Solving Enthelpy Problem: ΔH = 12.9 kJ

[jools111]

Homework Statement

I am working on a lab that has me a little confused:

I diluted 5.5g of NaOH into 200ml of water to disassociate the sodium ions. In the equation below you can see my temperature results.

Write the net ionic equation for the reaction, and note the value of ΔH.

NaOH ---> Na + OH

Homework Equations

ΔH = m•ΔT•Q
ΔH = (200 g + 5.5 g)(final temp - initial temp)(4.18 J/g°C)
ΔH = (205.5 g)(41 °C - 26 °C)(4.18 J/g°C)
ΔH = 12.9 kJ

The Attempt at a Solution

Because it's an exothermic reaction (It created heat...), I know ΔH should be negative. Why is it coming up positive? I'm lost here... I'm pretty sure I've done everything correctly. Thanks!

 

[Borek]

Don't worry. You have correctly calculated amount of heat that was absorbed by the solution - it was gained by the solution, so it was 'lost' by the NaOH. Just change the sign.

Beware, correct reaction equation is

NaOH -> Na⁺ + OH^-

you can't ignore ions/charges.

 

[jools111]

Don't worry. You have correctly calculated amount of heat that was absorbed by the solution - it was gained by the solution, so it was 'lost' by the NaOH. Just change the sign.

Beware, correct reaction equation is

NaOH -> Na⁺ + OH^-

you can't ignore ions/charges.

--
buffer calculator, concentration calculator
pH calculator, stoichiometry calculator

So would the correct answer to the question be negative? I'm a little confused. I understand the logic, but can I just arbitrarily change the symbol like that?