The gas-phase equilibrium between nitryl chloride (NO2Cl) and nitrosyl chloride (NOCl) was studied: — Thermodynamics and Equilibrium Chemistry Question
Problem Context
The gas-phase equilibrium between nitryl chloride (NO2Cl) and nitrosyl chloride (NOCl) was studied:
NO2Cl(g) + NO(g) ⇌ NOCl(g) + NO2(g) (4a)
The equilibrium constant for reaction 4a was measured as K4a = 1.12 × 104 at 298 K and K4a = 4.68 × 103 at 340 K.
Calculate ∆G°4a at both 298 K and 340 K.
Model Answer
∆G°4a = –RTln(K4a)
At 298 K, ∆G°4a = –(8.314 J mol-1 K-1)(298 K)•ln(1.12 × 104) = –23.10 kJ mol-1
At 340 K, ∆G°4a = –(8.314 J mol-1 K-1)(340 K)•ln(4.68 × 103) = –23.89 kJ mol-1
Calculate ∆H°4a and ∆S°4a (assuming that they are essentially independent of temperature).
Model Answer
Since ∆G° = ∆H° – T∆S°:
∆Hº – (298 K)∆S° = –23.10 kJ mol-1
∆Hº – (340 K)∆S° = –23.89 kJ mol-1
Solving the simultaneous equations gives
∆H° = –17.5 kJ mol-1
∆S° = 18.8 J mol-1 K-1
Which of the four gaseous compounds involved in this equilibrium has the lowest standard molar entropy S° at 298 K? Justify your choice.
Model Answer
NO(g) has the lowest molar entropy because all four compounds are in the gaseous state, but NO has the fewest atoms per molecule and hence the fewest degrees of freedom among which energy can be distributed. All other molecules have a bond between nitrogen and oxygen (like NO), but also have other bonds among which energy can be distributed.
The normal boiling point of NOCl is –6 °C. Consider reaction 4b that produces liquid NOCl rather than gaseous NOCl:
NO2Cl(g) + NO(g) ⇌ NOCl(l) + NO2(g) (4b)
At 298 K, how will each of the thermodynamic quantities ∆H°4b, ∆S°4b, and ∆G°4b compare to the corresponding thermodynamic quantity for reaction 4a (i.e., will the quantity for 4b be greater than, less than, or equal to the quantity for 4a)? Justify your answers.
Model Answer
The properties of reaction 4b can be determined from those of 4a plus those of the condensation reaction 4*:
NO2Cl(g) + NO(g) ⇌ NOCl(g) + NO2(g) (4a)
NOCl(g) ⇌ NOCl(l) (4*)
NO2Cl(g) + NO(g) ⇌ NOCl(l) + NO2(g) (4b)
∆H°4b = ∆H°4a + ∆H°4*. Since ∆H°4* is negative ,
∆H°4b < ∆H°4a
∆S°4b = ∆S°4a + ∆S°4*. Since ∆S°4* is negative ,
∆S°4b < ∆S°4a
∆G°4b = ∆G°4a + ∆G°4*. Since ∆G°4* is positive at 298 K ,
∆G°4b > ∆G°4a
In experimental practice, the study of equilibrium 4a is complicated by the fact that NO2(g) is also in equilibrium with N2O4(g) according to reaction 4c under conditions where reaction 4a attains equilibrium.
2 NO2(g) ⇌ N2O4(g) (4c)
Suppose that samples of NO2Cl and NO are introduced into a reaction vessel and the mixture is allowed to attain equilibrium according to reactions 4a and 4c at 320 K. If the volume of the reaction vessel is doubled, will the number of moles of NOCl(g) increase, decrease, or stay the same after the system reattains equilibrium? Justify your answer.
Model Answer
As the volume of the container is doubled, all of the partial pressures of the gases will undergo an instantaneous decrease by a factor of two. Reaction 4a has an equal number of gaseous reactants as products, so its position will not be directly affected by this change. However, reaction 4c has more moles of reactants than products, so that reaction will shift to increase the number of moles of NO2(g). By Le Chatelier's principle, this will shift reaction 4a to the left, so the number of moles of NOCl(g) will decrease.