Quick Chemistry Question.

Hi,

Can someone please tell me if I am doing this Hess’s Law question correctly?
i.e.
check diagram (arrows etc) and my calculations...

---

Calculate the enthalpy change in the reaction:

2KHCO3 (s) --------> K2CO3 (s) + CO2 (g) + H2O (l)

See Diagram

ΔHf data:

KHCO3 (s): -959 kJmol-1
K2CO3 (s): -1146 kJmol-1
CO2 (g): -394 kJmol-1
H2O (l): -286 kJmol-1

ΔHr + 2ΔH1 = ΔH2 + ΔH3 + ΔH4

ΔHr = ΔH2 + ΔH3 + ΔH4 - 2ΔH1

= [-1146 - 394 - 286] - 2(-959)

= -1826 + 1918

= 92kJmol-1

I'm not sure whether the answer is correct or not... :confused:

most likely incorrect I'm guessing.

Thanks!

Willc

Hi,

Can someone please tell me if I am doing this Hess’s Law question correctly?
i.e.
check diagram (arrows etc) and my calculations...

---

Calculate the enthalpy change in the reaction:

2KHCO3 (s) --------> K2CO3 (s) + CO2 (g) + H2O (l)

See Diagram

ΔHf data:

KHCO3 (s): -959 kJmol-1
K2CO3 (s): -1146 kJmol-1
CO2 (g): -394 kJmol-1
H2O (l): -286 kJmol-1

ΔHr + 2ΔH1 = ΔH2 + ΔH3 + ΔH4

ΔHr = ΔH2 + ΔH3 + ΔH4 - 2ΔH1

= [-1146 - 394 - 286] - 2(-959)

= -1826 + 1918

= 92kJmol-1

I'm not sure whether the answer is correct or not... :confused:

most likely incorrect I'm guessing.

Thanks!

Hess' law by manipulation of the equations...

2KHCO3 --> K2CO3 + CO2 + H2O

2k + H2 + 2C + 3O2 --> 2KHCO3 ΔH = 2(-959) = -1918
2k + C + 3/2O2 --> K2CO3 ΔH = -1146

SUBTRACT

H2 + C + 3/2O2 --> 2KHCO3 - K2CO3 ΔH = -772
H2 + 1/2O2 ---> H2O ΔH = -286

SUBTRACT

C + O2 --> 2KHCO3 - K2CO3 - H20 ΔH = -486
C + O2 --> CO2 ΔH = -394

SUBTRACT

ZERO ---> 2KHCO3 - K2CO3 - H20 -CO2 ΔH = -92

REARRANGE

K2CO3 + H20 + CO2 --> 2KHCO3 ΔH = -92

THEREFORE

2KHCO3 --> K2CO3 + H20 + CO2 ΔH = +92

In agreement with your answer :biggrin:

Reply 2

Willc

OP

Erm, I'm not sure I understand what you've done there...
I'll try and write it out first to see if I understand.

Would it be easier to use the equation:

ΔH = ΣΔHf (Products) - ΣΔHf (Reactants)

Do I assume that the left-hand side is the reactants? Because this method also gives me 92 kJmol^-1.

Also, is this just known simply as 'enthalpy change', since it is just decomposition rather than combustion or formation?

Is my diagram correct by the way?

Argh I hate this and Redox reactions :frown:

.

Thank you very much, and sorry for all the questions, I'm dumb lol.

As Hess' law is just another way of stating the law of conservation of energy then you can manipulate energetics equations at will and they are always valid (provided you use the four rules of number) - that's all I did.

Bit pushed 4 tme at the mo get back to you about the diagram if nobody else does :smile:

and yes, you're correct - your way is easier - I was just checking it by going another way!

As all the elements in standard states are assigned anenergy value of zero then to break apart the compound on the left hand side is the opposite of the enthalpy of formation (must be broken apart) and then formation of the products molecules is the actual value of the enthalpies of formation.

This makes the enthalpy change

ΔH = ΣΔHf (Products) - ΣΔHf (Reactants)

note that we use the actual values of the ΣΔHf (Products) and the negative of the ΣΔHf (Reactants)

does that make it clearer?

Reply 6

Willc

OP

charco

As all the elements in standard states are assigned anenergy value of zero then to break apart the compound on the left hand side is the opposite of the enthalpy of formation (must be broken apart) and then formation of the products molecules is the actual value of the enthalpies of formation.

Yup, I understand this part now.

charco

This makes the enthalpy change

ΔH = ΣΔHf (Products) - ΣΔHf (Reactants)

note that we use the actual values of the ΣΔHf (Products) and the negative of the ΣΔHf (Reactants)

does that make it clearer?

What do you mean by actual values of the "ΣΔHf (Products)"? Is that basically adding up the enthalpy of formation values of the products.
i.e.
(-1146) + (-394) + (-286) = -1826

and so, the negative of the "ΣΔHf (Reactants)" is simply

- 2(-959) = +1918

then put them together so that:

ΔH = -1826 + 1918
ΔH = 92

I think that's going to be as clear as it can be.

So, this 'Enthalpy Change' value that I have calculated, does this show that 92kJmol^-1 is needed to decompose the potassium hydrogen carbonate...because it is endothermic? (+ve value).

God help me in the exam. :mad:

Sounds like you've got it sorted

you can think of any reactant as having to be broken down into its elements which then must be reformed as the products...

Hence sum of the ΔHf of the left hand side (reactants) but with the opposite sign (because they're being broken up)

Then they are reformed into the products --> hence sum of the ΔHf of the products but with the correct sign this time

Simply add up the two sets of enthalpy sums to give you the reaction enthalpy

- ΣΔHf (Reactants) + ΣΔHf (Products) = Reaction enthalpy

of course the books usually just state the rearranged form

Reaction enthalpy = ΣΔHf (Products) - ΣΔHf (Reactants)

Which does not particularly help in the understanding of WHY this is the case...

Don't worry about the exam just remember that you have to break the reactants down to the elements in the first stage (sum the enthalpies of formation and change sign)and then build them back up to the products in the second stage (sum the enthalpies of formation)

Then add it all up!

Your final statement is quite correct - it's endothermic therefore energy is needed to decompose the KHCO3

Bond breaking = endothermic as you would expect

I made a mistake when I did the original calculation and got an exothermic value. I knew that it could not be the answer as simple inspection tells me that it is a decomposition and is overwhelmingly likely to be endothermic.

Always useful to consider the likelihood of ANY answer.

Reply 8

Willc

OP

Yea, I think I'v finally worked it out.
I'v just done some other questions from some past papers and they're correct too, so something's clicked. :wink:

Thank you charco, you've been most helpful. :smile:

Go for it

Quick Chemistry Question.

Related discussions

Articles for you