The standard enthalphies of formation of ozone, O3(g), and oxygen atoms are +142.70 and +249.17 kJ/mol. Calculate the reaction enthalpy for 2O2(g)-->O3(g)+O(g)
It's that equation again.
To calculate the reaction enthalpy for the given reaction, we need to use the enthalpies of formation.
The given reaction can be written as follows:
2O2(g) → O3(g) + O(g)
We can find the enthalpy change for this reaction by using the enthalpies of formation of the reactants and products. The enthalpy change (∆H) can be calculated using the equation:
∆H = ∑(∆Hf(products)) - ∑(∆Hf(reactants))
Where ∑(∆Hf) is the sum of the enthalpies of formation.
In this case, we have two reactants and two products. Let's calculate the enthalpy change step by step:
Reactants:
2O2(g): 2 mol of O2(g)
Products:
O3(g): 1 mol of O3(g)
O(g): 1 mol of O(g)
Now, we can calculate the enthalpy change:
∆H = [∆Hf(O3(g)) + ∆Hf(O(g))] - [2∆Hf(O2(g))]
∆H = [142.70 + 249.17] - [2(0)]
Since the standard enthalpy of formation for elemental oxygen (O2(g)) is zero, the equation simplifies to:
∆H = 142.70 + 249.17
∆H = 391.87 kJ/mol
Therefore, the reaction enthalpy for the reaction 2O2(g) → O3(g) + O(g) is 391.87 kJ/mol.
To calculate the reaction enthalpy for the given reaction, we need to use the enthalpies of formation of the reactants and products.
The given reaction is:
2O2(g) -> O3(g) + O(g)
Step 1: Write the balanced chemical equation
The balanced chemical equation shows that for every two moles of O2, one mole of O3 and one mole of O is produced. Therefore, the reaction coefficients are:
2O2(g) -> 1O3(g) + 1O(g)
Step 2: Calculate the reaction enthalpy using the enthalpies of formation
The reaction enthalpy can be calculated using the equation:
ΔHrxn = ΣnΔHf(products) - ΣnΔHf(reactants)
where:
ΔHrxn = reaction enthalpy
ΣnΔHf(products) = sum of the products' enthalpies of formation multiplied by their respective coefficients
ΣnΔHf(reactants) = sum of the reactants' enthalpies of formation multiplied by their respective coefficients
Using the given enthalpies of formation:
ΔHf(O3) = +142.70 kJ/mol
ΔHf(O) = +249.17 kJ/mol
The equation becomes:
ΔHrxn = (1 × ΔHf(O3)) + (1 × ΔHf(O)) - (2 × ΔHf(O2))
ΔHrxn = (1 × +142.70 kJ/mol) + (1 × +249.17 kJ/mol) - (2 × ΔHf(O2))
Calculating:
ΔHrxn = 142.70 kJ/mol + 249.17 kJ/mol - 2ΔHf(O2)
Now, we need to find the value of ΔHf(O2). Since it is not given directly, we can obtain it by rearranging the equation as:
2ΔHf(O2) = 142.70 kJ/mol + 249.17 kJ/mol - ΔHrxn
ΔHf(O2) = (142.70 kJ/mol + 249.17 kJ/mol - ΔHrxn)/2
Finally, substitute the value of ΔHf(O2) back into the equation to solve for ΔHrxn:
ΔHrxn = 142.70 kJ/mol + 249.17 kJ/mol - [(142.70 kJ/mol + 249.17 kJ/mol - ΔHrxn)/2]
Solving this equation will give the reaction enthalpy for the given reaction.