Enthalpy change/Temperature change question (Decomposition of KHCo3

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Tahera2013
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#1
Report Thread starter 8 years ago
#1
Hi everyone, I am meant to do a lab report for an experiment and then using the temperature changes construct a Hess's cycle.

I basically tipped KHCO3 and K2CO3 into two different cups of HCl acid. But I'm afraid my calculations doesn't seem right?

Can anyone help me see where i went wrong.
When calulating the the Q= MCΔT you get the answer and then divide it by a 1000 to convert it into Kj and then divide it by the moles but the value seems way too small!



Finding the enthalpy change for the decomposition of KHCO3 indirectly using Hess's law
Aim:


  • Be able to convert temperature changes to enthalpy changes
  • Be able to use Hess's law to find an enthalpy change indirectly

Safety:


  • Wear eye protection
  • Potassium Carbonate is irritant
  • 2.00 moldn-3 HCL is irritant

Equipment:


  • 2.00 moldm-3 HCL
  • Anhydrous K2CO3 Solid
  • KHCO3 solid
  • Burette
  • Mass Balance
  • Plastic cup
  • Spatula, funnel

Procedure:


  • Using a burette, measure 30.0cm3 of the HCL acid into the plastic cup


  • Measure the temperature of the acid after it has stood for a few minutes



  • Weight the test tube containing the K2CO3


  • Tip the K2CO3 solid into the plastic cup stir carefully with the thermometer and record the highest temperature


  • Reweigh the tube/


  • Repeat steps 1-4 using KHCO3 instead, however for step 4 the lowest temperature should be recorded.

Data:

Reaction A: K2CO3 + 2HCL --> 2KCL + CO2 + H20 Q= M x C (4.2jg) x ΔT

Reaction B: KHCO3 + HCL ---> KCL + CO2 + H20

Reaction C: 2KHCO3 ---------> K2CO3 + CO2 + H20



Reaction of KHCO3 with 30cm3 of 1M HCl (Exothermic)

Mass of KHCO3 used = 6.17g

Temperature drop = 13.oC

Q= MCΔT


  • 30x 4.2 x 13= -1638 Joules


  • 6.17g/100.1gmol-1 = 0.0616 moles


  • -1638/1000= -1.638


  • -1.638/0.0616 = -26.59kjmol-1


Reaction of K2CO3(s) with 30cm3 of 2M HCl (Endothermic)


Mass of K2CO3(s) used = 1.79g

Temperature rise = 7.5oC

Q= MCΔT



  • 30 x 4.2 x 7.5 = 945 Joules


  • 1.79/138.4gmol-1 = 0.0129moles


  • 945/1000= 0.945kjmol-1


  • 0.945/0.0129moles = -73.25kjmol

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Borek
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#2
Report 8 years ago
#2
Why do you think these results are wrong?

They can't be both right though - they should differ in sign, you listed them both as negative,

Won't hurt to check what is the limiting reagent in both cases, as if not everything reacted, your answer is off.

Finally, I wonder if you can't slightly better results remembering that the mass after reaction is not 30g - it is slightly higher (but not by the mass of the solid added).
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tom szemeti1
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#3
Report 1 year ago
#3
(Original post by Tahera2013)
Hi everyone, I am meant to do a lab report for an experiment and then using the temperature changes construct a Hess's cycle.

I basically tipped KHCO3 and K2CO3 into two different cups of HCl acid. But I'm afraid my calculations doesn't seem right?

Can anyone help me see where i went wrong.
When calulating the the Q= MCΔT you get the answer and then divide it by a 1000 to convert it into Kj and then divide it by the moles but the value seems way too small!



Finding the enthalpy change for the decomposition of KHCO3 indirectly using Hess's law
Aim:



  • Be able to convert temperature changes to enthalpy changes

  • Be able to use Hess's law to find an enthalpy change indirectly

Safety:



  • Wear eye protection

  • Potassium Carbonate is irritant

  • 2.00 moldn-3 HCL is irritant

Equipment:



  • 2.00 moldm-3 HCL

  • Anhydrous K2CO3 Solid

  • KHCO3 solid

  • Burette

  • Mass Balance

  • Plastic cup

  • Spatula, funnel

Procedure:



  • Using a burette, measure 30.0cm3 of the HCL acid into the plastic cup


  • Measure the temperature of the acid after it has stood for a few minutes



  • Weight the test tube containing the K2CO3


  • Tip the K2CO3 solid into the plastic cup stir carefully with the thermometer and record the highest temperature


  • Reweigh the tube/


  • Repeat steps 1-4 using KHCO3 instead, however for step 4 the lowest temperature should be recorded.

Data:

Reaction A: K2CO3 + 2HCL --> 2KCL + CO2 + H20 Q= M x C (4.2jg) x ΔT

Reaction B: KHCO3 + HCL ---> KCL + CO2 + H20

Reaction C: 2KHCO3 ---------> K2CO3 + CO2 + H20



Reaction of KHCO3 with 30cm3 of 1M HCl (Exothermic)

Mass of KHCO3 used = 6.17g

Temperature drop = 13.oC

Q= MCΔT



  • 30x 4.2 x 13= -1638 Joules


  • 6.17g/100.1gmol-1 = 0.0616 moles


  • -1638/1000= -1.638


  • -1.638/0.0616 = -26.59kjmol-1


Reaction of K2CO3(s) with 30cm3 of 2M HCl (Endothermic)


Mass of K2CO3(s) used = 1.79g

Temperature rise = 7.5oC

Q= MCΔT



  • 30 x 4.2 x 7.5 = 945 Joules


  • 1.79/138.4gmol-1 = 0.0129moles


  • 945/1000= 0.945kjmol-1


  • 0.945/0.0129moles = -73.25kjmol
one thing is you put a temperature rise as endo and temperature drop as exo, it's the other way round
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Tahera2013
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#4
Report Thread starter 9 months ago
#4
(Original post by tom szemeti1)
one thing is you put a temperature rise as endo and temperature drop as exo, it's the other way round
Where were you 7 years ago haha!

Many thanks!
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