[MissDee]

Given the following heats of formation, :delta: H_f, in kJ per mole, as obtained from a table of :delta: H_f data, calculate :delta: H for the reaction in Part f. Compare your answer with the result you obtained in Part e. NaOH(s), -425.6; Na⁺(aq), -240.1; OH^- (aq), -230.0

Results from Part e:  :delta: H = -43009.0923 J/mole
Results from Part f: NaOH(s)  Na⁺(aq) + HO^-(aq)

For the answer I got:  :delta: H = -44.5 kJ

Is this correct?

The formula my book gives me is :delta: H = -q_H2O. Does this mean that :delta: H always negative?

Thank you in advance.

[MissDee]

delta H = -qH2O doesn't mean delta H is always negative.  Sometimes the qH2O is negative in it's own right, which would mean you have the "negative of a negative" and thus a positive delta H

Understandable. So my answer would be a positive?

[enahs]

 :delta: H can be either positive or negative.

The  :delta: H = -q_H2O comes from how you perform the experiment.  You dissolve NaOH in water. From the 0th law of thermodynamics, if object a is in contact with object b and object c is in contact with object b, and object a is in thermal equilibrium with b, and b is in thermal equilibrium with c, then a and c are in thermal equilibrium.

So, you dissolve NaOH in water and measure the temperature of the water with a thermometer. You are therefor calculating qH₂O. Assuming a perfect situation, any energy in the form of heat (q) that causes a change in the water must have come from the NaOH (1st law). So, if the water takes in energy in the form of heat (gets hotter), it had to come from somewhere; in this case the NaOH process, meaning the NaOH gave off heat. If the water gives away energy (gets colder), then it had to go somewhere and it goes into the NaOH and it gets hotter in the process.

So, per your setup, the energy of your interested species is opposite of the water.  :delta: H can be negative or positive.

[rabolisk]

q_H2O in your case is positive, and delta H is negative.