Question

Suppose 1.26 g of C2(NO2)6(s) is placed in a 200.0 cm3 vessel where it is detonated.

The heat capacities, Cv, at 25 °C for the gases formed are as follows:

CO2: 28.5 J mol-1 K-1

NO: 21.5 J mol-1 K-1

NO2: 29.5 J mol-1 K-1

Assuming the heat capacities don’t change and all the heat liberated in the reaction remains in the container, what will be the pressure of the mixture of gases after detonation?

delta H for this problem would be -455.66 kJ

C2(NO2)6 (s) ----> 2NO2 (g) + 4NO (g) + 2CO2 (g) is the stoichiometry

Answer 1

Given 1.26 g of C2(NO2)6(s) is placed in a 200.0 cm3 vessel, the balanced equation is

C2(NO2)6 (s) ----> 2NO2 (g) + 4NO (g) + 2CO2 (g) delta Hrxn=-455.66 kJ,

then dHrxn for 1.26g C2(NO2)6
dHrxn = 1.26g C2(NO2)6 x (1 mol C2(NO2)6 / 300.08g C2(NO2)6) x (-451.1 kJ / mol C2(NO2)6) = -1.895 kJ.

Now calculate the moles of CO2, NO and NO2 formed.

then the moles
mol NO2 = 1.26g C2(NO2)6 x (1 mol C2(NO2)6 / 300.08g C2(NO2)6) x (2 mol NO2 / 1 mol C2(NO2)6) = 0.008398mol NO2

mol NO.. = 1.26g C2(NO2)6 x (1 mol C2(NO2)6 / 300.08g C2(NO2)6)x (4 mol NO / 1 mol C2(NO2)6) = 0.01680mol NO

mol CO2 = 1.26g C2(NO2)6 x (1 mol C2(NO2)6 / 300.08g C2(NO2)6)x (2 mol CO2 / 1 mol C2(NO2)6) = 0.008398mol CO2
Finally, from an energy balance, assuming the same final temperature, assuming we heat the mix to 140°C to make the rxn occur

then
Heat kJ = (m x Cv x (Tf - 140°C) CO2 + (m x Cv x (Tf - 140°C) NO + (m x Cv x (Tf - 140°C) NO2

1895 J = (0.008398mol NO2 x 29.5 J/mol°C x (Tf - 140°C)) + (0.01680mol NO x 21.5 J/mol°C x (Tf - 140°C)) + (0.008398mol CO2 x 28.5 J/mol°C x (Tf - 140°C))

simplifying.. and dropping the units since we can see that Tf will be °C

1895 = 0.2477 x (Tf - 140) + 0.3611 x (Tf - 140) + 0.2393 x (Tf - 140)

rearranging
Tf = (1895 / (0..2477 + 0.3611 + 0.2393)) + 140 = 2374°C = 2647K

once you have Tf, use PV = nRT => P = nRT/V to calculate P

P = nRT/V = (0.008398 + 0.01680 + 0.008398) x (0.08206 Latm/molK) x (2647K) / (0.2000L) = 36.5atm.