The standard free energy of formation of nitric oxide, NO, at 1000. K (roughly the temperature in an automobile engine during ignition) is 77.5 kJ/mol. Calculate the equilibrium constant for the reaction
dG = -RTlnK
To calculate the equilibrium constant, we need to use the relationship between the standard free energy change (∆G°) and the equilibrium constant (K).
The equation relating ∆G° and K is:
∆G° = -RT ln(K)
Where:
∆G° is the standard free energy change
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
ln(K) is the natural logarithm of K
Given:
∆G° = 77.5 kJ/mol = 77,500 J/mol (since 1 kJ = 1000 J)
T = 1000 K
Let's substitute these values into the equation and solve for ln(K):
77,500 J/mol = - (8.314 J/(mol·K)) × (1000 K) × ln(K)
ln(K) = -77,500 J/mol / ((8.314 J/(mol·K)) × (1000 K))
ln(K) = -9.32
Now, we need to find K by taking the exponential of ln(K):
K = e^ln(K)
K = e^(-9.32)
Calculating this, we find:
K ≈ 7.97 × 10^(-5)
Therefore, the equilibrium constant for the reaction is approximately 7.97 × 10^(-5).