The decomposition of dinitrogen tetroxide produces nitrogen dioxide:
N2O4 (g) ---> 2 NO2(g) ΔG°= 2.80 kJ/mol
Find the minimum partial pressure of N2O4 at which the reaction is spontaneous if P(NO2)=2.00 bar and T=298 K.
To find the minimum partial pressure of N2O4 at which the reaction is spontaneous, we can use the Gibbs free energy equation:
ΔG = ΔG° + RT ln(Q)
where ΔG is the Gibbs free energy change, ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and Q is the reaction quotient.
In this case, we want to find the minimum partial pressure of N2O4, so we can express the reaction quotient in terms of the partial pressures of N2O4 and NO2:
Q = (P(NO2))^2 / P(N2O4)
Since the reaction is at equilibrium, ΔG = 0. Therefore, we can solve for P(N2O4) by rearranging the equation:
0 = ΔG° + RT ln(Q)
-ΔG° = RT ln(Q)
ln(Q) = -ΔG° / RT
Substituting the given values, we have:
ln(Q) = (-2.80 kJ/mol) / (8.314 J/(mol·K) * 298 K)
ln(Q) = -1.191
Taking the exponential of both sides:
Q = e^(-1.191)
Now we can substitute the given value of P(NO2) into the reaction quotient equation:
2 = (P(NO2))^2 / P(N2O4)
Rearranging the equation, we can solve for P(N2O4):
P(N2O4) = (P(NO2))^2 / 2
Substituting the given value of P(NO2) and the calculated value of Q:
P(N2O4) = (2.00 bar)^2 / 2
P(N2O4) = 2.00 bar
Therefore, the minimum partial pressure of N2O4 at which the reaction is spontaneous is 2.00 bar.