thermochemical equation?

Question

I am stumped on this one...

The complete combustion of 1.00 mol of sucrose (table sugar), C12H22O11, yields 5.65 x 10 3(exp) kJ.
a). Write a balanced thermochemical equation for the combustion of sucrose.

Do I write it as an endothermic reaction because energy is absorbed (bonds are broken) in combustion? Or because it says combustion do I have to include O2 or H20 on the reactant side? Am I even on the right track? We weren't given any examples to work with in our notes and the textbook is of no use with an example. This question does not give you an equation to work with...or to balence or am I not seeing it?

b). Calculate the amount of energy that is released when 5.00 g of sucrose (about one teaspoon) is combusted.

I think for this one I can get my 2nd molar amount and then be able to find AH?

Thanks in advance for your input:))

Answer 1

Combustion is essentially the mixing of a compound (usually a hydrocarbon or carbohydrate) with oxygen, producing carbon dioxide and water.

The balanced formula for the combustion of sucrose:
C12H22O11 + 12 O2 --> 12 CO2 + 11 H2O

Combustion reactions are endothermic: They release heat. The heat content of the products is less than the heat content of the reactants (delta H is negative).

b) The molar mass of sucrose is:

12 (12.01 amu ) + 22 ( 1.01 amu ) + 11 ( 16.00 amu) = 342.34 g/mol.

n = m / M = 5.00 g / 342.34 g/mol
n = 0.0146 mol

(0.0145 mol) (5.65 x 10^3 kJ / 1mol) = 82.5 kJ is released.

Answer 2

You should write a balanced equation showing the sucrose + O2 yielding CO2 + H2O.

It is an exothermic reaction, as your problem clearly stated that the reaction yields energy. Make sure to put the energy on the products side of your balanced equation.

For the last part, calculate the molar mass of the sucrose. 5.00 grams will be a fraction of that molar mass. Then look at the stoichiometry of your balanced equation. How many moles of sucrose are invovled. Calculate the molar ratio that 5.00 grams represents and multiply by the KJ / mol

Answer 3

C12H22O11 + 12 02 ---> 12 CO2 + 11 H2O deltaH=5650 kJ

Your Balanced Thermochemical Equation

C12H22O11 in g=(1.00mol C12H22O11) (342.0g C12H22O11/ 1mol C12H22O11)
=342g C12H22O11

342g C12H22O11/ 5g C12H22O11 = 65.4

Energy used= (5650kJ / 65.4) = 86.4kJ

Im not too sure about part 2 thought

Answer 4

if the -6278kJ is the nice and comfortable temperature of the reaction as written, ie for the combustion of two moles benzene then combustion of two x seventy 8.11g produces 6278kj or 24.3 g produces (6278/(2 x seventy 8.11)) x 24.3 = 976.5 kJ