Consider the following reaction:
3C(s) + 4H2(g) -> C3H8(g); DH° = –104.7 kJ; DS° = –287.4 J/K at 298 K
What is the equilibrium constant at 400.0 K for this reaction?
To find the equilibrium constant (K) at a different temperature (400.0 K) for a given reaction, we need to use the Gibbs free energy equation:
ΔG° = ΔH° - TΔS°
Where:
ΔG° is the change in Gibbs free energy at standard conditions,
ΔH° is the change in enthalpy at standard conditions,
T is the temperature in Kelvin,
ΔS° is the change in entropy at standard conditions.
First, let's convert the entropy change from J/K to kJ/K by dividing by 1000:
ΔS° = -287.4 J/K / 1000 = -0.2874 kJ/K
Now, we can calculate the change in Gibbs free energy at standard conditions (ΔG°) using the given values:
ΔG° = ΔH° - TΔS°
ΔG° = -104.7 kJ - 298 K * (-0.2874 kJ/K)
ΔG° = -104.7 kJ + 85.7502 kJ
ΔG° ≈ -18.9498 kJ
Next, we can use the relationship between ΔG° and K to find the equilibrium constant K at the new temperature (400.0 K):
ΔG° = -RT * ln(K)
Where:
R is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K)),
T is the temperature in Kelvin,
ln denotes the natural logarithm.
Rearrange the equation to solve for K:
ln(K) = -ΔG° / (RT)
Now, plug in the values:
ln(K) = -(-18.9498 kJ) / (0.008314 kJ/(mol·K) * 400.0 K)
ln(K) ≈ 90.793
Finally, solve for K by taking the exponential of both sides (e^ln(K) = K):
K ≈ e^90.793
Using a calculator or software to compute e^90.793, we find:
K ≈ 1.4027 x 10^39
Therefore, the equilibrium constant (K) at 400.0 K for this reaction is approximately 1.4027 x 10^39.