2 NO(g) + 1 O2(g) <==> 2 NO2(g)
Calculate K, the equilibrium constant, from thermodynamic data at 289 K.
(Hint #1: Assume that S and H do not change with temperature.)
To calculate the equilibrium constant, K, from thermodynamic data at 289 K for the given reaction, we will use the equation:
ΔG° = -RT ln K
where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (289 K), and ln represents the natural logarithm.
Given the hint that S (entropy) and H (enthalpy) do not change with temperature, we can assume that the values for ΔH° and ΔS° are constant. We can obtain these values from standard thermodynamic tables.
Using the equation:
ΔG° = ΔH° - TΔS°
we can rearrange it to solve for ΔG°:
ΔG° = -R T ln K
ΔH° - T ΔS° = -R T ln K
ΔH° = -R T ln K + T ΔS°
Now we substitute the given values into the equation as follows:
ΔH° = - (8.314 J/(mol·K)) × (289 K) × ln K + T ΔS°
After substituting the values for R (8.314 J/(mol·K)), T (289 K), ΔS°, and solving for ΔH°, we will get the value for ΔH°.
Once we have obtained ΔH° and ΔS°, we can use these values to calculate K using the equation:
ΔG° = -RT ln K
Rearranging the equation:
ln K = -ΔG° / (RT)
Substituting the values for R (8.314 J/(mol·K)), T (289 K), and ΔG°, and solving for ln K, we can then calculate K by taking the exponent of both sides of the equation:
K = e^(ln K)