Question

What are the correctly balanced half reactions of

C₁₂H₂₂O₁₁ + 12O₂ ---> 12CO₂ + 11H₂O

Answer 1

Need less added O2
Still produce CO2 and H2O
example:
C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O

Consider the following reaction, with [T=298.15K, P=1 bar]

C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)

Step a: Let the reaction proceed in a vessel whose walls are adiabatically insulated but held at const P, this causes a temperature change from 298.15 K to T’, then C12H22O11(s) + 12O2(g) → 12CO2(g)[T’ and 1 bar] + 11H2O(l) For this step a: qp,a = 0 and ΔHa = 0 (adiabatic, const P, i.e. ΔHa = qp,a) But of course T changed, and since const.P, V will also change Step b: In the second step the system is brought to thermal equilibrium with the surroundings. The walls of the vessel switched to diathermal and heat flows from reaction vessel to a very large water bath in the surroundings. The temperature changes from T’ back to 298.15K 12CO2(g) [T’] + 11H2O(l) → 12CO2 (g) [T=298.15K] + 11H2O(l) For step b: heat flow from surroundings is: q p,b = ΔHb (again, since const P: ΔHb = qp,b) For the overall reaction, sum the steps: for standard state, T=298.15 K, P=1 bar: use superscript o ΔH o reaction = ΔH o a + ΔH o b = 0 + q p,b = qp Heats of formation Problem with this method above is that to solve for Hrxn need to do actual reaction, measure calorimeter properties, and when want to consider a new reaction, need to do the experiment again. Can we find a way to predict reaction enthalpies for any set of reactants and products? Since a cycle, and we only care about initial and final states, it would be most efficient to know just the enthalpy of each of the reagents and products, in the same standard conditions, and just sum them up on each side and take the difference. To do this we establish standard enthalpies (heats) of formation for all the reagents/products of interest and then take combinations of them.

Consider again, C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)

ΔH o rxn = ΔH o prod - ΔH o reag = 12H o m [CO2 (g)] + 11H o m [H2O (l)] –H o m [C12H22O11 (s)] – 12 H o m [O2 (g)] where m subscript denotes molar quantities.

Above equation worked out reaction enthalpy from ΔHf o of each component, reactant or product. Could view reaction as sum of other reactions (in this case just for products):

1. C12H22O11 (s) + 12O2 (g) → 12CO2 (g) + 11H2O (l)

2. 12 x [C(s) + O2 (g)  CO2 (g)] 3. 11 x [H2(g) + 1/2 O2(g)  H2O (l) ]

Each of these (1-3) is a combustion reaction, but if we sum (2+3), they provide the products. The sums of the enthalpies of combustion of C and of H2 (i.e. ΔHrxn for the reactions written) provide an enthalpy of formation for the products, CO2 and H2O.