answer this question (1 Viewer)

[kooltrainer]

the equation showing synthesis of ammonia is
N2(g) + 3H2(g) <=> 2NH3 (g) delta H =-92kJ/mol
in one production cycle, 28x10^-3 Kg of N2 react with 6x10^-3kg of H2.
6.8x10^-3 Kg of NH3 (g) are produced. how much heat is released?
show working out please..

 

[namburger]

184kj. just a guess =]

 

[kooltrainer]

could u show sum working out plz

 

[xiao1985]

6.8x10^-3 Kg of NH3 produced =

6.8 e6 g NH3 (MM = 17g/mol) = 4 e5 moles
molar heat = 92kJ/mole
heat produced = 3.68 e7 kJ = 3.68 e10 J

 

[Undermyskin]

What is e? Hey hey, don't try to confuse us! lolz.

I think this one should be simply worked out by finding the number of moles of NH3. And all the ^-3 kg is converted to g.

nNH3= 6.8/17= 0.3994 mol

From the equation, for every 2 moles of NH3 formed, 92 kJ is released
---> H= 0.3994*46= 18.37 kJ

 

[brenton1987]

What is e? Hey hey, don't try to confuse us! lolz.

e is 10^.

eg 2e3 = 2*10³ = 2000

Xiao: all of those e numbers should be negative or that is one hell of an amount of ammonia.

 

[Pwnage101]

From the equation, for every 2 moles of NH3 formed, 92 kJ is released

I'm still confused about this - many textbooks say delta h = -92kj/mol, however others do: N2 + 3H2 <---> 2NH3 +92kJ

so, yeh, could someone clarify this - is 46kJ released per mole of ammonia produced, or is it 92kJ?

thx

 

[brenton1987]

I'm still confused about this - many textbooks say delta h = -92kj/mol, however others do: N2 + 3H2 <---> 2NH3 +92kJ
so, yeh, could someone clarify this - is 46kJ released per mole of ammonia produced, or is it 92kJ?

d_fH kJ mol^-1 = -46

From SI Chemical Data, 5th Edition.

 

[Undermyskin]

The text books say that the Delta H you get is from each mol of the product formed based on the equation. i.e. if the equation above gives 1/2 N2, 3/2 H2 and 1NH3 with -92kJ/mol then for 1 mol of NH3 formed, 92kJ is released. However, as it says 2NH3 with -92kJ/mol, then for each mol of NH3 formed, 46kJ of heat is produced.

 

[Pwnage101]

However, as it says 2NH3 with -92kJ/mol, then for each mol of NH3 formed, 46kJ of heat is produced.

are u not contradicting yourself? When u say "-92kJ/mol" that means 92 kJ of energy released PER MOL, yet u r saying for each mol of NH3, 46kJ is released

i understand how u get that - 92/2 = 46, since 2 moles are produced, but if it states "delta h = -92kJ/mol" , that is saying for every mole of NH3 produced, 92kJ is released - u cant say "something is (2x)kJ/mol, but since there was 2 moles produced (x)kJ is released per mol"

kJ/mol means the amount released per mol, so even if an equation forms 2 moles or 5 moles or how ever many moles, it shouldnt change delta h - if wat u r sayin is true, then they should write "delta h = -46kJ/mol, and since 2 moels are produced in this equation, 92kJ is produced", but they don't

it's confusing
thx for tryin to explain, but yeh, still doesnt make sense

 

[appletooth]

are u not contradicting yourself? When u say "-92kJ/mol" that means 92 kJ of energy released PER MOL, yet u r saying for each mol of NH3, 46kJ is released

i understand how u get that - 92/2 = 46, since 2 moles are produced, but if it states "delta h = -92kJ/mol" , that is saying for every mole of NH3 produced, 92kJ is released - u cant say "something is (2x)kJ/mol, but since there was 2 moles produced (x)kJ is released per mol"

kJ/mol means the amount released per mol, so even if an equation forms 2 moles or 5 moles or how ever many moles, it shouldnt change delta h - if wat u r sayin is true, then they should write "delta h = -46kJ/mol, and since 2 moels are produced in this equation, 92kJ is produced", but they don't

it's confusing
thx for tryin to explain, but yeh, still doesnt make sense

I think Undermyskin is right, actually...

For the reaction N2(g) + 3H2(g) <=> 2NH3(g) + 92kJ, don't you just read the equation as it is? 1 mole of N2(g) reacts with 3 moles of H2(g) to produce 2 moles of NH3(g) and 92 kJ of energy for that reaction.

Therefore for every 2 moles of NH3(g) produced, you're also producing 92 kJ of energy.

Halve that, and for every 1 mole of NH3(g) produced, you're producing 46 kJ of energy.

Writing N2(g) + 3H2(g) <=> 2NH3(g) deltaH= –92 kJ/mol pretty much means the same thing. The value of delta H is applied to the entire equation, not just the ammonia gas.

 

[Pwnage101]

I think Undermyskin is right, actually...

For the reaction N2(g) + 3H2(g) <=> 2NH3(g) + 92kJ, don't you just read the equation as it is? 1 mole of N2(g) reacts with 3 moles of H2(g) to produce 2 moles of NH3(g) and 92 kJ of energy for that reaction.

Therefore for every 2 moles of NH3(g) produced, you're also producing 92 kJ of energy.

Halve that, and for every 1 mole of NH3(g) produced, you're producing 46 kJ of energy.

Writing N2(g) + 3H2(g) <=> 2NH3(g) deltaH= –92 kJ/mol pretty much means the same thing. The value of delta H is applied to the entire equation, not just the ammonia gas.

Yes, i understand what u r sayin in the first half of ur post (hence y i was confused in the 1st place), but the second half doesnt make sense - it would say "-92kJ/applied over entire equation" if it is what u r sayin - it does not say that, instead it says "-92kJ/mol"

this value (-92kJ/mol) as it is stated means for every mole of NH3 produced, 92kJ of energy is applied, or am i mistaken?

we cant change wat "kJ/mol" means - it means exactly that, no matter how many moles ar produced, the value for kj/mol will be the same!!!
---
take for eg:

x + y ----> xy (delta h = +50kJ/mol)

this can also be written as :

x + y + 50kJ ------> xy

thus 50kJ are needed for the profuction of 1 mole of product

however, say i have the equation not in simplest form:

3x + 3y --------> 3xy

delta h is still +50kJ/mol, however since we have 3 times the products, 150kJ will be needed for teh reaction to proceed as shown in equation 2 and 3 moles of product be produced

---

if delta h was just -92kJ (with no '/mol') then i would agree with u 100% but it doesn not, most textbooks say "delta h = -92kJ/mol"

If i take to be true what u say: "The value of delta H is applied to the entire equation" , then i dont understand y it is 'kJ/mol', unless we take the whole equation as the base value for 1 mole....

 

[appletooth]

Okay this discussion is starting to deviate from the original question asked, but I found something in my textbook that might clarify what I've said:

Taken from 'Pathways to Chemistry', p456 (the 1996 version)

There are different conventions for writing the enthalpy change for reactions. The following two examples illustrate this.

H2(g) + I2(g) --> 2HI(g) deltaH= –10 kJ/mol

deltaH is reported in the units of kJ/mol. In this case 10 kJ of heat is evolved per mole of H2(g) or per mole of I2(g). As 2 mol of HI(g) form then 5 kJ of heat is evolved per mole of HI(g).

2SO2(g) + O2(g) --> 2SO3(g) deltaH= –198 kJ

In this case deltaH is reported for the equation as written. Thus, 198 kJ of heat is evolved per mole of O2 reacting or 198 kJ of heat is evolved for each 2 mol of SO3(g) formed (99 kJ evolved per mole of SO3(g) formed).

Therefore, if the equation we have is:

N2(g) + 3H2(g) <=> 2NH3(g) deltaH= –92 kJ/mol

Then:

92 kJ of heat is evolved per mole of N2 reacting, or 92 kJ of heat is evolved for each 2 mol of NH3 formed (46 kJ evolved per mole of NH3(g) formed).

Hopefully that makes more sense now. I don't know why it appears to be so contradictory with the "deltaH = x kJ/mol" part. I agree with you that if you think it through the logic is kind of flawed, but I'm sure there's a reason why... I just don't know it.

 

[Pwnage101]

Okay this discussion is starting to deviate from the original question asked, but I found something in my textbook that might clarify what I've said:

Taken from 'Pathways to Chemistry', p456 (the 1996 version)



Therefore, if the equation we have is:

N2(g) + 3H2(g) <=> 2NH3(g) deltaH= –92 kJ/mol

Then:

92 kJ of heat is evolved per mole of N2 reacting, or 92 kJ of heat is evolved for each 2 mol of NH3 formed (46 kJ evolved per mole of NH3(g) formed).

Hopefully that makes more sense now. I don't know why it appears to be so contradictory with the "deltaH = x kJ/mol" part. I agree with you that if you think it through the logic is kind of flawed, but I'm sure there's a reason why... I just don't know it.

cheers
that makes sense now - taking the 'mol' part of 'kj/mol' to mean amount of moles of a REACTANT - all clarified, 46kJ of energy released per mole of NH3 created

i was a bit unsure, hence y i asked "this value (-92kJ/mol) as it is stated means for every mole of NH3 produced, 92kJ of energy is applied, or am i mistaken? ", and u answered that - i was mistaken, it was for every mole of N2 reacting

anyways
thx

 

[kooltrainer]

so all the N2 and H2 numbers are there to trick you?

 

[Undermyskin]

What the...?

Yes, you need to stick to the number of moles of products.

Er... what did I say wrong? Of course 46kJ/mol is what I mean. So???