January 19, 2022, 03:20:58 AM
Forum Rules: Read This Before Posting


Topic: Calorimetry  (Read 9618 times)

0 Members and 1 Guest are viewing this topic.

Offline highschoolhelp16

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Calorimetry
« on: April 15, 2012, 07:38:57 PM »
How do you calculate the amount of heat required to convert 100.0g of solid methanol at its normal freezing point to gaseous methanol at its normal boiling point? I have tried everything I can think of, and after 2hrs of ridiculous answers I need help.

Offline fledarmus

  • Chemist
  • Sr. Member
  • *
  • Posts: 1676
  • Mole Snacks: +203/-28
Re: Calorimetry
« Reply #1 on: April 15, 2012, 07:49:16 PM »
So what does "everything you can think of" consist of?

Offline highschoolhelp16

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Re: Calorimetry
« Reply #2 on: April 15, 2012, 07:52:18 PM »
I have tried q=m*s*deltaT, and I have done multiple equations involving the boiling and vaporization point. I tried (100.0g)(0.424ca/gc)(337.2k-175.5k)=6,877.28cal, i think that is in the right direction but I am unsure. it think that is the heat required to bring the methanol to its boiling point.

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Calorimetry
« Reply #3 on: April 15, 2012, 08:31:51 PM »
I think you are forgetting to include the enthalpy of fusion (energy needed for solid :rarrow: liquid at the melting temperature) and the enthalpy of vaporisation (energy needed for liquid :rarrow: gas at the boiling temperature) into your calculation. Your calculation of (100.0g)(0.424ca/gc)(337.2k-175.5k)=6,877.28cal only gives the energy needed to heat the liquid from 175.5 K to 337.2 K, it doesn't take into account the conversion of solid to liquid or liquid to gas.

Offline highschoolhelp16

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Re: Calorimetry
« Reply #4 on: April 15, 2012, 08:34:42 PM »
so how would i fit 3.16 KJ/mol into the equation?

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Calorimetry
« Reply #5 on: April 15, 2012, 08:35:37 PM »
Well, how many moles of methanol do you have?

Offline highschoolhelp16

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Re: Calorimetry
« Reply #6 on: April 15, 2012, 08:41:04 PM »
0.062 mol nh4no3

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Calorimetry
« Reply #7 on: April 15, 2012, 08:43:31 PM »
Methanol has molecular formula CH3OH.

Offline highschoolhelp16

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Re: Calorimetry
« Reply #8 on: April 15, 2012, 08:47:02 PM »
sorry I was looking at the wrong problem, its 0.16 mol CH3OH

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Calorimetry
« Reply #9 on: April 15, 2012, 08:51:46 PM »
I think you might like to check that again because I got a different answer. Remember you have 100 g of methanol. But anyways, once you have worked out the number of moles, you can find the energy needed for the phase transformation. I assume 3.16 kJ/mol is the enthalpy of fusion?

Offline highschoolhelp16

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Re: Calorimetry
« Reply #10 on: April 15, 2012, 08:57:03 PM »
its 3.12 mol, and yes the Hfusion is 3.16kJ/Mol

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Calorimetry
« Reply #11 on: April 15, 2012, 09:01:32 PM »
Ok, so if you have 3.12 moles and the enthalpy change is 3.16 kJ PER mole, then the energy change needed for solid to liquid transformation is? It is a similar process for finding the energy needed to go from liquid to gas, but you'll be using a different enthalpy value obviously. Then you need to add these energy values onto your calculation of (100.0g)(0.424ca/gc)(337.2k-175.5k) but you will need to convert to the same units.

Offline highschoolhelp16

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Re: Calorimetry
« Reply #12 on: April 15, 2012, 09:05:31 PM »
so is the energy change needed 9.86kJ/Mol?

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: Calorimetry
« Reply #13 on: April 15, 2012, 09:06:36 PM »
It will just be 9.86 kJ, the 'moles' unit cancel out when you multiply.

Offline highschoolhelp16

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Re: Calorimetry
« Reply #14 on: April 15, 2012, 09:10:43 PM »
so the equation should be (100.0g)(0.424cal/gc)(337.2K-175.5K)(9.86kJ)?

Sponsored Links