Calculate the equilibrium constant at 24 K for a reaction with ΔHo = 10 kJ and ΔSo = (-100) J/K. (Don't round unil the end. Using the exponent enlarges any round-off error.)
i got g to be 12.4. What else do i do?
To calculate the equilibrium constant (K) at 24 K using the values for ΔHo (standard enthalpy change) and ΔSo (standard entropy change), you need to use the equation:
ΔGo = ΔHo - TΔSo
Where:
ΔGo is the standard Gibbs free energy change
ΔHo is the standard enthalpy change
T is the temperature in Kelvin
ΔSo is the standard entropy change
Since you have ΔHo = 10 kJ = 10,000 J and ΔSo = -100 J/K, and temperature T = 24 K, you can substitute these values into the equation:
ΔGo = (10,000 J) - (24 K)(-100 J/K)
Now, multiply -100 J/K by 24 K:
ΔGo = 10,000 J + 2400 J
Add both values:
ΔGo = 12,400 J
However, the equilibrium constant (K) is related to ΔGo through the equation:
ΔGo = -RT ln(K)
Where:
R is the gas constant (8.314 J/(mol·K))
ln is the natural logarithm
K is the equilibrium constant
Rearranging the equation, we have:
K = e^(-ΔGo / (RT))
Substituting the values:
K = e^(-12,400 J / (8.314 J/(mol·K) * 24 K))
Perform the calculation inside the parentheses:
K = e^(-12,400 / (199.536))
Now, divide -12,400 by 199.536:
K = e^(-62.153)
Finally, evaluate e^(-62.153):
K ≈ 2.4988 x 10^(-28)
Thus, the equilibrium constant at 24 K, rounded to the appropriate number of significant figures, is approximately 2.4988 x 10^(-28).