The value of Kp for the reaction at 298.15 K is approximately 2.49 x 10^4, while at 1173.00 K it is approximately 1.29 x 10^-4.
To calculate the value of Kp for the reaction at 298.15 K and at 1173.00 K, we need to use the equation:
Kp = exp(-ΔG°/RT)
Where ΔG° is the standard Gibbs free energy change for the reaction, R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
First, we need to calculate the ΔG° for the reaction at each temperature using the thermodynamic data provided.
At 298.15 K:
ΔG° = ΔH° - TΔS°
ΔG° = (82.05 kJ/mol) - (298.15 K) * (219.9 J/mol-K) / 1000 J/kJ
ΔG° = 82.05 kJ/mol - 65.84 kJ/mol
ΔG° = 16.21 kJ/mol
At 1173.00 K:
ΔG° = ΔH° - TΔS°
ΔG° = (82.05 kJ/mol) - (1173.00 K) * (219.9 J/mol-K) / 1000 J/kJ
ΔG° = 82.05 kJ/mol - 258.02 kJ/mol
ΔG° = -175.97 kJ/mol
Now we can calculate the values of Kp for each temperature:
At 298.15 K:
Kp = exp(-ΔG°/RT)
Kp = exp(-(16.21 kJ/mol) / ((8.314 J/(mol·K)) * (298.15 K)))
Kp ≈ 2.49 x 10^4
At 1173.00 K:
Kp = exp(-ΔG°/RT)
Kp = exp((-(-175.97 kJ/mol)) / ((8.314 J/(mol·K)) * (1173.00 K)))
Kp ≈ 1.29 x 10^-4
The probable question maybe:-
Calculate the value of Kp for the reaction at 298.15 K and at 1173.00 K
Consider the reaction:
2N, (9) + O2(9) = 2N,0(g)
Thermodynamic data for N,O(g) are:
AH°, = 82.05 kJ/mol; S° = 219.9 J/mol -K; AG°; = 104.2 kJ/mol.