What is the value of K for this aqueous reaction at 298 K? A+B<==> C+D delta G= 23.41 kJ?
dG = -RTlnK\
-23,410J = 8.314*298*lnK
Solve for ln K.
[-23,410/(8.314*298)] = lnK
Then hit the ex button
K = about 8E-5
To determine the value of the equilibrium constant (K) for the given reaction, we need to use the relationship between ΔG (Gibbs free energy change) and K. The equation is as follows:
ΔG = -RT ln(K)
Where:
- ΔG is the Gibbs free energy change (in J)
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (298 K)
- K is the equilibrium constant
However, in the given question, ΔG is given in kJ, so we need to convert it to J by multiplying it by 1000:
ΔG = 23.41 kJ × 1000 = 23,410 J
Now we can substitute the values into the equation:
23,410 J = - (8.314 J/(mol·K)) × 298 K × ln(K)
Simplifying the equation:
ln(K) = -23,410 J / [-(8.314 J/(mol·K)) × 298 K]
ln(K) = 8.903
To find K, we need to take the exponential of both sides:
K = e^(8.903)
Using a calculator, we find that K ≈ 7.36 × 10^3. Hence, the value of K for this aqueous reaction at 298 K is approximately 7.36 × 10^3.