135k views
0 votes
What is the enthalpy change, ∆H for the following reaction 2 HI → I2 + H2

1 Answer

3 votes

Answer:

Step-by-step explanation:

To calculate the enthalpy change (∆H) for a chemical reaction, you need to know the enthalpies of formation (∆Hf) of the reactants and products involved. The enthalpy change can be determined using Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.

The balanced chemical equation is:

2 HI → I2 + H2

Now, let's look up the enthalpies of formation for each compound involved in the reaction:

∆Hf (HI) = -169.3 kJ/mol

∆Hf (I2) = 0 kJ/mol (elemental state, so ∆Hf = 0)

∆Hf (H2) = 0 kJ/mol (elemental state, so ∆Hf = 0)

Now, apply Hess's Law to calculate the enthalpy change:

∆H = Σ∆Hf(products) - Σ∆Hf(reactants)

∆H = [∆Hf (I2) + ∆Hf (H2)] - [2 * ∆Hf (HI)]

∆H = [0 + 0] - [2 * (-169.3)]

∆H = 0 + 338.6

∆H ≈ 338.6 kJ/mol

The enthalpy change (∆H) for the reaction 2 HI → I2 + H2 is approximately 338.6 kJ/mol.

User Doug Kimzey
by
8.4k points