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highschoolhelp16

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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.

fledarmus

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Re: Calorimetry
« Reply #1 on: April 15, 2012, 07:49:16 PM »
So what does "everything you can think of" consist of?

highschoolhelp16

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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.

UG

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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 liquid at the melting temperature) and the enthalpy of vaporisation (energy needed for liquid 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.

highschoolhelp16

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Re: Calorimetry
« Reply #4 on: April 15, 2012, 08:34:42 PM »
so how would i fit 3.16 KJ/mol into the equation?

UG

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Re: Calorimetry
« Reply #5 on: April 15, 2012, 08:35:37 PM »
Well, how many moles of methanol do you have?

highschoolhelp16

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Re: Calorimetry
« Reply #6 on: April 15, 2012, 08:41:04 PM »
0.062 mol nh4no3

UG

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Re: Calorimetry
« Reply #7 on: April 15, 2012, 08:43:31 PM »
Methanol has molecular formula CH3OH.

highschoolhelp16

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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

UG

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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?

highschoolhelp16

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Re: Calorimetry
« Reply #10 on: April 15, 2012, 08:57:03 PM »
its 3.12 mol, and yes the Hfusion is 3.16kJ/Mol

UG

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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.

highschoolhelp16

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Re: Calorimetry
« Reply #12 on: April 15, 2012, 09:05:31 PM »
so is the energy change needed 9.86kJ/Mol?

UG

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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.

highschoolhelp16

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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)?