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#### DrCMS

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #30 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

#### GrNz

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #31 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 !! .

#### SAMMY

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #32 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!!!!!!!!!!

#### SAMMY

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##### Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #33 on: January 10, 2011, 05:20:26 PM »
GrNZ you got the question