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#### GrNz

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##### Tricky Thermodynamics Problem. about calorimetry
« 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.
« Last Edit: December 17, 2010, 03:42:01 AM by GrNz »

#### DrCMS

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #1 on: December 17, 2010, 03:16:04 AM »

#### GrNz

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #2 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)
« Last Edit: December 17, 2010, 04:26:15 AM by GrNz »

#### GrNz

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #3 on: December 17, 2010, 07:34:17 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)

#### DrCMS

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #4 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?

#### GrNz

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #5 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.

#### DrCMS

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #6 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?

#### GrNz

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #7 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...

#### GrNz

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #8 on: December 18, 2010, 11:08:43 PM »
170+ viewer but still  No proper help

#### Borek

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #9 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?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### GrNz

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #10 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

#### Borek

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #11 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?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### GrNz

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #12 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...

#### thegirlbehindtheveil

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #13 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

#### thegirlbehindtheveil

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #14 on: January 08, 2011, 03:06:38 PM »