Question

The combustion of carbon, as graphite, in oxygen forms carbon dioxide. How much work is done when 5.00 grams of carbon are burned in the presence of excess oxygen at 22 °C and 1.00 atm? [1 L·atm = 101.325 J] (a) 0.0 J; (b) +22.8 J; (c) –22.8 J; (d) +534 J (e) none of the above;

Answer 1

C + O₂ -----------------------> CO₂

w = -P∆V, where the negative sign is the work done by the system, and is negative for the system is loosing energy.
P = Pa and V is m^3
so for every mol of C there is one mol of O₂

5 g C * 1 mol C / 12 g Sn = 0.42 mols C

so this gives 0.42 mols O₂

PV = nRT or V = (nRT / P)

(0.42 mol * 0.082 (L*atm)/(K*mol)* 295K) / 1 atm = 10.15 L O₂

Now the work done is

w = -10.15 L * 1 atm = -10.15 L Atm * 101.325J = -1028 J

Answer : e