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Topic: Enthalpy of Neutralization  (Read 1626 times)

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Offline Diamonds

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Enthalpy of Neutralization
« on: January 30, 2013, 01:33:12 AM »
"Which combination of solutions of HCl and NaOH would produce the largest ΔT?"

a. 50 mL of 1 M HCl with 50 mL of 1 M NaOH
b. 50 mL of 2 M HCl with 50 mL of 2 M NaOH
c. 100 mL of 1 M HCl with 50 mL of 2 M NaOH
d. 100 mL of 1 M HCl with 10 mL of 1 M NaOH

So I know that this is an enthalpy of neutralization problem. Because it is given in kJ/mol, I know that the higher T will be the one with the more moles involved.
a. 50 mmol and 50 mmol
b. 100 mmol and 100 mmol
c. 100 mmol and 100 mmol
d. 100 mmol and 100 mmol.

So now a. is out.

It seems logical to me that the one that is made of higher concentrations will produce a larger ΔT, but if you were to ask me that I would say "just because."
What is the relevant equation that I am missing which includes concentration, and not just number of moles, that I can use to mathematically prove to myself that the one made of stronger concentrations (b) is the one that will provide the largest ΔT?

Thanks so much in advance!

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Re: Enthalpy of Neutralization
« Reply #1 on: January 30, 2013, 04:12:29 AM »
Amount of heat evolved is directly proportional to the number of moles reacting.

Number of moles is CV, so you can combine two equations and calculate amount of heat as a function of concentration (but you won't be able to get rid of the volume).

See if you can combine it with q=mcΔT (you can assume density and heat capacity to be these of pure water to make calculations easier). Correctly done it will give you ΔT as a function of total volume and number of moles reacting.
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