[Shanth]

Firstly, is the enthalpy change of neutralization for 2KHCO3 + 2HCl ----> 2KCl + 2H2O + 2CO2  is the same as for KHCO3 + HCl ----> KCl + H2O + CO2?.


....and how would u calculate deltaHr for the reaction :2KHCO3(s) ------> K2CO3(s) + CO2(g) + H2O(l)  when your given deltaHn for KHCO3 + HCl ----> KCl + H2O + CO2 and K2CO3 + 2HCl ---------> 2KCl + H2O + CO2

[curiouscat]

Firstly, is the enthalpy change of neutralization for 2KHCO3 + 2HCl ----> 2KCl + 2H2O + 2CO2  is the same as for KHCO3 + HCl ----> KCl + H2O + CO2?.

No. It isn't same.

[magician4]

ref. the second part of your question, hint:

if given

(I) A + B  C + D ; ΔHr(reaction I)

it directly follows that

(II) C + D  A + B ; ΔHr(reaction II) = - ΔHr(reaction I)

so, you can "revert" an equation including the ΔHr belonging to easily, which at times is quite usefull in Hess type calculations


regards

Ingo

 

[Shanth]

So, wil it be 2xdelta Hn of k2co3 and 1xdelta Hn? Cuz that's what I did today in my board practicals and my friends say it's the other way around. And do u have an idea of how many marks I would lose for that?

[magician4]

so, as everything already is ex post facto , here's what you should have done:

first, you take

(I) KHCO₃ + HCl ----> KCl + H₂O + CO₂ ; ΔH_f(I)

and multiply it by two:

(a) 2 KHCO₃ + 2 HCl  2 KCl + 2 H₂O + 2 CO₂ ; 2 * ΔH_f(I)
(so you already have "2 KHCO₃" leftside, exactly like you'd need it for your target equation)

then you take the next equation, and revert it:
(II) K₂CO₃ + 2 HCl  2 KCl + H₂O + CO₂ ; ΔH_f(II)
 
(b) 2 KCl + H₂O + CO₂  K₂CO₃ + 2 HCl ; - ΔH_f(II)


then, you add up (a) and (b)

(a) 2 KHCO₃ + 2 HCl  2 KCl + 2 H₂O + 2 CO₂ ; 2 * ΔH_f(I)
+
(b) 2KCl + H₂O + CO₂  K₂CO₃ + 2HCl ; - ΔH_f(II)
=============================
2 KHCO₃ + 2 HCl + 2KCl + H₂O + CO₂    2 KCl + 2 H₂O+ 2 CO₂ + K₂CO₃ + 2HCl ; 2 * ΔH_f(I) - ΔH_f(II)

...and at last "clean up" , i.e. remove everything that's left/right identical ( and therefore marked in red above):

(c) 2 KHCO₃    H₂O+ CO₂ + K₂CO₃  ; 2 * ΔH_f(I) - ΔH_f(II)


there you are

regards

Ingo

[Shanth]

Thank you very much Ingo. Sorry for asking too many questions but since this method is different to the method I was taught (hess's law triangle), could u explain to me when do u have to revert the equations and why do u  have to add them both (is it always adding or is there subtracting too?) this method looks safer than the triangle one where one can get quite confused!

[magician4]

many ways to skin the cat...


and this is true for systematics ref. calculations using Hess' law, also.

Very early on I personally did discover that too much thinking about when to subtract , when to add, is kind of overdoing for me.
so , consequentely, I brought everything down to "just adding".

I'm looking at my starting equations "what is where" , and compare it to "where  should it be by the end of the day" ( in this case: KHCO₃ and K₂CO₃ were my substances of orientation, and the target equation gave me the answers)

then, if necessary, I rearrange accordingly, or multiply to get the desired stoechiometric factors (ALWAYS keeping in mind to maipulate the ΔH_f 's accordingly)

... and then I just add up.
Usually, with just one addition, everything falls in place as desired.
I like my way


regards

Ingo