(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

When 2.0 g of NaOH were dissolved in 53.0 g water in a calorimeter at 24.0 degrees C, the temperature of the solution went up to 33.7 degrees C.

2. Relevant equations

Enthalpy change is equal to the opposite sign of the heat of water.

3. The attempt at a solution

For enthalpy change of the reaction, I got -2.1 x 10^{3}joules.

To find enthalpy change for the solution of 1 g NaOH in water, it would just be -2.1 x 10^{3}divided by 1 g, right?

And to find enthalpy change for the solution of one mole NaOH, it'd be -1.2 x 10^{3}divided by 1 mole, right?

Now, the equation of the NaOH dissolving in the water would be like this.

NaOH ---> Na + OH

Is this right?

It then asks me to look at enthalpies of formation as given in thermodynamic tables. I am to calculate enthalpy change for the reaction as written in the equation and I should compare it with the enthalpy change for one mole NaOH.

I'm looking at the table and NaOH is -425.6 kJ/mol. The Na is -240.1 kJ/mol and the OH is -230.0 kJ/mol.

What now? Isn't the NaOH supposed to match up with what I got previously for one mole?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Thermochemistry Help

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**