Calculating Heat of NaOH Dissolution in Water

[WahooMan]

Homework Statement

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

Homework Equations

q = c * m * delta t

-(c * m * delta t) = (c * m * delta t)

c = specific heat, m = mass, t = temperature

The Attempt at a Solution

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

a) Calculate q H20.

I used the equation q = cmt, so

q = (4.18)(53.0)(9.70) = 2150J

b) Find the change of enthalpy for the reaction as it occurred in the calorimeter.

= -2150J

c) Find the change of enthalpy for the solution of 1.00g NaOH in water.

For this I just halved -2150J and got -1080 (3 significant figures) but I'm not sure if this is correct.

d) Find the change of enthalpy for the solution of one mole NaOH in water.

For this I found the mass of one mole of NaOH, which is 40.00g. Then I multiplied this by my answer to part c and got -43200J/mol

e) Using enthalpies of formation as given in thermodynamic tables, calculate the change of enthalpy for the reaction NaOH yields Na^+ + OH^-

I looked at a thermodynamic data table and found the delta H value of Na^+ to be -240100J/mol, OH^- to be -230000J/mol, and NaOH to be -425600J/mol. This obviously isn't what I got for my answer in part d. What have I done wrong? Please Help!

 

[Borek]

You are mistaking enthalpy of dissolution with enthalpy of formation.