is already 10^-7M. Adding, say 10^-9M of acid will bring it to 1.01 x 10^-7M, a very small change that will also undergo subsequent change as per the self-ionisation of water equilibrium [though the change would be more insignificant than this], and pH will be very close to 7].