- #1
WahooMan
- 22
- 0
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!
Last edited: