Calculate Delta G for H2O(g) + 1/2 O2(g) <-> H2O2(g) at 600. K using the following data:
H2(g) + O2(g) <-> H2O2(g) k = 2.3 X 10^6 at 600. K
2H2(g) + O2(g) <-> 2H2O(g) k = 1.8 X 10^37 at 600. K
To calculate Delta G for the reaction H2O(g) + 1/2 O2(g) <-> H2O2(g) at 600 K, we need to use the relationship between Delta G and equilibrium constant (K).
First, we'll write the balanced equation for the given reaction:
H2O(g) + 1/2 O2(g) <-> H2O2(g)
Now, we'll write the equilibrium expression using the given equilibrium constant for the reaction:
K = [H2O2(g)] / [H2O(g) × O2(g)^(1/2)]
Next, we'll use the equilibrium constant for the reaction 2H2(g) + O2(g) <-> 2H2O(g) to determine the equilibrium constant for the reverse reaction (H2O(g) + 1/2 O2(g) <-> H2O2(g)).
Since the coefficient of the reverse reaction is half of the forward reaction, the equilibrium constant for the reverse reaction is the square root of the equilibrium constant for the forward reaction.
So, K_reverse = √(1/K_forward) = √(1/1.8 × 10^37)
Now, we can calculate the Delta G for the given reaction using the equation: Delta G = -RT ln(K_reverse)
Where:
R = gas constant = 8.314 J/(mol·K)
T = temperature in Kelvin = 600 K
Substituting the values into the equation, we get:
Delta G = - (8.314 J/(mol·K)) × (600 K) × ln(√(1/1.8 × 10^37))
Calculating this expression will yield the value of Delta G for the given reaction at 600 K.