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!