Thermochemistry Help: Calculating Enthalpy Change of NaOH Dissolving in Water

[Ryuk1990]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

For enthalpy change of the reaction, I got -2.1 x 10³ joules.

To find enthalpy change for the solution of 1 g NaOH in water, it would just be -2.1 x 10³ divided by 1 g, right?

And to find enthalpy change for the solution of one mole NaOH, it'd be -1.2 x 10³ 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?

 

[nate808]

Hello there,

Thank you for posting your question on the forum. Your calculations and equations are correct so far. To find the enthalpy change for the solution of 1 g NaOH in water, you would divide the enthalpy change by the amount of NaOH used (2.0 g). Similarly, to find the enthalpy change for the solution of one mole NaOH, you would divide the enthalpy change by the number of moles of NaOH used (0.05 moles). This is because enthalpy change is an extensive property, meaning it depends on the amount of substance used in the reaction.

As for the comparison with the enthalpy of formation, you are correct that the enthalpy change for the reaction should match up with the enthalpy of formation of NaOH (-425.6 kJ/mol in this case). However, the values you have calculated so far are for the enthalpy change of the solution, not the reaction. To calculate the enthalpy change for the reaction, you need to take into account the enthalpies of formation for both products (Na and OH) as well as the reactant (NaOH). The equation for the reaction would be:

NaOH (s) + H2O (l) -> Na+ (aq) + OH- (aq)

Using the enthalpies of formation for each species, you can calculate the overall enthalpy change for the reaction. The enthalpy change for the reaction should match up with the enthalpy of formation for NaOH, as you mentioned.

I hope this helps clarify things for you. Keep up the good work with your calculations and equations!