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

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Discussion Overview

The discussion revolves around the heat of reaction for the neutralization reactions involving KOH and HCl, as well as NaOH and H2SO4. Participants explore the calculations related to the heat released during these reactions, including the interpretation of enthalpy changes and the significance of net ionic equations.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states the heat of reaction for KOH and HCl is -54 kJ/mol and asks for the heat involved in the reaction with NaOH and H2SO4.
  • Another participant suggests that the net ionic reaction should be written first to understand the process better.
  • Some participants propose that all neutralization reactions between strong acids and bases yield the same heat of neutralization, specifically -54 kJ/mol.
  • There is a discussion about the interpretation of the heat produced in relation to the number of moles of water formed, with some arguing that it is incorrect to say "moles of water produce" heat without clarifying the context of the reaction.
  • One participant emphasizes the importance of the sign of the enthalpy change, noting that a negative sign indicates an exothermic reaction.
  • There is confusion regarding the provided answer choices for the heat of reaction, with some participants questioning the validity of -27 kJ as a possible answer.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the heat of reaction and the significance of the net ionic equations. There is no consensus on the correct answer to the heat of reaction for the second reaction, and multiple competing views remain regarding the calculations and interpretations presented.

Contextual Notes

Some participants highlight the need for clarity in the definitions and conventions used in thermodynamics, particularly regarding the signs of enthalpy changes and the context of heat production in chemical reactions.

chikis
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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.
 
Last edited:
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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?
 
Last edited:
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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