Jump to content

calorimetry


Ice-cream

Recommended Posts

Can any1 help me with this?

 

In a coffee-cup calorimeter, 50ml of 0.1M AgNO3 and 50ml of 0.1M HCl are mixed to yield the following reaction:

 

Ag+(aq) + Cl-(aq) --> AgCl(s)

 

The 2 solutions were initially at 22.6 degrees C and the final temperature is 23.4 degrees C. Calculate the heat that accompanies this reaction in kJ/mol of AgCl formed. Assume that the combined solution has a mass of 100g and a specific heat capacity of 4.18 J/degrees C g.

 

I used q = ms(detaT) = 100 x 4.18 x 0.8 = 334.4J. I then thought since there are 0.01 mols altogether, then that means for 0.01 moles it's 334.4J, so 1 mole is 33440J. but that's the wrong answer. I think it's the moles part that's wrong - i'm not sure if it's ok to find moles of AgNO3, moles of HCl and then just add them up...

 

then answer's -66kJ/mol

Link to comment
Share on other sites

Im just a highschool student myself so im sure people around here know more than i do but

 

the mole ratio is 1:1 --> 1 i thought. so you do not add the 0.05L*0.1 M = 0.005 moles together, i believe.

 

To get the molar heat of reaction, try 334.4 /0.005 !

 

hope thats a help, maybe someone else can explain it more clearly to you than i could the reason for doing this.

Link to comment
Share on other sites

  • 3 weeks later...

q=mc(delta T)

q= 100 X 4.18 X 0.8

q= 334.4 J (0.334 KJ)

 

moles= (Molarity X volume) / 1000

moles = 0.005

 

so, 0.3344 / 0.005 = 66.88

 

therefore, -66.88 KJmol-1 (delta H is negative as heat is given out so it is exothermic)

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.