Group Work 4 - How much Heat can you get from a Marshmallow?
ANSWERS
1) The experimental method for measuring the energy content of food uses calorimetry, specifically combustion by reaction with oxygen gas in a bomb calorimeter. Oxygen gas will completely oxidize all carbon atoms to their highest oxidation state, +4, forming carbon dioxide.
Write the combustion reaction of sucrose . (Note the 2 products are both gases.)
C12H22O11 + 12 O2(g) à 12 CO2(g) + 11 H2O(g)
2) Here are the experimental results:
à one marshmallow (7.5 g) combusted at 25 deg C increased the temperature from 25.0 to 37.1 deg C.
What is the sign of the heat change for the calorimeter?
Since the temperature increases, the sign of the heat change fo rthe calorimeter is +
What is the sign of heat change of the marshmallow combustion reaction?
Since heat was lost from the combustion reaction of sucrose, the sign of the heat change for the calorimeter is -.
q(surroundings) = - q(system)
so: q(calorimeter) = - q(reaction)
3) The heat capacity of the bomb calorimeter was measured and found to be 9.43 kJ/ deg C.
What is the meaning of the “heat capacity” of the bomb calorimeter?
The heat capacity refers to how much heat (energy) is required to raise the temperature of something by one degree. So the “heat capacity” of the bomb calorimeter is the a number expressing how its temperature will change as energy is added or removed as heat.
4) Evaluate the heat change, q, for the combustion reaction.
q(calorimeter) = (Tf – Ti) Cp = 12.1 deg x 9.43 kJ/deg C = 114.1 kJ
q(reaction) = -114.1 kJ
5) Evaluate the heat per gram of marshmallow combustion.
Mass heat = q/g = -114.1 kJ / 7.5 g = -15.2 kJ/g
6) Evaluate the molar heat of a marshmallow combustion.
molar heat = q/#mol = -114.1 kJ / (7.5 g/342.3g/mol) = -5207.7 kJ/mol
7) Dietary energy is expressed not in joules or in kJ, but in calories. 1 calorie = 4.184 joules.
And dietary calories , Cal (for human nutritional information) are equal to 1000 calories.
1 Cal = 1000 cal = 4184 joules.
How many dietary calories are in a marshmallow based on your experimental calorimetry data?
-114.1 kJ x 1000 J/kJ x 1 cal /4.184 J x 1 Cal/1000 cal = 27 Calories
8) One can calculate a theoretical enthalpy change for a reaction, DH(rxn), using standard enthalpies of formation, DH(f). In this method , one obtains the molar standard enthalpies for each species in the combustion reaction C12H22O11 + 12 O2(g) à 12 CO2(g) + 11 H2O(g)
using Hess’s law and DH(rxn) = S[DHf(prdts)] - S[DHf(rgts)]
The change of enthalpy of a reaction is independent of the path of the reaction.
sum[Hf(prdts)] - sum[Hf(rgts)] = 12 DHf(CO2) + 11 DHf(H2O) - DHf(sucrose) – 12 DHf(O2)]
9) Use data in Appendix B (page A11) in your text to calculate the standard combustion enthalpy for sucrose.
12 DHf(CO2) + 11 DHf(H2O) - DHf(sucrose) – 12 DHf(O2)] =
12 x-393.51 + 11 x-241.82 - (-2222.1) – 12 x 0.0 = -5160 kJ/mol
(units in kJ/mol
10) How does the experimental combustion enthalpy compare with the theoretical enthalpy calculation? In order to properly compare combustion enthalpies from the two methods, the qv (heat change at constant volume as measured in the bomb) must be changed to qp (heat change at constant pressure) using:
qp = qv + RTdelta n where R= 8.314 J / mol K
and delta n = #mol gas(f) - #mol gas(i) and T is in Kelvin
= -5160 kJ/mol + 8.314 J/mol K x 298 x 11 mol since delta n = 23 mol - 12 mol = 11 mol
= - 5180.4 kJ/mol
Beyond Groupwork questions –
1. a) Imagine that you have done a combustion reaction of another sugar, galactose and obtained a combustion molar heat of -2526 kJ/mol. Show how you could use standard enthalpies of formation, DH(f) to calculate the standard heat of formation of this sugar galactose. Galactose is C6H12O6
[many folks were confused about what I wanted here. I wanted you to do the calulation as set up below, showing how the combustion enthhalpy could be used to calculate a standard heat of formtaion, assuming that one DIDN'T find it in a table already]
C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
delta H(combustion ) = delta Hf (prdts) - delta Hf (rgts)
-2526 kJ/mol for combustion = delta Hf (prdts) - delta Hf (rgts)
-2526 kJ/mol for combustion = 6(-393) + (-241) - delta Hf (galactose) - delta Hf (O2)
-2526 + (393) + 6(241) = - delta Hf (galactose)
- 1286 kJ/mol = delta Hf (galactose)
2. How much could a marshmallow warm you? What temperature would youe experience by eating one marshmallow? Assume your body has 8 L of fluid and assume the fluid is allwater.
(from groupwork) 114 kJ/ 1 marshmallow x 1000 J/ kJ = 8 L x 1000 mL x 1 g/mL x 4.184 J/deg C x delta T (deg C)
3.4 deg C = delta T
3. Another method to obtain the enthalpy content of a compound is to sum up the bond energies.
Using data in Table 7.3 (p 462) of Chapter 7, calculate the DH(rxn) for the combustion of sucrose.
balanced equation for combustion: C12H22O12 + 12 O2 --> 12 CO2 + 11 H2O
use: delta H (rxn) = bond enethalpies (rgts) - bond enethalpies (prdts)
SUM[bond enethalpies (rgts) ] kJ/mol
14 C-H bonds @ 414kJ/mol + 10 C-C bonds @ 347kJ/mol + 14 C-O bonds @ 351kJ/mol + 8 O-H bonds @460 kJ/mol + 12 O=O bonds @ 498.7 kJ/mol = 23, 845 kJ/mol
SUM[bond enethalpies (prdts) ] kJ/mol
24 C=O @ 799 kJ/mol* + 22 O-H@ 460 kJ/mol = 29,296 kJ/mol
* see footnote to bond enthalpies table in text
delta H (rxn) = 23,845 - 29,296 = -5451 kJ/mol
3. How does your result from (2) compare to the results you obtained in groupwork using experimental calorimetric data and standard enthalpies of formation? If they are not similar values, can you explain why?
-5451 (from bond enthalpy calculation ) vs -5208 (from calorimetry measurement) isn't too far off. The difference certainly can be mainly attributed to the fact that bond enthalpies assume all molecules are in teh gas phase; whereas the bomb calorimetry experiment astarted with solid sucrose.