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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: GrNz on December 17, 2010, 02:49:39 AM

Title: Tricky Thermodynamics Problem. about calorimetry
Post by: GrNz on December 17, 2010, 02:49:39 AM
Q1     (a) Calorimetry results for a typical beer (3.5 % alcohol by volume, or 2.8% alcohol by mass) show a fuel value (energy content) of 1.1 kJ g-1.

i)   Use the data given to calculate enthalpy  of combustion of ethanol, C2H5OH(l).                  

ii)   Hence, calculate the fuel value of the ethanol, in kJ per gram of beer.


iii)   What accounts for the remaining energy content of the beer?

Substance
   ΔfH°/kJ mol-1
   Concentration in beer

Carbohydrate
   
   
1.2 % by mass
Protein
      0.3 % by mass
C2H5OH (l)
   -278   2.8% by mass

CO2 (g)
   -394   
H2O (l)
   -286

   




       
 








(b)   The manufacturers of a new engine want to know how efficient it is.  They think that the petrol used to run the engine can be approximated well by octane and have asked you to tell them how much energy is available from burning octane, according to the following equation:
C8H18(l)  +  12.5O2(g)  →  8CO2(g)  +  9H2O(l)
(i)   Use the equation and the data given to calculate values of ΔrHo298 and ΔrUo298 for this reaction.                    
(ii) Comment on the relative magnitudes of the values you obtain.

Substance   ΔfHo298  /kJ mol-1
Octane C8H18(l)      −249.9
Carbon dioxide CO2(g)      −393.5
Water H2O(l)      −285.8

Next, the chief scientist points out that the chemical equation as written does not exactly mimic the operating conditions of the engine.  The exhaust gases are hot and contain water in the vapour phase; octane is also in the vapour phase (boiling point 399K).

(iii)   Estimate ΔrHo  for the reaction at 1000 K.  The enthalpy of vaporisation of water, ΔvapHo, is +40.7 kJ mol-1 and  the enthalpy of vaporisation of octane, ΔvapHo, is +41.5 kJ mol-1.               

Substance   Cp, m  /J K-1 mol-1
Octane C8H18(l)      187.8
Oxygen O2(g)      29.4
Carbon dioxide CO2(g)      37.1
Water H2O(l)      75.3


Q2
     (a) Choose the substance with the greater molar entropy in each of the following     pairs, and give a brief reason for each answer:
 (i) O2 (g) (0.5 atm, 298 K); O2 (g) (1.0 atm, 298 K)   
  (ii) butan-1-ol C4H9OH (l) (298K); diethyl ether C4H10O (l) (298K)                                


    (b) 4 moles of an ideal gas are compressed isothermally and reversibly from
   150 dm3 to 75 dm3 at 298 K.
   (i) Calculate the entropy change of the system.                      
    (ii) What is the entropy change of the surroundings?          
   (iii) What is the entropy change of the system if the process is carried out    irreversibly?         

   (c) Use the following data to estimate the normal boiling point (in K) of bromine.
   Br2(l) :   Sm0 =  152.2 J mol-1 K-1.
          Br2(g) :  Sm0 = 245.4 J mol-1 K-1;  ΔfH0 = 30.91  kJ mol-1.   
                                          
(d)  1 mole of liquid water is frozen at a temperature of  −5 oC. The molar heat capacity of  liquid water is 75.3 J K−1 mol−1 and  the molar heat capacity of ice is 37.6 J K−1 mol−1. The molar enthalpy of fusion of ice is 6.01 kJ  mol−1 at 273K.
  (i) Calculate the entropy change for the freezing process at −5 oC. Hint:
  split the process into
   H2O(l) (-5 oC) → H2O(l) (0 oC)  → H2O(s) (0 oC) → H2O(s) (-5 oC)  
  (ii) Calculate the entropy change of the surroundings. Hint: First calculate the enthalpy of fusion of ice at –5 oC.    (5 marks)
  (iii)  Deduce whether the process is spontaneous.   
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: DrCMS on December 17, 2010, 03:16:04 AM
Your attempts are?
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: GrNz on December 17, 2010, 03:35:27 AM
Q1
Calculate the ∆H for the combustion of ethanol according to the equation:

∆H = ∑∆Hө (products) - ∆Hө (reactants) ff
= (2x–394 + 3x–286) - (-278 + 0) = (-1646) - (-278) =   - 1368 kJ mol-1

ii) Mr of C2H5OH= 46.02

-1368 kJ mol-1/46.02 = 29.73 kJ g-1

2.8% of mass = 2.8/100 X 29.73 = 0.83 kJ g-1

ii) remaining energy content of the beer is provided by  carbohydrate and Protein..

b)
i) ∆rH298 = ∑∆fHө298 (products) - ∆Hfө 298 (reactants)
8x-393.5 +9x-285.8-(-249.9) = -5470.3 j

ii) I don't know how to work ∆rU298 and couldn't asnwer (ii)
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: GrNz on December 17, 2010, 07:34:17 AM
Your attempts are?

Q1
Calculate the ∆H for the combustion of ethanol according to the equation:

∆H = ∑∆Hө (products) - ∆Hө (reactants) ff
= (2x–394 + 3x–286) - (-278 + 0) = (-1646) - (-278) =   - 1368 kJ mol-1

ii) Mr of C2H5OH= 46.02

-1368 kJ mol-1/46.02 = 29.73 kJ g-1

2.8% of mass = 2.8/100 X 29.73 = 0.83 kJ g-1

ii) remaining energy content of the beer is provided by  carbohydrate and Protein..

b)
i) ∆rH298 = ∑∆fHө298 (products) - ∆Hfө 298 (reactants)
8x-393.5 +9x-285.8-(-249.9) = -5470.3 j

ii) I don't know how to work ∆rU298 and couldn't asnwer (ii)
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: DrCMS on December 17, 2010, 11:15:25 AM
Do you know the relationship between H and U? 
How does the volume of the system change from the initial conditions to the final conditions?
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: GrNz on December 17, 2010, 12:32:13 PM
Do you know the relationship between H and U? 
How does the volume of the system change from the initial conditions to the final conditions?

∆U= H +w . [assuming  assuming the reaction takes place in constant presser]. q= H

i have provided all the information as i was given. i don't think the volume of the system changes from initial condition to final conditions.
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: DrCMS on December 17, 2010, 01:13:33 PM
Do you know the relationship between H and U? 
How does the volume of the system change from the initial conditions to the final conditions?

∆U= H +w . [assuming  assuming the reaction takes place in constant presser]. q= H

i have provided all the information as i was given. i don't think the volume of the system changes from initial condition to final conditions.

Look at the moles of gas at the start and end of the reaction if you assume it takes place at constant pressure can the volume also stay constant? 
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: GrNz on December 18, 2010, 07:16:18 PM
Do you know the relationship between H and U? 
How does the volume of the system change from the initial conditions to the final conditions?

∆U= H +w . [assuming  assuming the reaction takes place in constant presser]. q= H

i have provided all the information as i was given. i don't think the volume of the system changes from initial condition to final conditions.

Look at the moles of gas at the start and end of the reaction if you assume it takes place at constant pressure can the volume also stay constant? 
i am sorry i lost you... could you tell me which gas are you on about? please...
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: GrNz on December 18, 2010, 11:08:43 PM
170+ viewer but still  No proper help :( :-[
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: Borek on December 19, 2010, 08:48:51 AM
Most of the views are from bots, not from users.

i am sorry i lost you... could you tell me which gas are you on about? please...

Which question are you solving? What is the reaction equation?
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: GrNz on December 19, 2010, 01:09:08 PM
Most of the views are from bots, not from users.

i am sorry i lost you... could you tell me which gas are you on about? please...

Which question are you solving? What is the reaction equation?

Those answer were for  Quesion no 1....

equation is C2H5OH(l)+3O2(g) :rarrow:2 CO2(g)+3H2O(l)

Q1
Calculate the ∆H for the combustion of ethanol according to the equation:

∆H = ∑∆Hө (products) - ∆Hө (reactants) ff
= (2x–394 + 3x–286) - (-278 + 0) = (-1646) - (-278) =   - 1368 kJ mol-1

ii) Mr of C2H5OH= 46.02

-1368 kJ mol-1/46.02 = 29.73 kJ g-1

2.8% of mass = 2.8/100 X 29.73 = 0.83 kJ g-1

ii) remaining energy content of the beer is provided by  carbohydrate and Protein..

b)
i) ∆rH298 = ∑∆fHө298 (products) - ∆Hfө 298 (reactants)
8x-393.5 +9x-285.8-(-249.9) = -5470.3 j
and Rest i Couldn't get my head around  :(
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: Borek on December 19, 2010, 01:24:48 PM
Take a look at the reaction equation - what is volume change? How many moles of gas before reaction? How many after?
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: GrNz on December 19, 2010, 02:18:14 PM
Take a look at the reaction equation - what is volume change? How many moles of gas before reaction? How many after?

ohh i see !!! 3 moles of gas produce 2 moles of gas. Now i see.. but  i am clue less now where and how am i suppose to use that information? could you please explain it to me sir...
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: thegirlbehindtheveil on January 07, 2011, 08:02:19 PM
hey so how would you work out 1b i, ii and iii please? i do not know how to them. thank you
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: thegirlbehindtheveil on January 08, 2011, 03:06:38 PM
Your attempts are?

please help with this question im stiuc
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: DrCMS on January 08, 2011, 03:30:20 PM
hey so how would you work out 1b i, ii and iii please? i do not know how to them. thank you

i) Use the reaction equation and the given heats of formation for the reactants and the products to calculate the heat of reaction.
ii) Use the reaction equation to calculate the volume change for the reaction which you can then use to calculate the work done by the reaction which with the heat of reaction from i above allows you to calculate U.
iii) Use the heats of vaporisation of water and octanol to modify the answer for i to account for the reaction occurring in the vapour phase at 1000K
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: SAMMY on January 08, 2011, 08:00:28 PM
Please can someone finish the last bits off:::

II know 3 moles produce 2 moles so how do we work out delta U ???
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: DrCMS on January 09, 2011, 06:37:20 AM
@ SAMMY and thegirlbehindtheveil

Go and look at the relationship between H and U.  Then look up how to calculate the work done by a system.
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: SAMMY on January 09, 2011, 08:07:58 AM
Yes but what equation do i use??

Delta H = Delta u + Delta (PV) is that correct??

Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: thegirlbehindtheveil on January 09, 2011, 08:13:24 AM
if we are working out ∆U

the the equation we use is:

∆U= q + w (?)

as we know q is the heat and we have just worked that out and now we need to calculate the work done to find the value of
∆U

but heres where im confused

do we use

w= pex∆v

to work out work done???

Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: SAMMY on January 09, 2011, 08:19:50 AM
Do we not use this formula??

Calculate Work, W. Use energy balance,
∆U = Q - W, Q = 0
n*Cv*∆T = -W


Calculate ∆U,
∆U = -W

Is this correct??
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: thegirlbehindtheveil on January 09, 2011, 08:50:36 AM
n*Cv*∆T = -W????/

i don't know thats why im confused which one do we use  ???
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: SAMMY on January 09, 2011, 08:56:13 AM
Does anyone know which formula we use?????

Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: SAMMY on January 09, 2011, 09:25:26 AM
DrCMS please help
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: thegirlbehindtheveil on January 09, 2011, 01:45:04 PM
what equation do we use then???


because this is what ive got so far

 if we are working out ∆U

the the equation we use is:

∆U= q + w (?)

as we know q is the heat and we have just worked that out and now we need to calculate the work done to find the value of
∆U

but heres where im confused

do we use

w= pex∆v

to work out work done???
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: thegirlbehindtheveil on January 09, 2011, 02:01:07 PM
im still confused?
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: SAMMY on January 09, 2011, 02:17:42 PM
Me2..anyone out there that can help??
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: DrCMS on January 09, 2011, 04:02:37 PM
From wikipedia  http://en.wikipedia.org/wiki/Enthalpy

The enthalpy of a system is defined as:

    H = U + pV

where

    H is the enthalpy of the system (in joules),
    U is the internal energy of the system (in joules),
    p is the pressure at the boundary of the system and its environment, (in pascals), and
    V is the volume of the system, (in cubic meters).

For this question the equation to use is ΔH = ΔU + W = ΔU + Δ(pV).  Taking 1 mole of starting material you can calculate Δ(pV) and using the value for ΔH calculated from part i solve for ΔU.


Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: thegirlbehindtheveil on January 09, 2011, 04:39:35 PM
Taking 1 mole of starting material- do you mean octane?

i know Δ (pV) = (pV) products − (pV) reactants.

is this right?
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: SAMMY on January 09, 2011, 04:54:34 PM
Hi..Im struggling here...when you figure it then please can you post it on here..

I think its no of moles for reactant minus no of moles products?

does that make sense??
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: DrCMS on January 09, 2011, 05:28:31 PM
Taking 1 mole of starting material- do you mean octane?

i know Δ (pV) = (pV) products − (pV) reactants.

is this right?

Yes
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: GrNz on January 09, 2011, 08:59:51 PM
I'll Help if any one of you tell me which question you guys are trying to solve....  as i also need to check weather my answer are right or wrong and we are running out of time  :-[.. i m pretty much sure i did it  right !! :).
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: SAMMY on January 10, 2011, 10:27:05 AM
Hiya

The question is below...reply back asap thanks..

Q1     (a) Calorimetry results for a typical beer (3.5 % alcohol by volume, or 2.8% alcohol by mass) show a fuel value (energy content) of 1.1 kJ g-1.

i)   Use the data given to calculate enthalpy  of combustion of ethanol, C2H5OH(l).                 

ii)   Hence, calculate the fuel value of the ethanol, in kJ per gram of beer.


iii)   What accounts for the remaining energy content of the beer?

Substance
   ΔfH°/kJ mol-1
   Concentration in beer

Carbohydrate
   
   
1.2 % by mass
Protein
      0.3 % by mass
C2H5OH (l)
   -278   2.8% by mass

CO2 (g)
   -394   
H2O (l)
   -286

(b)   The manufacturers of a new engine want to know how efficient it is.  They think that the petrol used to run the engine can be approximated well by octane and have asked you to tell them how much energy is available from burning octane, according to the following equation:
C8H18(l)  +  12.5O2(g)  →  8CO2(g)  +  9H2O(l)
(i)   Use the equation and the data given to calculate values of ΔrHo298 and ΔrUo298 for this reaction.                   
(ii) Comment on the relative magnitudes of the values you obtain.
Substance   ΔfHo298  /kJ mol-1
Octane C8H18(l)      −249.9
Carbon dioxide CO2(g)      −393.5
Water H2O(l)      −285.8

Next, the chief scientist points out that the chemical equation as written does not exactly mimic the operating conditions of the engine.  The exhaust gases are hot and contain water in the vapour phase; octane is also in the vapour phase (boiling point 399K).

(iii)   Estimate ΔrHo  for the reaction at 1000 K.  The enthalpy of vaporisation of water, ΔvapHo, is +40.7 kJ mol-1 and  the enthalpy of vaporisation of octane, ΔvapHo, is +41.5 kJ mol-1.               


Substance   Cp, m  /J K-1 mol-1
Octane C8H18(l)      187.8
Oxygen O2(g)      29.4
Carbon dioxide CO2(g)      37.1
Water H2O(l)      75.3


THere you go!!!!!!!!!!
Title: Re: Tricky Thermodynamics Problem. about calorimetry
Post by: SAMMY on January 10, 2011, 05:20:26 PM
GrNZ you got the question

can u assist please??