Neutralisation reaction - enthalpy change (1 Viewer)

[pwoh]

Does anyone know how to approach this question?

When 50 mL of 1.0 mol/L potassium hydroxide solution is added to 50mL of a 1.0 mol/L hydrochloric acid solution, 2.8 kJ of heat energy is given out. Calculate the enthalpy change for the reaction:

H+(aq) + OH-(aq) --> H2O(l)

 

[Deer]

I don't understand, why would they give you a different equation when they offer information on KOH and HCl?

 

[pwoh]

I don't understand, why would they give you a different equation when they offer information on KOH and HCl?

That's what I was wondering

Perhaps only the neutralisation reaction is relevant, the K and Cl are irrelevant?

But I have no idea how to approach the question. (It's from the Conquering Chemistry blackline masters)

 

[unholybuddha]

donno if im right but a neutralisation reaction is a proton transfer. so when hydrochloric acid reacts with potassium hydroxide, so a hydrogen ion from the hydrochloric acid is transferred to the hydroxide ion of the potassium hydroxide. this is what causes the change in enthalpy.

overall equation:
KOH(aq) + HCl(aq) -> KCl(aq) + H2O(l)
net ionic equation:
H+(aq) + OH-(aq) --> H2O(l)

using c=n/v
0.05mol of KOH and HCl are used, meaning 0.05mol of H+ and 0.05mol of OH- is used to produced 0.05mol of H2O. For the reaction 2.8kJ is released per 0.05mol of reactants as stated in the quetion.

therefore for enthalpy change, using some simple algebra ΔH = -(2.8kJ)/(0.05mol) = -56kJ/mol.
note: the enthalpy is negative because heat is released

i hope this is right, not 100% sure.

 

[pwoh]

That makes a lot of sense, thanks!

To clarify, is enthalpy change always measured per mol? I thought it was just heat change at a constant pressure.

 

[unholybuddha]

most of the time enthalpy change is measured in kJ/mol. dont take my word for it but