Experiment: Heat of Reaction help

[legking]

Okay, to begin with, I'll set the lab up for you. Apparently this is a pretty common lab, so I imagine most of you are familiar with it - I think I've heard it called a coffee-cup calorimeter lab before. I am to perform three reactions:

Reaction 1: Dissolve 1 tbsp (5.5) solid sodium hydroxide in 200 mL water; record temperature change.

Reaction 2: Mix 100 mL aqueous 1.0 mol/L NaOH with 100 mL aqueous 1.0 mol/L hydrochloric acid; record temperature change.

Reaction 3: Dissolve 1 tbsp (5.5 g) solid sodium hydroxide in 200 mL aqueous 1.0 mol/L hydrochloric acid; record temperature change.

Then I am to determine ΔH for each reaction, being the heat lost or gained by each reaction.

There are a series of questions I am to answer - some I've already answered but would like to verify, others I'm really lost on. I'll post them one or two at a time...

QUESTION 1: Write the net ionic equation for each reaction, and note the value of ΔH for each reaction.

MY ANSWER:

Now, I don't have any troubles crunching the numbers to determine ΔH, which is m•ΔT•Q. What I would like to know, though, is how I'm supposed to come up with a negative result. I mean, I know ΔH for each reaction is supposed to be negative because all the reactions are exothermic, but m, ΔT and Q are all positive numbers.

For example, ΔH for reaction 1 would be
ΔH = m•ΔT•Q
ΔH = (200 g + 5.5 g)(final temperature - initial temperature)(0.00418 kj/cal)
ΔH = (205.5 g)(41 °C - 26 °C)(0.00418 kj/cal)
ΔH = 13 kJ

I could achieve a negative result by making ΔT initial - final, but I always thought Δ values were final - initial. It may seem trivial, but it's bugging me.

Anyway, as for my net ionic equations:

For reaction 1, ionic equation: NaOH(s) ----- Na+(aq) + OH-(aq)
For reaction 2, ionic equation: OH-(aq) + H+(aq) ----- H2O(l)
For reaction 3, ionic equation: NaOH(s) + H+(aq) ----- H2O(l) + Na+(aq)

Do these look alright? I only ask because of Question 4:

QUESTION 4: Add the net ionic equation for Reaction 1 to the net ionic equation for Reaction 2. How does the result compare with the net ionic equation for Reaction 3?

MY ANSWER

Adding reactions 1 and 2:
NaOH(s) + OH-(aq) + H+(aq)----- Na+(aq) + OH-(aq) + H2O(l)

Compared to reaction 3:
NaOH(s) + H+(aq) ----- H2O(l) + Na+(aq)

The result is identical with the exception of an additional hydroxide ion found on either side of the reaction from the combination of the first two reactions. However, since reaction 3 is basically reactions 1 and 2 combined, shouldn't the two be identical? Did I get my net ionic equations wrong?

Please help! I will reward you all with a link to the funniest video you'll see this week!

 

[wmhcan]

yes, it is right, you just have to canceled out the OH-(aq) on both side of the equation, and for the negative value, the value you got is for the final solution, which means if you reverse your equation, it is positive, if you want to write it like "NaOH(s) ----- Na+(aq) + OH-(aq)", then it is negative.