Question

2H2O2(l) -> 2H2O(l) + O2(g)

Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K.

Answer 1

The equilibrium constant (K) for the reaction 2H2O2(l) -> 2H2O(l) + O2(g) at 298.15K is approximately 5.09 ×
`10^ 38`.

To calculate the equilibrium constant for the given reaction at 298.15K, we can use the standard thermodynamic data. The equilibrium constant (K) is a measure of the extent to which a reaction proceeds towards products.

The standard Gibbs free energy change (∆G°) is related to the equilibrium constant (K) by the equation:

∆G° = -RT ln(K)

1. Find the standard Gibbs free energy change (∆G°) for the reaction:

∆G° = ∑∆G°f(products) - ∑∆G°f(reactants)

2. Look up the standard Gibbs free energy of formation (∆G°f) values for each species involved in the reaction in the standard thermodynamic data tables.

3. Calculate the ∆G° using the ∆G°f values:

∆G° = [2(-237.2 kJ/mol) + 0 kJ/mol] - [2(-120.4 kJ/mol)]

∆G° = -474.4 kJ/mol - (-240.8 kJ/mol)

∆G° = -233.6 kJ/mol

4. Use the equation ∆G° = -RT ln(K) to solve for the equilibrium constant (K):

ln(K) = -∆G° / RT

5. Substitute the known values:

∆G° = -233.6 kJ/mol

R = 8.314 J/mol·K

T = 298.15 K

6. Calculate the natural logarithm of K:

`ln(K) = -(-233.6 kJ/mol) / (8.314 J/mol·K) * 298.15 K`

7. Simplify the equation:

`ln(K) = 233.6 kJ / (8.314 J/mol·K) * 298.15 K`

8. Convert kJ to J:

`ln(K) = 233,600 J / (8.314 J/mol·K) * 298.15 K`

9. Simplify further:

ln(K) = 88.6

10. Take the exponential of both sides to solve for K:

K =
`e^ln(K)`

K =
`e^88.6`

11. Calculate the value of K:

K ≈ 5.09 ×
`10^38`

Therefore, the equilibrium constant (K) for the reaction 2H2O2(l) -> 2H2O(l) + O2(g) at 298.15K is approximately 5.09 ×
`10^38`.