Difficult Dissolution/Neutralization Question

[PVnRT81]

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?

 

[Saitama]

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.

 

[PVnRT81]

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:

NaOH_s --> Na⁺_aq + OH^-_aq ΔH=-53.4 kJ

NaOH_aq + HCl_aq --> NaCl_s + 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.

 

[Saitama]

Thank you for your reply.

I have tried doing that below:

NaOH_s --> Na⁺_aq + OH^-_aq ΔH=-53.4 kJ

NaOH_aq + HCl_aq --> NaCl_s + 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.

 

[DeeNos]

I would first start by understanding the concept of molar heat of neutralization and how it relates to the reaction between sodium hydroxide and hydrochloric acid. The molar heat of neutralization is the heat released or absorbed when one mole of an acid and one mole of a base react to form one mole of water and a salt.

Next, I would use the data given to calculate the molar heat of neutralization of solid sodium hydroxide by a solution of hydrochloric acid. This can be done by subtracting the molar heat of dissolution of solid NaOH (-53.4 kJ/mol) from the molar heat of neutralization of a solution of NaOH by a solution of HCl (-54 kJ/mol). This will give you the molar heat of neutralization for the reaction between NaOH and HCl.

It is also important to note that the molar heat of neutralization is a constant value for a specific acid-base reaction, so the molar heat of neutralization calculated using the given data can be applied to any other similar reaction between NaOH and HCl.

In order to fully understand and solve this question, I would also recommend reviewing the principles of thermodynamics and enthalpy, as well as any relevant equations and concepts related to neutralization reactions. Additionally, seeking assistance from a teacher or fellow peer may also be beneficial in providing further guidance and clarification.