The standard heat of combustion of liquid 2,3,3-trimethylpentane [C8H18] is repo

Question

The standard heat of combustion of liquid 2,3,3-trimethylpentane [C8H18] is reported in a table of physical properties to be -4850 kJ/mol. A footnote indicates that the reference temperature for the reported value is 25 degrees C, and the presumed combustion products are CO2 (g) and H2O (g). (a) In some words, briefly explain what all that means.
(b) There is some question about the accuracy of the reported value, and you have been asked to determine the heat of combustion experimentally. You burn 2.010 grams of the hydrocarbon with pure oxygen in a constant-volume calorimeter and find that the net heat released when the reactants and products [CO2 (g) and H2O (g)] are all at 25 degrees C is sufficient to raise the temperature of 1 kg of liquid water by 21.34 degrees C. Write an energy balance to show that the heat released in the calorimeter equals n(C8H18) times delta Uc and calculate delta Uc (kJ/mol). Then calculate delta Hc. By what percentage of the measured value does the tabulated value differ from the measured one?
(c) Use the result of part B to estimate delta Hf (heat of formation) for 2,3,3-trimethylpentane. Why would the heat of formation of 2,3,3-trimethylpentane probably be determined this way rather than directly from the formation reaction? The standard heat of combustion of liquid 2,3,3-trimethylpentane [C8H18] is reported in a table of physical properties to be -4850 kJ/mol. A footnote indicates that the reference temperature for the reported value is 25 degrees C, and the presumed combustion products are CO2 (g) and H2O (g). (a) In some words, briefly explain what all that means.
(b) There is some question about the accuracy of the reported value, and you have been asked to determine the heat of combustion experimentally. You burn 2.010 grams of the hydrocarbon with pure oxygen in a constant-volume calorimeter and find that the net heat released when the reactants and products [CO2 (g) and H2O (g)] are all at 25 degrees C is sufficient to raise the temperature of 1 kg of liquid water by 21.34 degrees C. Write an energy balance to show that the heat released in the calorimeter equals n(C8H18) times delta Uc and calculate delta Uc (kJ/mol). Then calculate delta Hc. By what percentage of the measured value does the tabulated value differ from the measured one?
(c) Use the result of part B to estimate delta Hf (heat of formation) for 2,3,3-trimethylpentane. Why would the heat of formation of 2,3,3-trimethylpentane probably be determined this way rather than directly from the formation reaction?

Explanation / Answer

(a)

The standard heat of combustion for 2,3,3-trimethylpentane is reported as -4850 kJ/mol. This means that when we burn 1 mol of this compound in presence of oxygen, 4850 kJ of energy would get released.

Footnote says that reference temperature is 25 C and presumed products are CO2 and H2O.

That means the heat is assumed to be evolved at room temperature and the reactants and products are all at 25 C . When there is a complete combustion, only CO2 and H2O are formed But sometimes due to insufficient oxygen supply, the hydrocarbons are partially combusted forming CO as one of the products. Since that is not formed here, we can assume the compound is burned completely.

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(b)

Temperature of 1 kg water rises by 21.34 C.

Let's find how much heat is absorbed by the water.

The equation used for this is

Q = m*C*Delta T

Here, m is mass of water = 1 kg = 1000 g

C is specific heat of water, which is 4.184 J/ g C

Delta T is change in temperature which is 21.34 C

Let's find Q substituting these value in above equation

Q = 1000 g * 4.184 J / g C * 21.34 C

Q = 89286 J

Heat absorbed by water is 89286 J = 89.29 kJ

The Heat released to calorimeter is -89.29 kJ

This is the amount of heat released at constant volume.

Therefore Qv = -89.29 kJ

Heat released at constant volume is Total Change in internal energy

Therefore, Heat released to calorimeter ( Qv) = n ( C8H18) * Delta Uc

-89.29 kJ = n (C8H18) Delta Uc

Molar mass of C8H18 is 114.2

Given mass of of C8H18 is 2.010 g

Moles of C8H18 = 2.010 g * 1 mol / 114.2 g = 0.01760 mol

-89.29 kJ = 0.01760 mol * Delta Uc

Delta Uc = -89.29 kJ / 0.01760 = -5073 kJ /mol

Delta Uc = -5073 kJ / mol

Delta H at constant volume is calculated by following equation

Delta H = Delta Uc + Delta n* R* T

Here Delta Uc = -5073 kJ

R is gas constant , 0.008314 kJ/mol K

T is reaction temperature = 298 K

Delta n is change in moles of gaseous substances

Let's write the balanced equation for combustion of C8H18 to find change in moles of gaseous substances

C8H18 (l) + 25/2 O2 (g) ------------------> 8 CO2 ( g) + 9 H2O (l)

Delta n (gas) = 8 - 25/2 = -4.5 mol ........ O2 and CO2 are in (g) phase so we consider them only.

Substituting above these values in delta H equation, we get

Delta H = Delta Uc + Delta n * R * T

Delta H = -5073 + ( -4.5 mol) * 0.008314 kJ/ mol K * 298 K

Delta H = -5073 - 11.15 kJ

Delta H = -5084 kJ

% error = difference in measured values and tabulated value / measured value *100

% error = -5084 - ( -4850) / -5084 * 100

% error = -234 / - 5084 * 100

% error = 4.6 %

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(c)

The balanced reaction we have is

C8H18 (l) + 25/2 O2 (g) ------------------> 8 CO2 ( g) + 9 H2O (l)

Delta Hrxn = Delta Hf products - Delta Hf reactants

Delta Hf CO2 = -393.5 kJ/mol

Delta Hf H2O (l) = -285.8 kJ/mol

Delta Hf O2 = 0

Delta Hrxn = -5084 kJ

Substituting  these values, we get

-5084 kJ = [ 8 * -393.5 kJ + 9 * -285.8 kJ ] - [ Delta Hf C8H18 + 0]

-5084 kJ = ( -3148 - 2572 ) - Delta Hf C8H18

-5084 kJ = -5720 kJ - Delta Hf C8H18

-5084 kJ + 5720 kJ = - delta Hf C8H18

Delta Hf C8H18 = -636 kJ/mol

Delta Hf for C8H18 is determined this way, because the formation reaction only considers the overall formula C8H18 . But C8H18 occurs in various isomeric forms. When the reaction mentioned below is carried out, we are not sure which isomer is exactly formed.

8 C + 9 H2 -----------> C8H18

Above method uses the actual isomer and therefore the value that we get is more accurate

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