N204 (l) Delta Hf (kJ/mol) = -20
E cell S (J/mol.K) = 209
N204(g) Delta Hf (kJ/mol) = 10
E cell S (j/mol.k) = 304.1
Given the following data, determine the temperature at which n2o4 (l) and n204 (g) can exist in equilibrium?
a. 42 C
b. 3 C
c. -231 C
d. 30 C
e. -178 C
To determine the temperature at which N2O4 (l) and N204 (g) can exist in equilibrium, we can use the Gibbs free energy equation:
ΔG = ΔH - TΔS
where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
1. Convert the given temperatures to Kelvin:
a. 42°C = 42 + 273 = 315K
b. 3°C = 3 + 273 = 276K
c. -231°C = -231 + 273 = 42K
d. 30°C = 30 + 273 = 303K
e. -178°C = -178 + 273 = 95K
2. Calculate ΔG for the reaction N2O4 (l) ⇌ N204 (g) at each temperature:
For 42°C (315 K):
ΔG = ΔH - TΔS
ΔG = (-20 kJ/mol) - (315 K)(209 J/mol·K / 1000) = -20 - 65.835 = -85.835 kJ/mol
For 3°C (276 K):
ΔG = ΔH - TΔS
ΔG = (-20 kJ/mol) - (276 K)(209 J/mol·K / 1000) = -20 - 57.684 = -77.684 kJ/mol
For -231°C (42 K):
ΔG = ΔH - TΔS
ΔG = (10 kJ/mol) - (42 K)(304.1 J/mol·K / 1000) = 10 - 12.8182 = -2.8182 kJ/mol
For 30°C (303 K):
ΔG = ΔH - TΔS
ΔG = (10 kJ/mol) - (303 K)(304.1 J/mol·K / 1000) = 10 - 92.3023 = -82.3023 kJ/mol
For -178°C (95 K):
ΔG = ΔH - TΔS
ΔG = (10 kJ/mol) - (95 K)(304.1 J/mol·K / 1000) = 10 - 28.8995 = -18.8995 kJ/mol
3. The temperature at which N2O4 (l) and N204 (g) can exist in equilibrium is the temperature at which ΔG = 0. Therefore, we need to find the temperature at which ΔG is closest to 0 in magnitude.
Based on the calculated values of ΔG:
- At 42°C (315 K), ΔG = -85.835 kJ/mol
- At 3°C (276 K), ΔG = -77.684 kJ/mol
- At -231°C (42 K), ΔG = -2.8182 kJ/mol
- At 30°C (303 K), ΔG = -82.3023 kJ/mol
- At -178°C (95 K), ΔG = -18.8995 kJ/mol
The temperature at which ΔG is closest to 0 is -231°C (42 K). Therefore, the correct answer is option c. -231 C.