- #1
kenshi64
- 34
- 0
Hi! Thanks for your interest in this this post :D!
So two reactions I performed were: A) Adding 50 cm3 water to 0.025 moles CuS04 and measuring ΔT.
B) Adding 50 cm3 water to 0.025 moles CuS04.5H2O and measuring ΔT.
Well turns out the calculations were wrong, since I apparently didn't get kJ/mol but rather mol/40 or so since I didn't take 1 mole of each element.
I want to know whether this was right since my calculations don't yield answers close to the real value otherwise.
∆H reaction = (49.736*4.186*-0.228) /0.025moles= -2 kJ/mol
∆H2 = (51.85*4.186*(-0.929)) /0.025moles= -8 kJ/mol
-2kJ and -8kJ are totally wrong! So I was told about the whole multiply the answer by 40 business.
That gives us: -80-(-320)=-240 kJ which is still not close to the real answer of -11.7 kJ(rough).
I used a temperature sensor, a calorimeter too! What went wrong and how can I correct this? Thanks for your patience this was a lengthy post I hope you can help me solve this problem! :)
Cheers!
So two reactions I performed were: A) Adding 50 cm3 water to 0.025 moles CuS04 and measuring ΔT.
B) Adding 50 cm3 water to 0.025 moles CuS04.5H2O and measuring ΔT.
Well turns out the calculations were wrong, since I apparently didn't get kJ/mol but rather mol/40 or so since I didn't take 1 mole of each element.
I want to know whether this was right since my calculations don't yield answers close to the real value otherwise.
∆H reaction = (49.736*4.186*-0.228) /0.025moles= -2 kJ/mol
∆H2 = (51.85*4.186*(-0.929)) /0.025moles= -8 kJ/mol
-2kJ and -8kJ are totally wrong! So I was told about the whole multiply the answer by 40 business.
That gives us: -80-(-320)=-240 kJ which is still not close to the real answer of -11.7 kJ(rough).
I used a temperature sensor, a calorimeter too! What went wrong and how can I correct this? Thanks for your patience this was a lengthy post I hope you can help me solve this problem! :)
Cheers!