A reaction has a standard free-energy change of –11.70 kJ mol–1 (–2.796 kcal mol–1). Calculate the equilibrium constant for the reaction at 25 °C. Im so lost please help
To calculate the equilibrium constant for a reaction at a given temperature, you need to use the relationship between the standard free energy change (∆G°) and the equilibrium constant (K) known as the Gibbs free energy equation:
∆G° = -RT ln(K)
Where:
∆G° is the standard free energy change (in J/mol or kJ/mol)
R is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K))
T is the temperature in Kelvin (K)
ln is the natural logarithm function
In order to solve for K, we need to convert the temperature from Celsius to Kelvin.
Given that the temperature is 25 °C, we add 273 to convert it to Kelvin:
T = 25 °C + 273 = 298 K
Now we can plug the values into the equation and solve for K:
∆G° = -11.70 kJ/mol
R = 0.008314 kJ/(mol·K)
T = 298 K
-11.70 kJ/mol = -0.008314 kJ/(mol·K) × 298 K × ln(K)
Let's solve this equation step by step:
1. Divide both sides of the equation by (-0.008314 kJ/(mol·K) × 298 K):
(-11.70 kJ/mol) / (-0.008314 kJ/(mol·K) × 298 K) = ln(K)
2. Calculate the value on the left-hand side:
-11.70 kJ/mol / (-0.008314 kJ/(mol·K) × 298 K) ≈ 13.9
3. Take the natural logarithm of both sides:
ln(13.9) = ln(K)
4. Use a calculator to find the natural logarithm of 13.9:
ln(13.9) ≈ 2.63
5. Finally, solve for K by taking the inverse of the natural logarithm:
K ≈ e^(ln(13.9))
K ≈ e^(2.63)
K ≈ 13.9 (rounded to two decimal places)
Therefore, the equilibrium constant (K) for the reaction at 25 °C is approximately 13.9.