How do I find the equilibrium constant for 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g)

at 825 degrees celcius,
assuming that delta H and delta S knot do not depend on temperature,
and Delta H is -908 KJ and Delta S is 1206 KJ?

To find the equilibrium constant (K) for the given reaction at 825 degrees Celsius, you need to use the equation:

ln(K) = (-ΔH°/RT) + (ΔS°/R)

Where:
- ΔH° is the standard enthalpy change for the reaction (-908 kJ)
- ΔS° is the standard entropy change for the reaction (1206 J/K)
- R is the ideal gas constant (= 8.314 J/(mol*K))
- T is the temperature in Kelvin (825°C + 273.15 = 1098.15 K)

Step 1: Convert ΔH° to J/mol:
ΔH° = -908 kJ = -908,000 J/mol

Step 2: Plug in the values into the equation:
ln(K) = (-908,000 J/mol) / (8.314 J/(mol*K) * 1098.15 K) + (1206 J/K) / (8.314 J/(mol*K))

Step 3: Simplify the equation:
ln(K) = -121.09 + 145.34

Step 4: Calculate the value of ln(K):
ln(K) = 24.25

Step 5: Solve for K:
K = e^(ln(K)) = e^(24.25)

Using a calculator, K ≈ 5.36 × 10^10

Therefore, the equilibrium constant (K) for the reaction at 825 degrees Celsius is approximately 5.36 × 10^10.