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Topic: Calculate enthalpy change  (Read 23785 times)

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kivine

  • Guest
Calculate enthalpy change
« on: September 12, 2004, 03:07:10 AM »
Heres the question :

Calculate the amount of heat evolved and the enthalpy change for the following reaction ( assume that 4.3 J are required to raise he temperature of 1 cm3 of all solutions by 1 degree celsius )

------------------------Here is my answer-------------------------------

Empty weighing Boat          = 2.4332g

Zinc Powder + weighing boat       = 3.4349g

Weighing Boat + remains of zinc Powder    = 2.445g

Zinc Powder added             = 3.4349 - 2.445
                                  = 0.9899g



      Mass of Zinc added (gram)       =    ______0.9899___ ___
      Initial Temperature (°C)      =          _______22°C_______
      Highest Temperature (°C)      =          ____ __34°C _______



Zn (s) + CuSO4 (aq) ? Cu (s) + ZnSO4 (aq)


Number of moles of zinc - 0.9899/65.39
             = 0.015138400367028597644899831778559


Enthalpy change – [(4.3 J x 30cm3 x 12°C) / Number of moles of zinc] /1000
                   = 102.25651075872056633028442539892
          = 102.26 KJ/Mol


Since reaction is exothermic : -102.26KJ/Mol


--------------------------------------


Expected answer : -217 kJ/mol



--------------------------------------


My working should be wrong.anyone?
« Last Edit: September 12, 2004, 03:08:49 AM by kivine »

Demotivator

  • Guest
Re:Calculate enthalpy change
« Reply #1 on: September 12, 2004, 02:02:22 PM »
The calculations looks right.  Are you sure the total volume is 30cm3? Are you missing the heat capacity of boat or container?  Is this a lab experiment or a "paper" problem?

kivine

  • Guest
Re:Calculate enthalpy change
« Reply #2 on: September 12, 2004, 09:46:06 PM »
lab experiment.I'm new to this topic. but why is there a need to calculate the heat capacity of boat or container.

The copper sulphate is 30ml = 30cm3 ???

Demotivator

  • Guest
Re:Calculate enthalpy change
« Reply #3 on: September 12, 2004, 10:55:54 PM »
Well, everything that the solution is in contact with allows heat from the solution to transfer to it , meaning  the container gets warm too. The heat you calculated was only from the solution which is probably why the value is too low. Some of the heat flows to the container. Heat in the container = delta T x  heat capacity of container. The heat capacity is usually given so you don't need to calculate it.
But if the lab instructions did not call for it, then perhaps the heat capacity of container was assumed low enough not to contribute significantly, I don't know.
Another problem might be some heat could be lost to the air escaping measurement.
« Last Edit: September 12, 2004, 11:15:11 PM by Demotivator »

kivine

  • Guest
Re:Calculate enthalpy change
« Reply #4 on: September 13, 2004, 08:05:47 AM »
The question asked for heat evolved and enthalpy change. What is the difference, any ideas? or as  Demotivator mentioned
--{Heat in the container = delta T x  heat capacity of container}--  >>> heat evolved

Demotivator

  • Guest
Re:Calculate enthalpy change
« Reply #5 on: September 13, 2004, 10:16:08 AM »
In this case (conditions of constant pressure) the enthalpy and heat are the same number. But The enthalpy carries a negative sign to indicate that heat is evolved.

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