C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) ΔH°comb = −1270 kJ/molrxn Ethanol, C2H5OH, will combust in — Thermodynamics Chemistry Question

O₂ is reduced; its oxidation number changes from 0 in O₂ to −2 in CO₂ and H₂O.

1 point is earned for the correct answer with an appropriate justification.

n = PV / RT = ((1.03 atm) * (18.0 L)) / ((0.08206 L atm mol^-1 K^-1) * ((273.15 + 21.7) K)) = 0.766 mol CO₂

1 point is earned for the correct calculation.

0.766 mol CO₂ * (1 mol C₂H₅OH / 2 mol CO₂) * (46.07 g C₂H₅OH / 1 mol C₂H₅OH) * (1 mL / 0.79 g) = 22 mL

1 point is earned for the correct stoichiometry for moles of C₂H₅OH and CO₂.
1 point is earned for the correct volume of C₂H₅OH.

q = n * ΔH°comb = 0.766 mol CO₂ * (1 mol rxn / 2 mol CO₂) * (-1270 kJ / 1 mol rxn) = -486 kJ
The amount of heat released is 486 kJ.

1 point is earned for a correct calculation (negative sign is not required).

q_air = −q_rxn = +486 kJ = +486,000 J
q = mcΔT ⇒ ΔT = q / mc = 486,000 J / ((5.56 × 10^4 g) * (1.005 J/(g °C))) = 8.70°C
T_final = 21.7°C + 8.70°C = 30.4°C

1 point is earned for the correct final temperature.

Hydrogen bonding.
Ethanol can form hydrogen bonds between its molecules, whereas dimethyl ether cannot. The attractions between molecules are stronger in ethanol than in dimethyl ether; therefore, the boiling point of ethanol is higher.

1 point is earned for identifying hydrogen bonding as the relevant force.
1 point is earned for linking the greater strength of the intermolecular forces to a higher boiling point.

Decrease the volume of the reaction container. There are fewer moles of gaseous product than moles of gaseous reactants; therefore, an increase in pressure due to a decrease in volume would favor the formation of product.

Lower the temperature. A lower temperature favors an exothermic reaction, leading to the conversion of reactants into product.

1 point is earned for EACH valid answer with justification (2 points total).