Two rigid containers in thermal equilibrium at 298 K connected by a valve are isolated from the surr — Analytical Chemistry Chemistry Question
Mixing ideal gases
Two rigid containers in thermal equilibrium at 298 K connected by a valve are isolated from the surroundings. In one of the containers, 1.00 mol of He(g) and 0.50 mol of A(g) are present at 1.00 atm. In the other container, 2.00 mol of Ar(g) and 0.50 mol of B2(g) are present at 1.00 atm.
Predict whether the entropy will increase or decrease when the valve separating the two containers is opened assuming that no chemical reaction takes place.
Model Answer
VI → VI + VII
VII → VI + VII
Mixing of the gases increases entropy.
Predict whether the entropy will increase or decrease, stating all factors that will have contribution, if a chemical reaction takes place according to the following equation when the valve separating the two containers is opened.
A(g) + ½ B2(g) → BA(g) ∆H°298 = –99.0 kJ
Model Answer
Three factors should be considered;
i. mixing → ∆S > 0
ii. reaction
• ∆ng decreases → ∆S < 0
0.5 mol A + 0.25 mol B2 yields 0.5 mol BA Thus, ∆ng= – 0.25 mol
• Absolute entropy of BA should be greater than A ⇒ ∆S > 0
The change in entropy may be positive or negative but must be not very significant.
iii. As the reaction is exothermic, the heat absorbed should be absorbed by container and the gases present and thus temperature increases ⇒ ∆S > 0
It can be concluded that ∆Soverall > 0
Assuming that all the gases present are ideal, calculate the final pressure at the end of the reaction. The total heat capacity of two containers is 547.0 J⋅K–1.
Model Answer
q = ∆E, ∆H = ∆E + ∆ng RT, ∆ng = 1 – 1 – 1/2 = –1/2
For production of 1.0 mol AB:
∆E = ∆H - ∆ngRT = –9.90 ⋅ 10^4 – (–1/2) × 8.314 × 298 = –9.78 ⋅ 10^4 J = –97.8 kJ
A(g) + ½ B2 (g) → AB (g)
0.50 0.50 –
– 0.25 0.5
q = 0.50 × (–97.8) = – 48.9 kJ
Heat absorbed by the container and the gases present is qab = 4.89 ⋅ 10^4 J
nCv(He) (T – 298) + nCv(Ar)(T – 298) + nCv(AB)(T – 298) + nCv(B2) (T – 298) + Ccont(T – 298) = 4.89 ⋅ 10^4 J
For monatomic ideal gases Cv = 3/2 R
For diatomic ideal gases Cv = 5/2 R
(1.0 × 3/2 × 8.314 + 2.0 × 3/2 × 8.314 + 0.5 × 5/2 × 8.314 + 0.25 × 5/2 × 8.314 + 547) × (T - 298) = 4.89 ⋅ 10^4 J
T = 380 K
p = totaln RT / V
ntotal = nHe + nAr + nAB + nB2 = 1.00 + 2.00 + 0.50 + 0.25 = 3.75 mol
p = (3.75 × 8.314 × 380) / 97.74 = 121.2 Pa