# Chemical Forums

## Chemistry Forums for Students => High School Chemistry Forum => Topic started by: highschoolhelp16 on April 15, 2012, 07:38:57 PM

Title: Calorimetry
Post by: highschoolhelp16 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.
Title: Re: Calorimetry
Post by: fledarmus on April 15, 2012, 07:49:16 PM
So what does "everything you can think of" consist of?
Title: Re: Calorimetry
Post by: highschoolhelp16 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.
Title: Re: Calorimetry
Post by: UG 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.
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 08:34:42 PM
so how would i fit 3.16 KJ/mol into the equation?
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 08:35:37 PM
Well, how many moles of methanol do you have?
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 08:41:04 PM
0.062 mol nh4no3
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 08:43:31 PM
Methanol has molecular formula CH3OH.
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 08:47:02 PM
sorry I was looking at the wrong problem, its 0.16 mol CH3OH
Title: Re: Calorimetry
Post by: UG 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?
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 08:57:03 PM
its 3.12 mol, and yes the Hfusion is 3.16kJ/Mol
Title: Re: Calorimetry
Post by: UG 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.
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 09:05:31 PM
so is the energy change needed 9.86kJ/Mol?
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 09:06:36 PM
It will just be 9.86 kJ, the 'moles' unit cancel out when you multiply.
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 09:10:43 PM
so the equation should be (100.0g)(0.424cal/gc)(337.2K-175.5K)(9.86kJ)?
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 09:12:23 PM
or is 9.86kJ the energy required to convert the 100g from solid to boiling point?
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 09:14:59 PM
No, 9.86 kJ is the energy needed to convert solid to liquid. Now you need to find how much energy is needed to transform the liquid into gas, because the question states that you want gaseous methanol. I have stated in my previous post how to do this.
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 09:19:01 PM
so the energy needed to convert from liquid to gas should be 110.136kJ?
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 09:20:25 PM
Yes (I am assuming you used the correct Hvap value)
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 09:21:46 PM
so do I need to put these values into my original equation of (100.0g)(.424cal/gc)(337.2k-175.5k)? or was that just completely off track?
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 09:23:28 PM
Yes you add the two values onto whatever you got from (100.0g)(.424cal/gc)(337.2k-175.5k) but you will need to change the units so that they match, either convert kJ into cal or vice versa.
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 09:28:54 PM
So if I get 3,684.58kJ from that equation, I should add 110.14kJ, and 9.86kJ, to get 3768.58kJ total.
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 09:37:44 PM
So if I get 3,684.58kJ from that equation
That seems to be an awfully large number...(check again) but if thats what you got, then adding 110.14 kJ and 9.86 kJ onto it is correct.
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 09:40:52 PM
could the equation be wrong. Im using q=mass*specific heat*delta T
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 09:46:15 PM
Well initally you got 6,877.28 cal, how many kJ is this?
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 09:48:53 PM
I got 3684.58 kJ
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 10:01:14 PM
In thermochemistry 1 calorie is about 4.184 J

http://en.wikipedia.org/wiki/Calorie
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 10:03:41 PM
so 28.75kJ?
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 10:08:07 PM
Well going by your equation (100.0g)(.424cal/gc)(337.2k-175.5k), this should give 6856.08 cal. Then converting to kJ gives 28.69 kJ. But maybe you did some rounding somewhere.
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 10:11:13 PM
You're right I made a calculation error. so 28.69kJ is the heat required to convert 100.0g of solid methanol to gaseous methanol?
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 10:12:40 PM
No, remember you got to add 110.14kJ and 9.86kJ
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 10:13:52 PM
so 148.69 is the total heat required?
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 10:15:37 PM
Yes. 148.69 kJ
Title: Re: Calorimetry
Post by: highschoolhelp16 on April 15, 2012, 10:16:10 PM
Thank you so much for your help I really appreciate it :)
Title: Re: Calorimetry
Post by: UG on April 15, 2012, 10:19:32 PM
:) I'm glad we finally sorted that one out!