Difficult Dissolution/Neutralization Question

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Discussion Overview

The discussion revolves around calculating the molar heat of neutralization of solid sodium hydroxide (NaOH) by hydrochloric acid (HCl), utilizing provided data on the molar heat of dissolution and neutralization. Participants explore the application of Hess' Law to combine reactions and determine the enthalpy change.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about how to utilize the given data for the problem.
  • Another participant suggests that the problem may involve adding or subtracting the reactions and recommends writing down the relevant reactions.
  • A participant presents the reactions and applies Hess' Law, calculating the total enthalpy change as -107.4 kJ, but expresses uncertainty about the correctness of this approach.
  • Another participant affirms the calculation, stating that canceling the ions is valid since NaOH(aq) represents the dissociated ions.

Areas of Agreement / Disagreement

There is some agreement on the approach of using Hess' Law, but uncertainty remains regarding the correctness of the calculated enthalpy change and the treatment of the ions in the reactions.

Contextual Notes

Participants have not fully resolved the implications of canceling ions in the context of Hess' Law, and there may be assumptions about the states of the compounds that are not explicitly stated.

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Homework Statement



Find the molar heat of neutralization of solid sodium hydroxide by a solution of hydrochloric acid given the following data:

Molar heat of dissolution of solid NaOH: -53.4 kJ/mol

Molar heat of neutralization of a solution of NaOH by a solution of HCL = -54 kJ/mol

2. The attempt at a solution

I'm really stumped by this question.

I don't really understand how I can work with these two data.

Can anyone hint as to where I can begin?
 
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PVnRT81 said:

Homework Statement



Find the molar heat of neutralization of solid sodium hydroxide by a solution of hydrochloric acid given the following data:

Molar heat of dissolution of solid NaOH: -53.4 kJ/mol

Molar heat of neutralization of a solution of NaOH by a solution of HCL = -54 kJ/mol

2. The attempt at a solution

I'm really stumped by this question.

I don't really understand how I can work with these two data.

Can anyone hint as to where I can begin?

Hi PVnRT81!

I am not sure but I guess this is an "add or subtract the reactions" problem. Start by writing down the two given reactions and then write down the reaction you have to find the enthalpy for.
 
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Pranav-Arora said:
Hi PVnRT81!

I am not sure but I guess this is an "add or subtract the reactions" problem. Start by writing down the two given reactions and then write down the reaction you have to find the enthalpy for.

Thank you for your reply.

I have tried doing that below:

NaOHs --> Na+aq + OH-aq ΔH=-53.4 kJ

NaOHaq + HClaq --> NaCls + H2O l ΔH= -54 kJ

After adding them up using Hess' Law, I get the ΔH to be -107.4 kJ.

I am not sure if this is right since I canceled out the Na and OH ions in the first equation with NaOH in the second equation.
 
PVnRT81 said:
Thank you for your reply.

I have tried doing that below:

NaOHs --> Na+aq + OH-aq ΔH=-53.4 kJ

NaOHaq + HClaq --> NaCls + H2O l ΔH= -54 kJ

After adding them up using Hess' Law, I get the ΔH to be -107.4 kJ.

I am not sure if this is right since I canceled out the Na and OH ions in the first equation with NaOH in the second equation.

That looks right to me. :)

You can cancel the ions and NaOH(aq) as NaOH(aq) is basically those dissociated ions.
 

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