Consider the formation of N2O5(g) by this reaction. 2NO2(g) + 1/2O2(g) → N2O5(g) For this reaction, — Thermodynamics Chemistry Question
Problem Context
Consider the formation of N2O5(g) by this reaction.
2NO2(g) + 1/2O2(g) → N2O5(g)
For this reaction, ∆H o = –55.1 kJ and ∆S o = –227 J·K–1
Additional data are given in the Table.
Calculate these values.
i. ∆Hf o of N2O5(g)
Model Answer
∆Hrxn o = ∆H f (N 2 O5 ) o − 2∆H f (NO 2 ) o
–55.1 kJ = ∆H f (N 2 O5 ) o – 2 mol (33.2 kJ·mol–1)
∆H f (N 2 O5 ) o = +11.3 kJ·mol–1
ii. S o of N2O5(g)
Model Answer
∆Srxn o = So (N 2O 5 ) – 2So (NO2 ) + So (O2 )
–227.0 J·K–1 = So (N 2 O 5 ) – [2(–239.7 J·mol–1·K–1) + 1/2(205.1 J·mol–1·K–1)]
So (N 2 O 5 ) = 355.4 J·mol–1·K–1
iii. ∆G o of the given reaction at 25 °C
Model Answer
∆Go = ∆Ho – T∆So
∆Go = –55.1 kJ – (298 K)(–0.227 kJ·K–1)
∆Go = 12.5 kJ
iv. Kp of the given reaction at 25 °C
Model Answer
∆Go = –RT ln Kp
12500 J = (–8.314 J / mol ⋅ K) (298 K) ln Kp
ln Kp = –5.045 and Kp = 6.44 × 10–3
State and explain
i. whether this reaction is spontaneous at 25 °C.
Model Answer
This reaction is not spontaneous at 25 °C. The value of ∆G o is positive.
ii. how the numerical value of Kp would be affected by an increase in temperature.
Model Answer
An increase in temperature will cause ∆G o to become more positive because the value of ∆S o is negative. Therefore, the numerical value of Kp will decrease.
iii. how the relative amounts of reactant and product molecules would be affected by an increase in temperature.
Model Answer
An increase in temperature will cause the relative amount of reactants to increase and products to decrease. This can be explained by noting that the value of ∆Hrxn is negative, which means that adding heat will shift the reaction to the left. Another argument is that as the temperature increases, the value of the equilibrium constant Kp will decrease, also shifting the reaction to the left.
iv. why the S o values differ for NO2(g) and O2(g) at 25 °C.
Model Answer
The S o values for NO2(g) and O2(g) at the same temperature are not the same because NO2, with 3 atoms per molecule, has more possible arrangements than O2, with only 2 atoms per molecule. This leads to a higher value for entropy, although not very much higher. The molar mass of NO2 is also higher than for O2.