Consider the reaction
N2 + 3H2 in equilibrium with 2 NH3 .
At 25°C Ho = -92.22 kJ and So = -198.53 J/K. Using this information, calculate the equilibrium constant for the reaction at 226°C. (R = 8.314 J/K)
Enter your answer using TWO significant figures. You may use scientific notation with an "e" if necessary (e.g., 1.2e34).
To calculate the equilibrium constant (K) for the reaction at 226°C, we can use the equation:
ΔHo = -RTln(K)
Where:
ΔHo is the standard enthalpy change for the reaction,
R is the ideal gas constant (R = 8.314 J/K),
T is the temperature in Kelvin (25°C = 298 K),
K is the equilibrium constant for the reaction.
Given:
Ho = -92.22 kJ (which needs to be converted to J),
So = -198.53 J/K,
R = 8.314 J/K, and
T = 226°C (which needs to be converted to Kelvin).
First, let's convert Ho from kJ to J:
Ho = -92.22 kJ * (1000 J/1 kJ)
Ho = -92,220 J
Next, let's convert the temperature from Celsius to Kelvin:
T = 226°C + 273.15 K
T = 499.15 K
Now, let's plug the values into the equation:
-92,220 J = -(8.314 J/K) * 499.15 K * ln(K)
Simplifying the equation, we get:
ln(K) = -92,220 J / (8.314 J/K * 499.15 K)
ln(K) = -22.1
Now, we can solve for K by taking the exponential of both sides:
K = e^(-22.1)
Calculating this value using a calculator, we get:
K ≈ 1.8e-10
Therefore, the equilibrium constant (K) for the given reaction at 226°C is approximately 1.8e-10, rounded to two significant figures.