Calculate the standard enthalpy change for the reaction
OF2 (g) + H2O (l) → O2 (g) + 2HF (g).
Standard enthalpies of OF2, H2O, and HF are 20 kJ mol-1, -285 kJ mol-1, and -270 kJ mol-1 respectively.
To calculate the standard enthalpy change for the reaction, we use the equation:
ΔH = ΣnΔH(products) - ΣmΔH(reactants)
Where ΔH is the standard enthalpy change, ΣnΔH(products) is the sum of the products multiplied by their respective standard enthalpies, and ΣmΔH(reactants) is the sum of the reactants multiplied by their respective standard enthalpies.
Given that the reactants are OF2 and H2O, and the products are O2 and 2HF, we can substitute the values:
ΔH = [1*(0 kJ/mol) + 2*(-270 kJ/mol)] - [1*(20 kJ/mol) + 1*(-285 kJ/mol)]
Simplifying the equation, we get:
ΔH = -540 kJ/mol - [-265 kJ/mol]
= -540 kJ/mol + 265 kJ/mol
= -275 kJ/mol
Therefore, the standard enthalpy change for the reaction OF2 (g) + H2O (l) → O2 (g) + 2HF (g) is -275 kJ/mol.
To calculate the standard enthalpy change for a reaction, you need to use the standard enthalpies of formation (∆Hf°) of the reactants and products. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.
In this case, the reactants are OF2 (g) and H2O (l), and the products are O2 (g) and 2HF (g). Let's first calculate the enthalpy change for each compound:
∆Hf°[OF2] = 20 kJ mol-1
∆Hf°[H2O] = -285 kJ mol-1
∆Hf°[O2] = 0 kJ mol-1 (since O2 is an element in its standard state)
∆Hf°[HF] = -270 kJ mol-1
Now we can apply the Hess's law, which states that the total enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps.
The balanced equation for the reaction is:
OF2 (g) + H2O (l) → O2 (g) + 2HF (g)
The enthalpy change can be calculated as follows:
∆H°reaction = Σ(n * ∆Hf°products) - Σ(n * ∆Hf°reactants)
where n is the stoichiometric coefficient of each compound.
Using the given standard enthalpies of formation, we substitute the values:
∆H°reaction = (1 * ∆Hf°[O2]) + (2 * ∆Hf°[HF]) - (1 * ∆Hf°[OF2]) - (1 * ∆Hf°[H2O])
∆H°reaction = (1 * 0 kJ mol-1) + (2 * -270 kJ mol-1) - (1 * 20 kJ mol-1) - (1 * -285 kJ mol-1)
∆H°reaction = 0 kJ mol-1 - 540 kJ mol-1 - 20 kJ mol-1 + 285 kJ mol-1
∆H°reaction = -275 kJ mol-1
Therefore, the standard enthalpy change for the given reaction is -275 kJ mol-1.