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.