Question

I am currently working on a lab for my general chem class and we are finding...

I am currently working on a lab for my general chem class and we are finding enthalpy. I believe I found the correct Delta Hs, but moving fowards to find enthalpy of solution salt, enthalpy of formation of the salt, enthalpy of formation anion, and enthalpy of formation of the cation I am very lost. If I could be walked through how to solve my first trial so I can solve my second trial on my own would be AMAZING. The answer I got for Delta Hs was 1.25 kj/mole

For my first trial:

Mass of salt = 4.955g

Moles of salt= 0.085 moles

Mass of water = 58.001 g

Inital temperature of salt and water = 21.56 degrees C

Final temperature of salt and water = 21.13 degrees C

Heat flow of water = -1.04 x 10^2 J

Heat flow of salt = -1.84 J

Homework Answers

Answer #1

In general, ∆H = m x s x ∆T, where m is the mass of the reactants, s is the specific heat of the product, and ∆T is the change in temperature from the reaction.

Mass of salt = 4.955g

Moles of salt= 0.085 moles

Mass of water = 58.001 g

Inital temperature of salt and water = 21.56 degrees C

Final temperature of salt and water = 21.13 degrees C

Heat flow of water = -1.04 x 10^2 J

Heat flow of salt = -1.84 J

       we can calculate the total mass of the reactants as follows

4.955 x( 0.085) + 58.001 x (1) = 58.42 g

Now ∆T = Final temperature of salt - Inital temperature of salt

             = 21.13-21.56 = -0.43

∆H = m x s x ∆T

      = 58.42 x (-1.04 x 10^2 + (-1.84))x(-0.43)

       = 58.42 x105.84x(-0.43)

     =- 2658 kj/mole

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions