Heat of Reaction: KOH + HCl --> KCL + H2O, NaOH + H2SO4 --> Na2SO4 + H2O

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The discussion focuses on calculating the heat of reaction for the neutralization of KOH with HCl and NaOH with H2SO4. The heat of reaction for KOH and HCl is given as -54 kJ/mol, indicating an exothermic reaction. Participants emphasize the importance of writing the net ionic equations to understand the reactions better, which simplifies the calculation of heat involved. The confusion arises around the interpretation of the heat produced, with clarification that -54 kJ represents heat evolved during the formation of water in the reaction. Ultimately, the correct heat of reaction for the second equation is debated, with participants suggesting that the provided answer of -27 kJ is incorrect.
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Homework Statement

Given that for the reaction KOH(aq) + HCL(aq) ----> KCL(aq) + H2O(l) H = - 54kjmol^-1.
What is the quantity of heat involved in the reaction? 2NaOH(aq) + H2SO4(aq) ----> Na2SO4(aq) + 2H2O(l)?



Homework Equations


Heat of reaction Hrn =heat content of product Hp - heat content of reactant Hr. It can be written as Hrn= Hp-Hr

The Attempt at a Solution


Since Hrn=Hp-Hr. I don't know what to do again in other to get the value of Hp and Hr so that I can take the difference and Hrn will appear from no where.
 
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You don't need "heat content" of reactants/products.

First things first - write net ionic reaction.
 
This is a neutralization reaction between a strong acid and a strong base. Any neutralization reaction between a strong base and a strong acid yields the same heat of neutralization, -54 KJ/mol.

To better understand this, as Borek said, try to write the net ionic equation for both the reaction. You will see the magic then.
 
Borek said:
You don't need "heat content" of reactants/products.

First things first - write net ionic reaction.
For the reaction KOH(aq) + HCL(aq) ----> KCL(aq) + H2O(l)
Net ionic equation
H++OH---->H2O
For the reaction:
2NaOH(aq) + H2SO4(aq) ----> Na2SO4(aq) + 2H2O(l)
Net ionic equation
2OH-+2H+
---->2H2O
So what is the heat of reaction?
 
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If burning 1 candle produces x kJ, how many kJ will be produced when you burn 2 candles?
 
Borek said:
If burning 1 candle produces x kJ, how many kJ will be produced when you burn 2 candles?

If 1 mole of water produces -54kJ, then 2 moles of water will produce -108kJ. Could that be what you are tying to ask?
 
That's not what I am trying to ask, that's what you are trying to understand :-p

It is not "moles of water produce 108kJ", it is "108kJ evolved while producing 1 mole of water in the neutralization reaction". But otherwise you are right. Note the sign!
 
Borek said:
That's not what I am trying to ask, that's what you are trying to understand :-p

It is not "moles of water produce 108kJ", it is "108kJ evolved while producing 1 mole of water in the neutralization reaction". But otherwise you are right. Note the sign!

What does the sign have to the with the computation?
 
Earlier you wrote:

chikis said:
1 mole of water produces -54kJ

That's incorrect. If anything, when 1 mole of water is made in neutralization reaction, 54 kJ are produced. That means standard enthalpy of reaction is -54 kJ - because by convention enthalpy change for exothermic reaction is negative (lost by the system, not gained by the surroundings).

When you write "-54 kJ is produced" it suggests system gained 54 kJ, and the reaction was endothermic.
 
  • #10
Why do I have to write the net ionic equation before solving?
 
  • #11
You have to understand first what is actually going on before trying to intercept the situation. You don't eat food without seeing what is served to you, do you?
 
  • #12
AGNuke said:
You have to understand first what is actually going on before trying to intercept the situation.
You don't eat food without seeing what is served to you, do you?

No I don't, unless am blind.
 
  • #13
Same here. Try to actually see what reaction is going on. This case, just simple neutralization, formation of water by strong acid and strong base.

Always try to write down the reaction when solving the problem. This makes the problem visually easy and enables you to solve it rather easily, now that you are able to see the reaction.
 
  • #14
Borek said:
That's not what I am trying to ask, that's what you are trying to understand :-p

It is not "moles of water produce 108kJ", it is "108kJ evolved while producing 1 mole of water in the neutralization reaction". But otherwise you are right.
Note the sign!

I Now understand what you meant properly now. Because the sign is negative, the reaction is exothermic, therefore heat in kJ is evolved when 1 mole of water is produced. Isn't it?
 
  • #15
Here are the possible answers which the authors of the question set for the problem A.-108kj B. -54kj C. -27kj D. +27kj E. +54kj.
They went and choose C. = -27kj instead of A. = -108kj. I don't blame them as such, maybe that mistake is due to typo error. Don't you think the same?
 
  • #16
27 kJ (regardless of the sign) doesn't make any sense.
 
  • #17
Borek said:
27 kJ (regardless of the sign) doesn't make any sense.

I don't understand what you mean.
 
  • #18
Neither 27 kJ nor -27 kJ is a correct answer.
 
  • #19
Borek said:
Neither 27 kJ nor -27 kJ is a correct answer.

Thanks for the asistance. I do appreciate!
 
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