What is the value of 𝐾 for this aqueous reaction at 298 K?
A+B↽−−⇀C+DΔ
𝐺°=11.53 kJ/mol
dGo = -RT*lnK
To determine the value of 𝐾 for the given reaction at 298 K, we can use the equation:
Δ𝐺° = -RT ln K
Where:
- Δ𝐺° is the standard Gibbs free energy change, which is given as 11.53 kJ/mol.
- R is the gas constant, which is 8.314 J/(mol·K).
- T is the temperature in Kelvin, which is 298 K.
- K is the equilibrium constant that we want to find.
Let's calculate K step by step.
1. Convert Δ𝐺° from kJ/mol to J/mol:
Δ𝐺° = 11.53 kJ/mol × 1000 J/1 kJ = 11,530 J/mol
2. Convert R from J/(mol·K) to kJ/(mol·K):
R = 8.314 J/(mol·K) ÷ 1000 J/1 kJ = 0.008314 kJ/(mol·K)
3. Substitute the values into the equation and solve for K:
11,530 J/mol = -0.008314 kJ/(mol·K) × 298 K × ln K
4. Rearrange the equation and solve for ln K:
ln K = 11,530 J/mol ÷ (-0.008314 kJ/(mol·K) × 298 K) = -13.489
5. Take the natural exponential of both sides to solve for K:
K = e^(-13.489) ≈ 2.295 × 10^(-6)
Therefore, the value of K for the given aqueous reaction at 298 K is approximately 2.295 × 10^(-6).
To find the value of 𝐾 for this aqueous reaction at 298 K, we need to use the relationship between 𝐺° and 𝐾. The relationship is given by the equation:
Δ𝐺° = -𝑅𝑇 ln 𝐾
Where:
Δ𝐺° is the standard Gibbs free energy change for the reaction
𝑅 is the gas constant (8.314 J/(mol·K))
𝑇 is the temperature in Kelvin
𝐾 is the equilibrium constant
First, we need to convert the given value of Δ𝐺° from kJ/mol to J/mol:
Δ𝐺° = 11.53 kJ/mol × 1000 J/kJ = 11530 J/mol
Next, we substitute the known values into the equation and solve for 𝐾:
11530 J/mol = -8.314 J/(mol·K) × 298 K × ln 𝐾
Simplifying the equation:
ln 𝐾 = (-11530 J/mol) / (8.314 J/(mol·K) × 298 K)
Now, we can use the natural logarithm (ln) function to find 𝐾. Using a calculator, calculate the value of ln 𝐾. Then, take the inverse of ln (using the e^x function on a scientific calculator) to find 𝐾:
𝐾 = e^(ln 𝐾)
Keep in mind that the actual value of 𝐾 will depend on the specific reaction and its reaction stoichiometry.